Test Bank For Essentials of Radiographic Physics and Imaging, 4th Edition James Johnston All Chapter

Test Bank For Essentials of Radiographic Physics and Imaging, 4th Edition 2026 James Johnston Chapter 1-17 Chapter 01: Introduction to the Imaging Sciences MULTIPLE CHOICE 1. X-rays were discovered a. November 8, 1805 b. November 8, 1875 c. November 8, 1895 d. November 8, 1985 ANS: C X-rays were discovered November 8, 1895.

2. Barium platinocyanide was the material in Dr. Roentgen’s laboratory that a. covered the cathode ray tube b. fluoresced when the cathode ray tube was energized c. was used to produce the radiograph of Bertha Roentgen’s hand d. protected the people in the room from the x-rays ANS: B A piece of cardboard covered with barium platinocyanide fluoresced when the tube was energized, leading to further investigation.

3. Wilhelm Roentgen’s lab was located in a. Wurzburg b. Zurich c. Paris d. Boston ANS: A Dr. Roentgen’s lab was located at the University of Wurzburg in Wurzburg, Germany.

4. The first radiograph produced by Dr. Roentgen was of a. his own hand b. his daughter’s hand c. his son’s hand d. his wife’s hand ANS: D The first radiograph was taken December 22, 1895, of his wife, Bertha’s, hand.

5. Exposure times for very early radiographs ranged from a. 1 second to 5 seconds

b. 1 minute to 15 minutes c. 20 minutes to 2 hours d. 2 hours to 5 hours ANS: C Exposure times for early radiographs took from 20 minutes to 2 hours to produce an image.

6. Acute radiodermatitis was a. the radiation burn resulting from excessive exposure to x-rays b. common among early patients and operators of x-ray equipment c. a delayed reaction to excessive x-ray exposure d. all of these ANS: D Early on, the excessive radiation exposure to many operators and patients resulted in radiation burns, a delayed response to the exposure.

7. Who brought attention to the dangers of x-rays? a. Wilhelm Roentgen. b. Bertha Roentgen. c. Crookes. d. Thomas Edison.

ANS: D Thomas Edison, the famous American inventor, suffered a radiation burn and brought attention to the dangers of x-rays.

8. An example of how x-rays were used for entertainment or business gain in a dangerous manner was the a. fluoroscopic shoe fitter b. x-ray stove polish c. x-ray headache tablets d. x-ray golf balls ANS: A Although the stove polish, headache tablets, and golf balls used ―x-ray‖ in their names, the shoe fitter actually exposed shoppers to radiation.

9. Mass, length, and time are considered a. fundamental quantities b. derived quantities c. radiologic quantities d. none of these ANS: A Mass, length, and time are the most basic or fundamental quantities.

10. Velocity, acceleration, and work are a. fundamental quantities b. derived quantities c. radiologic quantities

d. none of these ANS: B Along with force, momentum and power, velocity, acceleration, and work are derived from the fundamental quantities.

11. Exposure, dose, and dose equivalent are a. fundamental quantities b. derived quantities c. radiologic quantities d. none of these ANS: C Along with the measure of radioactivity, dose, dose equivalent, and exposure are radiologic quantities.

12. The metric system is also known as the a. British system b. System International (SI) c. System of Units (SU) d. French system ANS: B The metric system is also known as the System International (SI).

13. In the SI system the unit of measure for mass is a. pound b. gram c. kilogram d. ton ANS: C The SI system uses kilogram to quantify mass.

14. In the SI system the unit of measure for length is a. meter b. kilometer c. foot d. mile ANS: A The SI system uses meter to quantify length.

15. In the SI system the unit of measure for time is a. minute b. second c. hour d. day ANS: B The SI system uses second to quantify time.

16. In the British system the unit of measure for mass is a. pound b. gram c. kilogram d. ton ANS: A The British system uses pound to quantify mass.

17. In the British system the unit of measure for length is a. meter b. kilometer c. foot d. mile ANS: C The British system uses foot to quantify length.

18. In the British system the unit of measure for time is a. minute b. second c. hour d. day ANS: B The British system uses second to quantify time.

19. _______________ is equal to distance traveled divided by the time needed to cover that distance. a. Work b. Momentum c. Velocity d. Acceleration ANS: C Distance traveled divided by the time needed to cover that distance is the formula to derive velocity.

20. Meters per second squared (m/s2) is the unit of measure of a. velocity b. momentum c. force d. acceleration ANS: D Meters per second squared (m/s2) is the unit of measure of acceleration.

21. Newton is the unit of measure of a. velocity b. momentum c. force

d. acceleration ANS: C Force is measured in Newtons.

22. Kilograms-meters per second (kg-m/s) is the unit of measure of a. velocity b. momentum c. force d. acceleration ANS: B Kilograms-meters per second (kg-m/s) is the unit of measure of momentum.

23. Joule is the unit of measure of a. power b. force c. work d. momentum ANS: C Joule is the unit of measure of work.

24. Watt is the unit of measure of a. power b. force c. work d. momentum ANS: A Watt is the unit of measure of power.

25. Fd (force  distance) is the formula to determine a. power b. force c. work d. momentum ANS: C Fd (force  distance) is the formula to determine work.

26. Work/time is the formula to determine a. power b. force c. work d. momentum ANS: A Work divided by the time over which it is done (work/t) is the formula for power.

27. The formula mv (mass  velocity) is used to determine

a. b. c. d.

power force work momentum

ANS: D Mass  velocity (mv) is the formula to determine momentum.

28. The formula ma (mass  acceleration) is for a. power b. force c. work d. momentum ANS: B Mass  acceleration (ma) is the formula to determine force.

29. What is the velocity of a javelin that travels 45 meters in 3 seconds? a. 0.067 m/s. b. 15 m/s. c. 67 m/s. d. 135 m/s.

ANS: B Velocity is determined by dividing the distance traveled (45 meters) by the time necessary to cover the distance (3 s); therefore 45 m/3 s or 15 m/s.

30. What is the acceleration of the javelin if the initial velocity is 0, the final velocity is 15 m/s and the time of travel is 3 seconds? a. 1 m/s2. b. 5 m/s2. c. 10 m/s2. d. 15 m/s2.

ANS: B Acceleration is determined by subtracting the initial velocity (0 m/s) from the final velocity (15 m/s) and then dividing that amount by the time it took (3 seconds), resulting in 5 m/s2.

31. How much force is needed to move a 30-kg piece of equipment at a rate of 3 m/s2? a. 10 N. b. 30 N. c. 60 N. d. 90 N. ANS: D Force is determined by multiplying mass (30 kg) by acceleration (3 m/s2) and is measured in Newtons. 30 kg  3 m/s2 = 90 N. 32. What is the momentum of a 30-kg object traveling at 2.5 m/s? a. 12 kg-m/s. b. 75 kg-m/s.

c. 150 kg-m/s. d. 187.5 kg-m/s.

ANS: B Momentum is determined by multiplying mass (30 kg) by its velocity (2.5 m/s), resulting in 75 kg-m/s.

33. How much work is done if a force of 20 N is applied to move a patient 100 meters? a. 0.5 J. b. 5 J. c. 200 J. d. 2000 J. ANS: D Work = Fd, in this case 20 (force) multiplied by 100 (distance) over which it’s moved, resulting in 2000 Joules.

34. If it takes 2 minutes to perform 360 J of work, what is the power? a. 3 W. b. 9 W. c. 180 W. d. 720 W. ANS: A Power is determined by dividing the work done (360 J) by the time it takes to do the work (2 minutes or 120 seconds). 360/120 = 3 Watts.

35. What is the velocity of a baseball that travels 15 meters in 2 seconds? a. 7.5 N. b. 7.5 m/s2. c. 7.5 J. d. 7.5 m/s.

ANS: D Velocity is determined by dividing the distance traveled (15 meters) by the time necessary to cover the distance (2 s); therefore 15 m/2 s or 7.5 m/s. The unit of measurement for velocity is meter/second (m/s).

36. If a basketball goes from being stationary to a velocity of 18 m/s in 3 seconds, what is its acceleration? a. 6 N. b. 6 m/s2. c. 6 m/s. d. 6 W. ANS: B Acceleration is determined by subtracting the initial velocity (0 m/s) from the final velocity (18 m/s) and then dividing that amount by the time it took (3 seconds), resulting in 6 m/s2.

The unit of measurement of acceleration is m/s2.

37. How much force is needed to move a 20-kg box whose acceleration is 5 m/s2?

a. b. c. d.

100 N. 100 W. 100 m/s2.

100 m/s.

ANS: A Force is determined by multiplying mass (20 kg) by acceleration (5 m/s2) and is measured in Newtons. 20 kg  5 m/s2 = 100 N. The unit of measurement of force is the Newton (N).

38. What is the momentum of the 20 kg box that is traveling 10 m/s? a. 200 m/s2. b. 200 W. c. 200 kg-m/s. d. 200 J. ANS: C Momentum is determined by multiplying mass (20 kg) by its velocity (10 m/s), resulting in 200 kg-m/s. Momentum is measured in kg-m/s.

39. How much work is done if 5 N of force is used to lift a box 3 meters high? a. 15 W. b. 15 kg-m/s. c. 15 N/s. d. 15 J. ANS: D Work is determined by multiplying force (5 N) by distance (3 m) over which it’s moved, resulting in 15 Joules. The Joule (J) is the unit of measurement of work.

40. If 240 J of work is done in 1 minute, how much power is consumed? a. 4 J. b. 4 W. c. 4 kg-m/s. d. 4 m/s.

ANS: B Power is determined by dividing the work done (240 J) by the time it takes to do the work (1 minutes or 60 seconds). 240/60 = 4 watts. The unit of measurement of power is the watt (W).

41. The property of an object with mass that resists a change in its state of motion is a. momentum b. power c. energy d. inertia ANS: D Inertia is the property of an object with mass that resists a change in its state of motion.

42. The principle of inertia was first described by a. Wilhelm Conrad Roentgen

b. Sir Isaac Newton c. Thomas Alva Edison d. Crookes ANS: B The principle of inertia was first described by Sir Isaac Newton in the 17th century.

43. Newton’s first law of motion states that, unless acted on by an external force, an object at rest a. will stay at rest b. will move very slowly c. will accelerate very quickly d. none of these ANS: A Newton’s first law of motion was that a body at rest will remain at rest unless an external force is applied.

44. The ability to do work is a. power b. energy c. inertia d. momentum ANS: B Energy is the ability to do work.

45. Energy in a stored state is a. kinetic energy b. energy of motion c. potential energy d. power ANS: C Potential energy is energy in a stored state; it can do work by virtue of position.

46. Kinetic energy is a. stored energy b. energy being expended c. the same as potential energy d. power ANS: B Kinetic energy is energy being used or expended.

47. Electromagnetic, chemical, electrical, and thermal are all types of a. waves b. equipment c. force d. energy ANS: D

Energy comes in many types, including electromagnetic, chemical, electrical, and thermal.

48. Einstein’s formula, E = MC2, demonstrates the relationship between a. matter and energy b. energy and electricity c. electricity and mass d. mass and electromagnetic energy ANS: A E = MC2 demonstrates the relationship between matter (M) and energy (E).

49. The standard radiologic unit that quantifies radiation intensity is the a. rem b. Becquerel c. gray d. roentgen ANS: D The roentgen quantifies radiation intensity.

50. The standard radiologic unit that quantifies the biological effect of radiation on humans and animals is the a. Becquerel b. rad c. roentgen d. sievert ANS: B The rad quantifies the biological effect of radiation on humans and animals.

51. The standard radiologic unit that quantifies occupational exposure or dose equivalent is the a. rem b. rad c. roentgen d. Becquerel ANS: A The rem quantifies occupational exposure or dose equivalent.

52. The ____________ is the SI unit equivalent to the rad. a. rem b. roentgen c. gray d. Becquerel ANS: C The gray (Gy) is the SI unit equivalent to the rad.

53. 1 rad = ______________. a. 10-2 Gy b. 10-1 Gy

c. 10 Gy d. 102 Gy ANS: A One rad is equal to 10-2 Gy or 1/100 Gy. 54. 1 rem = ______________. a. 10-2 Sv b. 10-1 Sv c. 10 Sv d. 102 Sv ANS: A One rem is equal to 10-2 Sv or 1/100 Sv. 55. The ____________ is the SI unit equivalent to the rem. a. rad b. roentgen c. Becquerel d. sievert ANS: D The sievert (Sv) is the SI unit equivalent to the rem.

56. 1 roentgen = ______________. a. 2.58  104 C/kg b. 2.58  103 C/kg c. 2.58  10-3 C/kg d. 2.58  10-4 C/kg ANS: D One roentgen (R) is equal to 2.58  10-4 C/kg (Coulombs per kilogram).

57. The SI radiologic unit that addresses the different biological effects of different types of ionizing radiation is the a. rad b. roentgen c. sievert d. gray ANS: C The sievert/rem addresses the different biological effects of different types of ionizing radiation.

58. The shortened form of the standard radiologic quantity curie is a. Cr b. Ci c. Ce d. Cu ANS: B

The Ci is the shortened form of curie.

59. The ____________ is the SI unit equivalent to the Curie. a. roentgen b. becquerel c. sievert d. gray ANS: B The Becquerel is the SI unit equivalent to the Curie.

60. What is the SI equivalent of 3 Ci? a. 3 Bq. b. 111 Bq. c. 1.11  1010. d. 1.11  1011.

ANS: D To convert Curies to Becquerels multiply the Curie value by 3.7  1010 (37,000,000,000), therefore 3  (3.7  1010 ) = 1.11  1011 Bq. 61. Which of the following is an expression of the relative risk to humans of exposure to ionizing radiation? a. tissue weighting b. effective dose c. dose equivalent d. absorbed dose ANS: B Effective dose is an expression of the relative risk to humans of exposure to ionizing radiation.

62. The tube head assembly consists of a. x-ray tube b. tube stand c. collimator d. all of these ANS: D The tube head assembly consists of the x-ray tube, collimator, and tube stand.

63. The positive electrode of the x-ray tube is the a. diode b. cathode c. anode d. canode ANS: C The positive electrode of the x-ray tube is the anode.

64. The negative electrode of the x-ray tube is the

a. b. c. d.

diode cathode anode canode

ANS: B The negative electrode of the x-ray tube is the cathode.

65. In a typical radiographic room the anode is located a. over the head end of the table b. over the foot end of the table c. in the middle of the table d. away from the table ANS: A In a typical radiographic room the anode is located over the head end of the table.

66. To help dissipate the heat produced during x-ray production, the x-ray tube housing is filled with a. air b. water c. refrigerant d. oil ANS: D Oil is found within the tube housing, surrounding the tube, to help dissipate heat.

67. The device that restricts the x-ray beam to the area of interest is the a. tube housing b. collimator c. mirror d. crosshair ANS: B The collimator, located beneath the tube housing, restricts the x-ray beam to the area of interest.

68. The purpose of the mirror inside the collimator is to a. restrict the x-ray beam b. allow the patient to see himself or herself c. focus the x-ray beam d. reflect the light source ANS: D Located within the collimator, the mirror reflects the light source onto the patient to show the x-ray field size and crosshairs.

69. Lead shutters are part of the a. tube housing b. tube stand c. collimator

d. x-ray tube ANS: C Adjustable lead shutters are found in the collimator and allow the x-ray beam to be restricted to the anatomic area of interest.

70. The floor mount, floor-ceiling mount, and the overhead tube assembly are types of a. tube stands or mounts b. x-ray tube designs c. collimator devices ANS: A The floor mount, floor-ceiling mount, and the overhead tube assembly are types of tube stands or mounts.

71. In the hospital setting, the most widely used tube stand or mount is the a. floor mount b. floor-ceiling mount c. overhead tube assembly d. under-table tube assembly ANS: C The most widely used tube stand-mount is the overhead tube assembly because of its versatility.

72. The device that is located just under the x-ray table and holds the image receptor in place is the a. Bucky assembly b. grid assembly c. floating assembly d. exit assembly ANS: A The Bucky assembly holds the image receptor in place and also includes a grid.

73. Upright radiographic examinations can easily be done with a a. floating table b. table Bucky c. fluoroscope d. wall unit ANS: D The wall unit has a vertical Bucky assembly that makes upright examinations much easier.

74. The electricity applied to the tube during x-ray production is controlled at the a. tube housing b. collimator c. control panel d. table top ANS: C

Setting the kilovoltage and milliamperage appropriate for the radiographic examination is done at the control panel.

75. A radiographic unit that can be taken to the patient’s bedside is considered a. mobile equipment b. permanently installed equipment c. bedside equipment d. stationary equipment ANS: A Mobile equipment is on wheels and can be taken to the patient’s bedside when he or she is too ill to travel to the radiology department.

76. What principle articulates the radiographer’s responsibility to minimize radiation exposure to the patient, oneself and others? a. ASRT b. ALARA c. ARRT d. HVL ANS: B It is the radiographer’s responsibility to minimize radiation dose to the patient, oneself, and others in accordance with the As Low As Reasonably Achievable (ALARA) Principle.

77. Which of the following is NOT a cardinal principle for minimizing radiation dose? a. distance b. shielding c. collimation d. time ANS: C Central to minimizing radiation dose to oneself and others are the cardinal principles of shielding, time, and distance.

78. Which of the following is NOT one of the ―tools/tasks‖ of radiation protection? a. decrease collimation b. increase kVp and decrease mAs c. avoid duplicate exams ANS: A Decreasing the x-ray field size (increased collimation), using higher kVp along with lower mAs, and avoiding duplicate exams are tools/tasks the radiographer can use to minimize patient radiation protection.

TRUE/FALSE 1. Mass does not change with gravitational force.

ANS: T Although weight is based on the effect of gravitational force, mass is not.

2. When 3 kilograms of frozen water is melted, it produces 3 kilograms of water.

ANS: T Mass does not change when the substance changes form.

3. Weight, measured in pounds, is not affected by gravitational force.

ANS: F Weight, as opposed to mass, changes when the gravitational force changes (earth versus moon, for example).

4. A floating table top is typical of today’s x-ray tables.

ANS: T Today’s x-ray table has a floating table top with electromagnetic locks.

5. Permanently installed radiographic equipment can never be replaced because it is permanent.

ANS: F Although it is called permanent, this type of equipment can be removed and replaced, but it does take a week or so.

6. Screening for pregnancy is not a task the radiographer would perform before a radiographic procedure on a female patient.

ANS: F Screening for pregnancy is another important task to minimize unnecessary exposure to a developing fetus.

7. Radiographers can develop good work habits by developing a mental checklist for radiographic procedures and performing them the same way every time.

ANS: T Radiography is a practice where being a ―creature of GOOD habits‖—is a good thing.

Develop a mental checklist for radiographic procedures and perform them the same way every time.

Chapter 02: Structure of the Atom Johnston/Fauber: Essentials of Radiographic Physics and Imaging, 4th Edition MULTIPLE CHOICE 1. The earliest atomic theory is generally associated with a. Bohr b. Leucippus c. Democritus d. Dalton ANS: B Although his theory was somewhat vague, Leucippus is most often the earliest person associated with atomic theory.

2. The word atom is derived from the Greek word atomos, meaning a. nuclear b. small c. indivisible d. invisible ANS: C The Greek word atomos means ―indivisible.‖ 3. In the early 1800s English chemist John Dalton theorized that a. elements form compounds b. atoms are unique to each element in size and mass c. a chemical reaction results from atoms being rearranged d. all of these ANS: D As a result of his experiments, John Dalton theorized that elements form compounds, atoms are unique to each element in size and mass, and that a chemical reaction results from atoms being rearranged.

4. Discovery of the electron is attributed to a. Dalton b. Bohr c. Thomson d. Rutherford ANS: C Joseph John ―J.J.‖ Thomson determined that the electron was a negatively charged part of the atom.

5. The ―plum pudding model‖ is associated with a. Bohr b. Rutherford c. Dalton d. Thomson

ANS: D Based on the physical arrangement of raisins in a plum pudding, Thomson described the atom and surrounding negatively charged particles (electrons).

6. The earliest atomic theory based on an arrangement similar to the solar system is attributed to a. Bohr b. Rutherford c. Dalton d. Thomson ANS: B Rutherford developed the theory that the atom consisted of a very dense nucleus with small electrons rotating around, similar to the sun and planets.

7. The most commonly known modern atomic theory was developed by a. Bohr b. Rutherford c. Dalton d. Thomson ANS: A Niels Bohr refined Rutherford’s atomic theory, based on the solar system, into the most commonly known atomic theory today.

8. The three fundamental particles of the atom are the a. element, nucleus, and electron b. electron, nucleus, and proton c. neutron, electron, and proton d. nucleus, proton, and neutron ANS: C The three fundamental components of the atom are the proton, electron, and neutron.

9. The atomic nucleus contains a. protons and neutrons b. protons and electrons c. electrons and neutrons d. all of these ANS: A The atomic nucleus contains varying amounts of protons and neutrons, depending on the element.

10. The component of the nucleus that has a positive charge and mass is the a. electron b. neutron c. proton d. none of these ANS: C

The proton is the part of the nucleus that has a positive charge and mass.

11. The component of the nucleus that has mass but no electrical charge is the a. electron b. neutron c. proton d. none of these ANS: B The neutron is found in the nucleus; it is very similar to the proton but has no electrical charge (neutral).

12. The fundamental component of the atom that has the smallest mass is the a. electron b. neutron c. proton d. none of these ANS: A The electron has significantly less mass than the neutron or proton.

13. The mass of an atom is primarily due to the mass of the a. neutrons b. nucleus c. electrons d. protons ANS: B The nucleus, consisting of both protons and neutrons, accounts for the majority of the mass of an atom.

14. If an atom has more protons than electrons, it will a. have a negative charge b. have a positive charge c. be electrically neutral d. have neither a positive nor negative charge ANS: B An atom with more protons than electrons will have a positive charge.

15. If an atom has more electrons than protons, it will a. have a negative charge b. have a positive charge c. be electrically neutral d. have neither a positive nor negative charge ANS: A An atom with more electrons than protons will have a negative charge.

16. If an atom has the same number of electrons and protons it will a. have a negative charge

b. have a positive charge c. be electrically neutral d. none of these ANS: C Having the same number of protons and electrons will result in a neutral atom, having neither a negative nor a positive electrical charge.

17. When an atom becomes negatively or positively charged it is usually due to a change in the number of a. protons b. electrons c. neutrons d. all of these ANS: B In that there is a weaker bond, the addition or loss of electrons typically produces a charged atom.

18. A negative ion is a. an electron b. an atom with more protons than electrons c. an atom with more neutrons that electrons d. an atom with more electrons than protons ANS: D A negative ion is a charged atom with more electrons than protons.

19. A positive ion is a. a proton b. an atom with more protons than electrons c. an atom with more neutrons that electrons d. an atom with more electrons than protons ANS: B A positive ion is a charged atom with more protons than electrons.

20. The force that holds the protons and neutrons together in the nucleus is the a. nuclear binding energy b. electron binding energy c. atomic energy d. proton/neutron energy ANS: A The force that holds the protons and neutrons together in the nucleus is the nuclear binding energy.

21. If a particle strikes a nucleus with the same amount of energy as the atom’s nuclear binding energy a. the atom will become a positive ion b. the atom will become a negative ion

c. it can split the atom d. it can fuse the atom ANS: C If a particle strikes a nucleus with the same amount of energy as the atom’s nuclear binding energy, it can break the atom apart.

22. The electrons stay in orbit around the nucleus because of a. their attraction to the protons b. their attraction to the neutrons c. their attraction to the other electrons d. all of these ANS: A The electrons stay in orbit because of their attraction to the positively charged protons in the nucleus.

23. The electron binding energy does NOT depend on a. how close it is to the nucleus b. how many neutrons there are in the nucleus c. how many protons there are in the nucleus ANS: B The electron binding energy depends on how close it is to the nucleus and how many protons there are in the nucleus.

24. The electron binding energy is stronger when a. there are more protons and the electron is closer to the nucleus b. there are fewer protons and the electron is closer to the nucleus c. there are fewer protons and the electron is farther from the nucleus d. there are more protons and the electron is farther from the nucleus ANS: A The electron binding energy is greater when the electron is closer to the nucleus and there are more protons in the nucleus.

25. The electron shell closest to the nucleus is lettered a. ―E‖ b. ―H‖ c. ―K‖ d. ―M‖ ANS: C The innermost electron shell is the ―K‖ shell.

26. The L shell can hold _______ electrons. a. 1 b. 2 c. 4 d. 8 ANS: D

Based on the formula 2n2, the L (second) shell can hold 2  22 electrons, or 8.

27. The N shell can hold _______ electrons. a. 4 b. 8 c. 32 d. 64 ANS: C Based on the formula 2n2, the N (fourth) shell can hold 2  42 electrons, or 32.

28. Except for the K shell, the maximum number of electrons that can be in the outermost shell of an atom is a. 4 b. 8 c. 16 d. 32 ANS: B With the exception of the K shell, no more than 8 electrons can be in the atom’s outermost shell. This is called the octet rule.

29. If an atom has 15 electrons, which will be the outermost shell? a. ―L‖ b. ―M‖ c. ―N‖ d. ―O‖ ANS: B With 15 electrons, 2 will fill the K shell, 8 will fill the L shell, and 5 will fill the M shell.

30. The number of protons in an atom’s nucleus is reflected in its a. atomic number b. atomic mass number c. element d. compound ANS: A The atomic number indicates the number of protons in the nucleus.

31. The number of protons and neutrons in the atom’s nucleus is the a. atomic number b. atomic mass number c. element d. compound ANS: B The sum of the protons and neutrons in an atom’s nucleus is its atomic mass number.

32. The simplest form of the substances that form matter is the a. atomic number

b. atomic mass number c. element d. compound ANS: C The element, such as hydrogen or oxygen, is the simplest form of substances that form matter.

33. Two or more atoms that bond together form a(n) a. atomic number b. atomic mass number c. element d. compound ANS: D More than one atom bonded together, such as two atoms of H and one of O (H2O), form a compound.

34. In a neutral atom, the atomic number does NOT indicate the number of a. protons b. neutrons c. electrons ANS: B In a neutral atom, the atomic number indicates the number of protons (by definition) but also the number of electrons (which are equal to the number of protons).

35. An atom of helium (42He) has a. two protons b. four protons c. four neutrons d. four electrons ANS: A The atomic number, the number of protons, is the lower number, two.

36. An atom of oxygen (168O) has a. eight protons b. eight neutrons c. eight electrons d. all of these ANS: D The atomic mass number (16) less the atomic number (number of protons—8) equals the number of neutrons (8). The number of electrons equals the number of protons (8).

37. How many neutrons does 73Li (lithium) have? a. 3 b. 4 c. 7 d. 10

ANS: B Subtracting the atomic number (3) from the atomic mass number (7) determines the number of neutrons (4).

38. How many electrons does a neutral atom of carbon (126C) have? a. 3 b. 6 c. 12 d. 18 ANS: B A neutral atom has the same number of protons and electrons, in this case 6.

39. For the chemical element sodium (2211Na), the atomic number is a. eleven b. twenty-two c. thirty-three d. none of these ANS: A The atomic number, number of protons in the nucleus, is the lower number, 11.

40. For the chemical element sodium (2211Na), the atomic mass number is a. 11 b. 22 c. 33 d. none of these ANS: B The atomic mass number, which equals the number of protons and neutrons in the nucleus, is the upper number, 22.

41. Atoms with the same number of protons but different number of neutrons are a. isotopes b. isotones c. isobars d. isomers ANS: A As isotope is an atom that has the same number of protons but different number of neutrons as compared with the element.

42. Atoms with the same atomic number but different atomic mass numbers are a. isotopes b. isotones c. isobars d. isomers ANS: A Having the same atomic number (number of protons) and different atomic mass number (number of neutrons are different) results in an atom being classified as an isotope.

43. Atoms with the same number of neutrons but different number of protons are a. isotopes b. isotones c. isobars d. isomers ANS: B An isotone has the same number of neutrons but different number of protons.

44. Atoms with different number of protons but the same combined number of protons and neutrons are a. isotopes b. isotones c. isobars d. isomers ANS: C An isobar has a different number of protons but the atomic mass number (protons and neutrons) is the same.

45. Atoms with different atomic numbers but the same atomic mass numbers are a. isotopes b. isotones c. isobars d. isomers ANS: C Isobars have different number of protons (atomic number) but the atomic mass number (protons and neutrons) is the same.

46. Atoms with the same atomic number and atomic mass number but have different energy within their nuclei are a. isotopes b. isotones c. isobars d. isomers ANS: D The isomer has the same number of protons and neutrons but the energy level within the nucleus is different.

47. 2311Na is an _______________ of 2211Na. a. isotopes b. isotones c. isobars d. isomers ANS: A 2311Na is an isotope of 2211Na because it has the same number of protons (11) and different number of neutrons, as seen in the increased atomic mass number.

48. 13153I and 13254Xe are a. isotopes b. isotones c. isobars d. isomers ANS: B 13153I and 13254Xe are isotones because they have the same number of neutrons (131 – 53 = 132 – 54) but different number of protons (53 vs. 54).

49. 73Li and 74Be are a. isotopes b. isotones c. isobars d. isomers ANS: C 73Li and 74Be are isobars because they have the same atomic mass numbers (7) but different numbers of protons (3 vs. 4).

50. The periodic table of elements classifies by period and group. The period is the a. row b. column c. group d. type of element ANS: A The periodic table of elements includes seven periods, the rows of the table.

51. The periodic table of elements classifies by period and group. The group is the a. row b. column c. period d. type of element ANS: B The periodic table of elements includes eight groups, the columns of the table.

52. Atoms in each period have the same number of a. electrons in the outermost shell b. atomic mass number c. electrons d. electron shells ANS: D Atoms in each period have the same number of electron shells.

53. Atoms in each group have the same number of a. electrons in the outermost shell b. atomic mass number

c. electrons d. electron shells ANS: A Atoms in each group have the same number of electrons in the outermost shell, increasing from left to right.

54. A compound consists of a. at least two molecules b. at least two elements c. at least two different materials d. all of these ANS: B A compound is a molecule that consists of atoms of at least two different elements.

55. When the bond between two atoms is due to their sharing an outer-shell electron, this is called a a. molecular bond b. ionic bond c. compounding bond d. covalent bond ANS: D Covalent bonding is based on atoms sharing an outer-shell electron.

56. When the bond between two atoms is due to one atom giving up an electron and the other atom gaining an electron, it is called a a. molecular bond b. ionic bond c. compounding bond d. covalent bond ANS: B Ionic bonding is based on one atom giving up an electron (becoming a positive ion) and the other gaining an electron (becoming a negative ion) and then being attracted to each other.

TRUE/FALSE 1. The electrons rotate around the nucleus at a single energy level.

ANS: F The electrons rotate around the nucleus at different energy levels, based on their distance from the nucleus.

2. Electron shells are the hard coating around the electron.

ANS: F Electron shells are the defined energy levels around the atomic nucleus.

3. Each electron shell has a specific limit to the amounts of electrons it can hold.

ANS: T There is a specific limit to how many electrons each shell can hold.

4. The outermost shell of an atom can hold fewer than 8 electrons.

ANS: T Although there can be no more than 8 electrons in the outermost shell, there can be fewer than 8.

5. Each element has an unchanging number of protons.

ANS: T Each element (H, O, C, etc.) has an unchanging number of protons.

6. Elements can only occur naturally.

ANS: F Although there are 92 naturally occurring elements, more than a dozen have been created artificially.

7. The atoms of the elements at the top of the periodic table of elements are the most complex.

ANS: F The atoms at the elements at the bottom of the table have more electron shells and are more complex.

8. In the middle of the periodic table of elements there are elements that don’t fit exactly into one of the eight groups.

ANS: T The inner transitional metals, located in the middle of the table, do not fit into the eight groups.

9. All compounds are molecules and all molecules are compounds.

ANS: F All compounds are molecules, containing atoms of at least two elements, but not all molecules are compounds, occurring when multiple atoms of the same element combine.

10. An ionic bond results in an electrically charged molecule or compound.

ANS: F An ionic bond is the result of two charged atoms being attracted to each other, creating a neutral molecule or compound.

Chapter 03: Electromagnetic and Particulate Radiation Johnston/Fauber: Essentials of Radiographic Physics and Imaging, 4th Edition

MULTIPLE CHOICE 1. Electromagnetic theory was developed in the late 1800s by a. Bohr b. Rutherford c. Maxwell d. Planck ANS: C James Maxwell developed the electromagnetic theory, a significant advancement in physics.

2. Electromagnetic radiation a. has no mass b. has significant mass c. has varying amounts of mass d. none of these ANS: A One of the characteristics of electromagnetic radiation is that it has no mass.

3. What kind of disturbance does electromagnetic radiation NOT cause? a. electrical b. mechanical c. magnetic ANS: B Electromagnetic radiation causes both magnetic and electrical disturbances.

4. Different electromagnetic radiations are put in order by the a. electromagnetic table b. electromagnetic spectrum c. electromagnetic chart d. electromagnetic list ANS: B The electromagnetic spectrum places the different electromagnetic radiations in order based on their energy level.

5. Which of the following is not part of the electromagnetic spectrum? a. Visible light b. Microwaves c. Sound d. Gamma rays ANS: C Sound is not a form of electromagnetic radiation.

6. Which of the following electromagnetic radiations have the lowest energy level? a. X-rays b. Visible light

c. Microwaves d. Radio waves ANS: D Radio waves have the lowest energy level of all electromagnetic radiations.

7. Electromagnetic radiation is a form of energy that originates from a. electricity b. magnetism c. the atom d. something moving in space ANS: C Electromagnetic radiation is a form of energy that originates from the atom, either through orbital transitions or excess energy being emitted from atomic nuclei.

8. In the formula E = hf, E represents a. Planck’s constant b. frequency c. energy d. electricity ANS: C The E in E = hf stands for ―energy.‖ 9. In the formula E = hf, h represents a. Planck’s constant b. frequency c. energy d. electricity ANS: A The h in E = hf stands for ―Planck’s constant,‖ a specific mathematical value (equal to 4.135  10-15 eV-sec; 6.626  10-34 J-sec).

10. In the formula E = hf, f represents a. Planck’s constant b. frequency c. energy d. electricity ANS: B The f in E = hf stands for ―frequency.‖ 11. The formula E = hf demonstrates that a. as frequency increases, energy decreases b. as frequency increases, energy increases c. as energy increases, heat decreases d. as energy increases, heat increases ANS: B

The formula E = hf demonstrates the direct relationship between frequency and energy; as one increases, so does the other.

12. In the formula E = hf, energy is measured in a. volts b. amps c. frequency volts d. electron volts ANS: D In the formula E = hf, energy is measured in electron volts (eV).

13. When considering the wave properties of electromagnetic radiation, the distance from the peak of one wave to the next is a. wavelength b. amplitude c. frequency d. velocity ANS: A Wavelength is the distance from the peak of one wave to the next.

14. When considering the wave properties of electromagnetic radiation, the maximum height of a wave is a. wavelength b. amplitude c. frequency d. velocity ANS: B The maximum height of a wave is amplitude.

15. When considering the wave properties of electromagnetic radiation, the number of waves that pass by a given point per second is a. wavelength b. amplitude c. frequency d. velocity ANS: C Frequency is the number of waves that pass by a given point per second.

16. Microwaves travel at a velocity of a. 3  106 m/s b. 3  107 m/s c. 3  108 m/s d. 3  109 m/s ANS: C As do all the electromagnetic radiations, microwaves travel at the speed of light, 3  108 m/s.

17. As the wave’s frequency increases, the wavelength a. stays the same b. increases c. decreases d. it depends on the type of electromagnetic radiation ANS: C Wavelength and frequency are inversely related, so as frequency increases, wavelength decreases.

18. As a wave’s wavelength increases, its frequency a. stays the same b. increases c. decreases d. it depends on the type of electromagnetic radiation ANS: C Wavelength and frequency are inversely related, so as wavelength increases, frequency decreases.

19. Wavelength is generally measured in a. hertz b. feet c. centimeters d. meters ANS: D Wavelength is generally measured in meters.

20. Frequency is typically measured in a. hertz b. feet c. centimeters d. meters ANS: A Frequency is measured in hertz (cycles per second).

21. 1 megahertz (MHz) is equal to a. 100 hertz b. 1000 hertz c. 1,000,000 hertz d. 1,000,000,000 hertz ANS: C 1 megahertz (MHz) is equal to 1,000,000 hertz.

22. 1 kilohertz (kHz) is equal to a. 100 hertz b. 1000 hertz

c. 1,000,000 hertz d. 1,000,000,000 hertz ANS: B 1 kilohertz (kHz) is equal to 1000 hertz.

23. What characteristics of electromagnetic waves can the formula a. frequency and wavelength b. frequency and velocity c. velocity and wavelength

be used to calculate?

ANS: A The formula allows calculation of a wave’s frequency and wavelength. The velocity is a known quantity (speed of light).

24. When the formula a. wavelength b. frequency c. amplitude d. the speed of light

is changed to

, the c represents the constant symbol for:

ANS: D The c in the formula has replaced the v for velocity with the constant symbol for the speed of light, the velocity of all electromagnetic radiation.

25. What is the frequency of electromagnetic radiation if the wavelength is 1  10-4 m? a. 3  10-12 Hz. b. 3  10-4 Hz. c. 3  104 Hz. d. 3  1012 Hz. ANS: D Rearranging the formula

to solve for f,

, or 3  1012 Hz.

26. What is the wavelength of electromagnetic radiation if the frequency is 3  1010 m? a. 1  10-2 m. b. 1  102 m. c. 3  10-2 m. d. 3  102 m.

ANS: A Rearranging the formula

to solve for l,

or 1  10-2 m.

27. The inverse square law relates the intensity (of light/x-rays) to a. velocity b. time c. mass d. distance ANS: D

The inverse square law relates intensity to distance.

28. The process of removing an electron from an atom is a. annihilation b. atomization c. ionization d. none of these ANS: C Removal of an electron from an atom creates an ion and the process is called ionization.

29. An ion pair is a. an electron and proton b. a proton and neutron c. an atom and the electron that was removed from it d. an atom with an extra electron and an atom that is missing an electron ANS: C Following ionization, an ion pair consists of an atom and the electron that was removed from it.

30. X-rays and gamma rays differ in a. the energy source that produces them b. the effect they have on matter c. their energy level d. all of these ANS: A X-rays and gamma rays differ only in the energy source that produces them.

31. X-rays are produced a. from unstable atoms b. using fast-moving electrons c. using fast-moving atoms d. using fast-moving metals ANS: B X-rays are produced when fast-moving electrons strike metal (in the x-ray tube).

32. Gamma rays are produced a. from unstable atoms b. using fast-moving electrons c. using fast-moving atoms d. using fast-moving metals ANS: A Gamma rays are the result of excess energy being emitted from an unstable atom as it works to become stable.

33. Radio waves are used in a. computed tomography

b. ultrasound c. radiography d. magnetic resonance imaging ANS: D Radio waves of a specific frequency are used in magnetic resonance imaging to be absorbed and emitted from the nucleus of hydrogen atoms.

34. Microwaves heat up food because their energy causes the atoms and molecules to a. dissolve b. expand c. vibrate d. contract ANS: C The microwaves cause the atoms and molecules to vibrate, which then results in heat.

35. The electromagnetic radiation that passes between the television remote and television is a. visible light b. infrared light c. microwaves d. ultraviolet light ANS: B Infrared light passes between the television remote and television.

36. An object we see as white is _______________ all of the wavelengths of visible light. a. absorbing b. diffusing c. reflecting d. changing ANS: C A white object is reflecting all the wavelengths of visible light.

37. An object we see as black is _______________ all of the wavelengths of visible light. a. absorbing b. diffusing c. reflecting d. changing ANS: A A black object is absorbing all the wavelengths of visible light.

38. The electromagnetic radiation used in tanning beds is a. visible light b. infrared light c. microwaves d. ultraviolet light ANS: D Ultraviolet light is used in tanning beds.

39. Particulate radiation does NOT include: a. alpha particles b. x-ray particles c. beta particles ANS: B Alpha particles and beta particles constitute particulate radiation.

40. Alpha and beta particles are similar to x-rays and gamma rays in that they a. have no mass b. are part of the electromagnetic spectrum c. have the energy to ionize matter d. have characteristics of wavelength and frequency ANS: C Particulate radiation has the energy to ionize matter, similar to x-rays and gamma rays.

41. The general process of a radioactive element giving off excess energy and particles to regain stability is a. radioactivity b. radioactive decay c. ionization d. electron emission ANS: B Radioactive decay is the general process of a radioactive element giving off excess energy and particles to regain stability.

42. Radioactive elements a. have excess energy in their nuclei b. emit particles and energy c. are trying to become stable elements d. all of these ANS: D Radioactive elements, to regain stability, emit particles and energy from their nuclei.

43. Half-life is a. the rate at which a radioactive material decays b. half the time it takes for all the radioactivity to decay c. the rate at which particulate radiation is emitted from a radioactive atom’s nucleus d. none of these ANS: A Half-life is a way of describing the rate at which a radioactive material decays.

44. Half-life is defined as a. half the time it takes for all of the remaining atoms in an amount of a radioactive element to decay b. half the time it takes for half the remaining atoms in an amount of a radioactive

element to decay c. the time it takes for half the remaining atoms in an amount of a radioactive element to decay d. none of these ANS: C Half-life is the time it takes for half the remaining atoms in an amount of a radioactive element to decay.

45. The half-life of technetium-99m is 6 hours. How many unstable atoms will remain after 12 hours? a. One sixth of the original amount. b. One fourth of the original amount. c. One third of the original amount. d. Half of the original amount.

ANS: B After 6 hours half of the unstable atoms will remain; after 12 hours it is reduced by half again, resulting in one fourth of the original amount.

46. Alpha particles consist of a. two protons and two electrons b. two protons and two neutrons c. two electrons and two neutrons d. none of these ANS: B An alpha particle consists of two protons and two neutrons.

47. An alpha particle is the same as the nucleus of a(n) a. hydrogen atom b. oxygen atom c. carbon atom d. helium atom ANS: D The alpha particle’s two protons and two neutrons are identical to the makeup of the helium nucleus.

48. When compared with a beta particle, the alpha particle is a. much smaller b. much larger c. more penetrating d. none of these ANS: B The alpha particle is significantly larger than the beta particle.

49. When the alpha particle picks up two electrons as it passes through air it becomes a a. neutral atom of hydrogen b. radioactive atom of hydrogen

c. neutral atom of helium d. radioactive atom of helium ANS: C When the alpha particle (two protons and two neutrons) pick up two electrons, it becomes a neutral atom of helium.

50. The positively charged beta particle is a(n) a. electron b. alpha particle c. negatron d. positron ANS: D The positron is a positively charged beta particle.

51. As compared to an alpha particle, a beta particle a. has less mass b. has more mass c. has the same mass d. is less penetrating ANS: A A beta particle has significantly less mass than an alpha particle.

52. A negatively charged beta particle behaves the same as a(n) a. proton b. neutron c. electron d. positron ANS: C A negatively charged beta particle behaves exactly the same as an electron.

53. Which of the following is NOT a source of manmade ionizing radiation? a. x-rays b. radiopharmaceuticals used in nuclear medicine c. gamma rays d. all are mand made ionizing radiation ANS: C Manmade ionizing radiation used for medical purposes include x-rays and radiopharmaceuticals used in nuclear medicine. Gamma rays are a form of natural/background ioning radiation.

TRUE/FALSE 1. All electromagnetic radiation travels at the same speed.

ANS: T All electromagnetic radiation travels at the same speed, the speed of light.

2. Electromagnetic energy cannot travel through a vacuum.

ANS: F Electromagnetic energy can travel through a vacuum.

3. Electromagnetic radiation has properties of a wave and a particle.

ANS: T Called wave-particle duality, electromagnetic radiation has properties of a wave and a particle.

4. Higher energy radiation acts more like waves than like particles.

ANS: F Lower energy radiation acts more like waves; higher energy radiation acts more like particles.

5. All of the electromagnetic radiations are capable of ionizing matter.

ANS: F Only the highest energy electromagnetic radiations are able to ionize matter.

6. X-rays and gamma rays are the only electromagnetic radiations that are able to ionize matter.

ANS: T Of the electromagnetic radiations, only x-rays and gamma rays have enough energy to knock an electron from an atom’s shell.

7. All radioactive elements occur naturally.

ANS: F Many radioactive elements, such as radium, occur naturally but others are artificial.

8. The beta particle is an electron that is emitted from an outer shell.

ANS: F The beta particle is emitted from the nucleus.

9. We are exposed to non-medical ionizing radiation everyday categorized as natural or background radiation.

ANS: T We are exposed to ionizing radiation every day. These sources may be discussed as natural or background and manmade.

Chapter 04: The X-Ray Circuit Johnston/Fauber: Essentials of Radiographic Physics and Imaging, 4th Edition

MULTIPLE CHOICE 1. The study of stationary electric charges is a. electromagnetism b. electrodynamics c. electronics d. electrostatics ANS: D Electrostatics is the study of stationary (static) electric charges.

2. The study of electric charges in motion is a. electromagnetism b. electrodynamics c. electronics d. electrostatics ANS: B Electrodynamics is the study of electric charges in motion.

3. Electricity is typically thought to be the same as a. electromagnetism b. electrodynamics c. electronics d. electrostatics ANS: B Electricity is typically thought of as electric charges in motion, or electrodynamics.

4. The unit of measurement of electrical charges (SI) is a. Watt b. Joule c. Newton d. Coulomb ANS: D The Coulomb is the SI unit of measurement for electrical charge.

5. One Coulomb is equal to the electrical charge of 6.25  1018 a. electrons b. protons c. neutrons d. nuclei ANS: A One Coulomb is equal to the electrical charge of 6.25  1018 electrons.

6. A material that has a large number of free electrons is a a. good insulator b. good conductor

c. poor conductor d. All of these ANS: B A material that has a large number of free electrons is a good conductor.

7. Which of the following is NOT a correct law of electrostatics: a. like charges repel b. unlike charges attract c. electric charges reside only on the internal surface of conductors d. all of these are laws of electrostatics ANS: C Like charges repel and opposite charges attract. Electric charges reside only on the external surface of conductors.

8. As two charges get closer, the electrostatic force a. is directly proportional to the amount of distance between them b. is inversely proportional to the amount of distance between them c. is inversely proportional to the square of the distance between them d. is directly proportional to the square of the distance between them ANS: C As two charges get closer, the electrostatic force is inversely proportional to the square of the distance between them.

9. If a conductor is curved, the greatest amount of charge will be found a. at the beginning of the conductor b. at the end of the conductor c. at the area of the greatest curvature of the conductor d. along the straightest area of the conductor ANS: C With a curved conductor, the greatest electric charge will be found in the area of the most curvature.

10. Rubbing a rubber rod with fur is an example of electrification by a. contact b. friction c. induction ANS: B Rubbing a rubber rod with fur is an example of electrification by friction.

11. When there is movement of some of the electrons of an uncharged metallic object as it is brought into the electric field of a charged object, this is electrification by a. contact b. friction c. induction ANS: C

Electrification by induction occurs when a metallic object (without a charge) is brought into the electric field of a charged object. The result is that some of the electrons in the uncharged object move, creating electricity.

12. Electrification by induction requires that the two objects a. are in direct, constant contact with each other b. are not at all in contact with each other c. contact each other and then separate d. are as far apart as possible from each other ANS: B Induction of electricity only occurs when the objects are not in contact with each other but close enough for the uncharged object to be in the charged object’s electrical field.

13. To get electrons to flow along a conductor, there must be a. an excess of negative charges at one end of the wire b. an excess of negative charges at both ends of the wire c. an excess of positive charges at one end of the wire, and an excess of negative charges at the other d. an excess of positive charges at both ends of the wire ANS: C To make electrons flow along a conductor, there must be an electric potential—an excess of electrons at one end of the wire and an excess of positive charges at the other end.

14. The ability to do work because of a separation of charges is a. electric current b. electric potential c. resistance d. electrostatics ANS: B Electric potential is the ability to do work because of a separation of charges.

15. The flow of electrons along a conductor is a. electric current b. electric potential c. resistance d. electrostatics ANS: A Current is the flow of electrons along the conductor.

16. The property of a circuit that impedes the flow of electrons is a. electric current b. electric potential c. resistance d. electrostatics ANS: C Resistance is the property of a circuit that resists the flow of electrons.

17. The volt is the unit of measure for a. electric current b. electric potential c. resistance d. electrostatics ANS: B Electric potential is measured in volts.

18. The ampere is the unit of measure for a. electric current b. electric potential c. resistance d. electrostatics ANS: A Electric current is measured in amperes.

19. A volt is the potential difference that will maintain a current of _____ ampere in a circuit with a resistance of ______ ohm. a. 1; 2 b. 2; 1 c. 1; 1 d. 2; 2 ANS: C A volt is the potential difference that will maintain a current of one ampere in a circuit with a resistance of one ohm.

20. A volt is the ratio of work per unit of charge and is written a. Joules/Coulomb b. Coulomb/Joules c. Watt/Coulomb d. Coulomb/Watt ANS: A Joules/Coulomb describes a volt.

21. A 12-volt battery use 12 Joules of energy to move a. 1 Coulomb of charge b. 2 Coulombs of charge c. 12 Coulombs of charge d. 24 Coulombs of charge ANS: A Twelve Joules per 1 Coulomb is equal to 12 volts.

22. Kilovoltage peak is abbreviated as a. kP b. kPk

c. kVPk d. kVp ANS: D Kilovoltage peak, describing the voltage used during x-ray production, is abbreviated kVp. 23. The ampere is defined as ______ Coulomb passing by a point in ____ second. a. 1; 1 b. 2; 1 c. 1; 2 d. 2; 2 ANS: A An ampere is defined as 1 Coulomb passing by a point in 1 second.

24. A material that allows electrons to move is a(n) a. volt b. insulator c. conductor d. inductor ANS: C A conductor is a medium that allows electrons to move.

25. A material that resists the movement of electrons is a(n) a. ohm b. insulator c. conductor d. inductor ANS: B An insulator is a material that resists the flow of electrons.

26. Which of the following materials is a good conductor? a. Glass. b. Copper. c. Plastic. d. All of these are good conductors.

ANS: B Copper is an excellent conductor.

27. Glass is an example of a(n) a. insulator b. conductor c. material that allows the flow of electrons d. inductor ANS: A Glass is an insulator, resisting the flow of electrons.

28. A battery produces

a. b. c. d.

alternating current indirect current direct current current that changes direction

ANS: C A battery produces current that flows in only one direction, direct current.

29. Current that changes direction in the conductor is a. alternating current b. indirect current c. direct current d. weak current ANS: A Alternating current is constantly changing direction in the conductor.

30. The 60 Hz of electricity flowing into homes in the U.S. characterizes a. alternating current b. indirect current c. direct current d. weak current ANS: A 60 Hz of electricity flowing into homes in the U.S. refers to 60 cycles per second alternating current.

31. A(n) ______ is the electrical resistance between two points along a conductor that, when one volt is applied, produces 1 ampere of current. a. amp b. volt c. ohm d. Watt ANS: C By definition, an ohm is the electrical resistance between two points along a conductor that, when one volt is applied produces 1 ampere of current.

32. Ohm’s law can be written as a. amps = volts  ohms b. volts = amps  ohms c. ohms = volts  amps d. none of these ANS: B Ohm’s law is V = IR or volts = amps  ohms.

33. Ohm’s law can be written as a. I = V/R b. I = R/V c. R = I/V

d. R = IR ANS: A Ohm’s law can be written as V = IR or I = V/R or R = V/I. 34. When a conductor is longer: a. resistance decreases b. resistance increases c. resistance remains the same d. none of these ANS: B A longer conductor has more resistance than a shorter one.

35. When a conductor has a smaller cross-sectional area: a. resistance decreases b. resistance increases c. resistance remains the same d. none of these ANS: B A conductor has a smaller cross-sectional area and has more resistance.

36. When a metallic conductor is colder: a. resistance decreases b. resistance increases c. resistance remains the same d. none of these ANS: A A chilled metallic conductor has less resistance.

37. When a conductor is more twisted: a. resistance decreases b. resistance increases c. resistance remains the same d. none of these ANS: C As long as the length of the conductor remains the same, twisting the conductor does not affect resistance.

38. Which of the following is NOT a good electrical conductor? a. Gold. b. Copper. c. Glass. d. Water.

ANS: C Glass is not a good conductor of electricity; it is used as an insulator.

39. Which of the following is NOT a good insulator?

a. b. c. d.

Glass.

Gold.

Wood.

Ceramic.

ANS: B Gold is a good, although expensive, electrical conductor and would not be used as an insulator.

40. Rubber is a. an excellent conductor b. an excellent insulator c. a weak conductor d. a weak insulator ANS: B Rubber is a commonly used insulator because it works so well.

41. A material that will conduct electricity, but not as well as good conductors, but also will insulate, but not as well as good insulators are called a(n) a. excellent conductor b. weak conductor c. weak insulator d. semiconductor ANS: D A material that will conduct electricity, but not as well as good conductors, but also will insulate, but not as well as good insulators are called a semiconductor.

42. A closed pathway of wires and related elements is a(n) a. diode b. fuse c. rheostat d. electric circuit ANS: D An electric circuit is a closed pathway of wires and circuit elements allowing electricity to flow.

43. When a switch is turned on, the electric circuit a. is closed b. does not allow electrons to flow c. is opened d. none of these ANS: A Turning a switch on closes the circuit, allowing the flow of electrons.

44. A circuit with elements that are arranged like a ―bridge‖ or branching across a conductor is called a a. series circuit

b. parallel circuit c. bridge circuit d. series-parallel circuit ANS: B A parallel circuit has its circuit elements arranged like a ―bridge‖ or branching across a conductor.

45. A device that temporarily stores charge is the a. rheostat b. capacitor c. transformer d. fuse ANS: B The capacitor is an electronic device that stores an electric charge for a brief time.

46. An adjustable form of a resistor is the a. rheostat b. capacitor c. switch d. diode ANS: A The rheostat is a resistor that can be adjusted.

47. Which of the following does NOT open a circuit to keep the circuit and devices safe? a. Fuse. b. Resistor. c. Circuit breaker.

ANS: B Both the fuse and circuit breaker open the circuit when there is excessive current, protecting the circuit and associated devices.

48. Which of the following devices allow the electricity to flow in one direction only? a. Fuse. b. Switch. c. Transformer. d. Diode.

ANS: D The diode, or solid-state rectifier, allows electrons to flow in one direction only.

49. The device that regulates the amount of current flowing in the circuit is the a. battery b. resistor c. fuse d. transformer ANS: B By using a resistor to increase resistance, the amount of current is regulated.

50. The device that produces electrons through a chemical reaction is a a. transformer b. fuse c. resistor d. battery ANS: D The battery is a source of electrons, produced by a chemical reaction.

51. Which of the following devices can increase or decrease voltage by a set amount? a. Battery. b. Capacitor. c. Transformer.

ANS: C The transformer can increase or decrease voltage by a preset amount.

52. The purpose of grounding is to a. provide a source for electricity b. neutralize a charged object c. replace the circuit breaker d. none of these ANS: B Grounding is a protective measure that neutralizes charged objects.

53. The ability of a material to attract iron, cobalt, or nickel is called a. electromagnetism b. potential difference c. magnetism d. voltage ANS: C The ability of a material to attract iron, cobalt, or nickel is the definition of magnetism.

54. The lines of force in space associated with a magnet are called a. lines of flux b. dipoles c. domains d. lines of field ANS: A Lines of flux are the lines of magnetic force in space.

55. Inside the magnet, the lines of flux travel from a. north pole to south pole b. south pole to north pole c. east pole to west pole d. west pole to east pole ANS: B

The magnetic lines of flux travel from south pole to north pole inside the magnet.

56. Outside the magnet, the lines of flux travel from a. north pole to south pole b. south pole to north pole c. east pole to west pole d. west pole to east pole ANS: A The magnetic lines of flux travel from north pole to south pole outside the magnet.

57. Like poles of a magnet _________ each other and opposite poles ________ each other. a. attract; attract b. repel; repel c. attract; repel d. repel; attract ANS: D Like poles (north and north; south and south) repel each other. whereas opposite magnetic poles (north and south) attract each other.

58. The force of magnetic attraction varies a. directly with the square of the strength of the poles b. inversely with the square of the strength of the poles c. directly with the square of the distance between the poles d. inversely with the square of the distance between the poles ANS: D The force of magnetic attraction (or repulsion) varies inversely with the square of the distance between the poles.

59. Magnetic field strength is measured in a. tesla b. ohms c. Newtons d. Joules ANS: A The SI unit of measurement for magnetic strength is the tesla (T).

60. A material that is weakly repelled by a magnetic field is classified as a. nonmagnetic b. paramagnetic c. ferromagnetic d. diamagnetic ANS: D Diamagnetic materials are weakly repelled by a magnetic field.

61. A material that is strongly attracted to a magnetic field is a. nonmagnetic

b. paramagnetic c. ferromagnetic d. diamagnetic ANS: C Ferromagnetic materials are strongly attracted to a magnetic field.

62. A material that is weakly attracted to a magnetic field is a. nonmagnetic b. paramagnetic c. ferromagnetic d. diamagnetic ANS: B Paramagnetic materials are weakly attracted to a magnetic field.

63. Wood and plastic are considered: a. nonmagnetic b. paramagnetic c. ferromagnetic d. diamagnetic ANS: A Wood and plastic are nonmagnetic, not attracted to a magnetic field at all.

64. The idea that electricity flowing through a conductor produces a magnetic field was identified by a. Faraday b. Roentgen c. Tesla d. Oersted ANS: D Hans Oersted discovered that electricity flowing through a wire produced a magnetic field.

65. The idea that electricity could be induced in a wire by moving it through a magnetic field was discovered by a. Faraday b. Roentgen c. Tesla d. Oersted ANS: A Michael Faraday found that moving a wire through a magnetic field could induce a electric current in the wire.

66. According to Faraday’s Law of Electromagnet Induction, if the strength of the magnetic field is increased, _________ is also increased. a. current b. resistance c. voltage

d. all of these ANS: A According to Faraday’s Law of Electromagnet Induction, if the strength of the magnetic field is increased, electric current is also increased.

67. An electromagnet includes: a. a coil of wire b. an iron core c. a solenoid d. all of these ANS: D An electromagnet includes a coil of wire (solenoid) wrapped around an iron core.

68. Reducing the number of coils in the wire will result in a. increased electromagnetic induction b. reduced electromagnetic induction c. no effect on electromagnetic induction ANS: B Less voltage is induced when there are fewer coils in the wire.

69. Increasing the strength of the magnetic field results in a. increased electromagnetic induction b. reduced electromagnetic induction c. no effect on electromagnetic induction ANS: A More voltage is induced with a stronger magnetic field strength.

70. Slowing down the speed at which the conductor passes through the magnetic field results in a. increased electromagnetic induction b. reduced electromagnetic induction c. no effect on electromagnetic induction ANS: B Less voltage is induced when the speed at which the coil of wire passes through the magnetic field is reduced.

71. Mutual induction involves electricity being induced into a ______________. a. primary coil b. secondary coil c. moving magnetic field d. none of these ANS: B Mutual induction results in electricity being induced in the secondary coil of wire.

72. With mutual induction, to induce voltage in the secondary coil, there must be a. a physical connection between the two coils of wire b. alternating current in the primary coil of wire

c. direct current in the primary coil of wire ANS: B Mutual induction requires alternating current to be flowing in the primary coil of wire.

73. _______________ current is induced in the secondary coil when mutual induction takes place. a. alternating b. direct c. neither of the above d. both of the above ANS: A Alternating current is induced in the secondary coil during mutual induction.

74. With self-induction, the secondary current a. is in a secondary coil b. is in the original coil c. opposes the original current d. is in the original coil and opposes the original current ANS: D Self-induction involves only one coil of wire and produces current that opposes the original.

75. Self-induction is based on the concepts introduced in a. Ohm’s law b. Lenz’s law c. Oersted’s law d. Faraday’s law ANS: B Self-induction is based on the concepts of Lenz’s law.

76. A device that converts mechanical energy to electrical energy is a a. motor b. transformer c. generator d. electromagnet ANS: C The generator converts mechanical energy, such as wind or moving water, to electrical energy.

77. A device that converts electrical energy to mechanical energy is a a. motor b. transformer c. generator d. electromagnet ANS: A The motor converts electrical energy to mechanical energy.

78. A step-up transformer results in a. increased current b. increased voltage c. increased resistance ANS: B A step-up transformer results in increased voltage on the secondary side.

79. A step-down transformer results in a. increased current b. increased voltage c. increased resistance ANS: A A step-down transformer results in decreased voltage and increased current on the secondary side.

80. A step-up transformer results in a. decreased current b. decreased voltage c. decreased resistance ANS: A A step-up transformer results in increased voltage and decreased current on the secondary side.

81. A step-down transformer results in a. decreased current b. decreased voltage c. decreased resistance ANS: B A step-down transformer results in decreased voltage and increased current on the secondary side.

82. A transformer with 100 times more turns on the secondary side as compared with the primary a. is a step-up transformer b. is a step-down transformer c. is a transformer that works on self-induction d. will have 100 times the current on the secondary side ANS: A A transformer with 100 times more turns on the secondary wire is a step-up transformer, resulting in 100 times the voltage.

83. A transformer with 100 turns on the primary side and 50 turns on the secondary a. is a step-up transformer b. is a step-down transformer c. is a transformer that works on self-induction d. will have 100 times the current on the secondary side

ANS: B A transformer with 100 turns on the primary side and 50 on the secondary is a step-down transformer, resulting in one half the voltage and two times the current on the secondary side.

84. Closed-core and shell-type transformers a. are seldom used in x-ray equipment b. are less efficient types of transformers c. are simpler types of transformers d. are more sophisticated and incorporate a ferromagnetic core ANS: D Closed-core and shell-type transformers are more sophisticated, efficient transformers and are typically used in x-ray equipment that incorporate a ferromagnetic core.

85. The autotransformer a. uses self-induction b. uses two coils of wire c. does not use a magnetic core d. is not typically found in an x-ray machine ANS: A The autotransformer is used in x-ray equipment and operates based on self-induction, using one coil of wire.

86. The autotransformer is found on the a. primary circuit b. secondary circuit c. filament circuit d. none of these ANS: A The autotransformer is found on the primary circuit.

87. The mA meter is found on the a. primary circuit b. secondary circuit c. filament circuit d. none of these ANS: B The mA meter is found on the secondary circuit.

88. The rectifiers are found on the a. primary circuit b. secondary circuit c. filament circuit d. none of these ANS: B The rectifiers are found on the secondary circuit.

89. The main power switch is found on the a. primary circuit b. secondary circuit c. filament circuit d. none of these ANS: A The main power switch is found on the primary circuit.

90. The x-ray tube (not filaments) is found on the a. primary circuit b. secondary circuit c. filament circuit d. none of these ANS: B The x-ray tube (not filaments) is found on the secondary circuit.

91. The rheostat is found on the a. primary circuit b. secondary circuit c. filament circuit d. none of these ANS: C The rheostat is found on the filament circuit.

92. The primary side of the high-voltage transformer is found on the a. primary circuit b. secondary circuit c. filament circuit d. none of these ANS: A The primary side of the high-voltage transformer is found on the primary circuit.

93. The timer circuit is found on the a. primary circuit b. secondary circuit c. filament circuit d. none of these ANS: A The timer circuit is found on the primary circuit.

94. The step-down transformer is found on the a. primary circuit b. secondary circuit c. filament circuit d. none of these

ANS: C The step-down transformer is found on the filament circuit.

95. The secondary side of the high-voltage transformer is found on the a. primary circuit b. secondary circuit c. filament circuit d. none of these ANS: B The secondary side of the high-voltage transformer is found on the secondary circuit.

96. The filaments within the x-ray tube are found on the a. primary circuit b. secondary circuit c. filament circuit d. none of these ANS: C The filaments inside the x-ray tube are found on the filament circuit.

97. The line compensator is typically wired to the a. timer circuit b. primary side of the high-voltage transformer c. autotransformer d. step-down transformer ANS: C The line compensator is typically wired to the autotransformer.

98. The device referred to as the kVp selector is the a. primary side of the high-voltage transformer b. secondary side of the high-voltage transformer c. step-down transformer d. autotransformer ANS: D The autotransformer adjusts based on the kVp set at the operating console.

99. The only adjustable transformer in the x-ray circuit is the a. primary side of the high-voltage transformer b. secondary side of the high-voltage transformer c. autotransformer d. step-down transformer ANS: C The autotransformer is the only transformer in the x-ray circuit that is adjustable because it does not have a fixed number of turns on both the primary and secondary sides.

100. The timer located on the secondary circuit is the a. synchronous timer

b. mAs timer c. digital timer d. none of these ANS: B The only timer located on the secondary circuit is the mAs timer.

101. The timer that is based on a motor is the a. synchronous timer b. mAs timer c. digital timer d. electronic timer ANS: A The synchronous timer is based on the synchronous motor.

102. AEC stands for a. automatic x-ray control b. applied exposure control c. automatic exposure control d. automatic exposure centering ANS: C AEC stands for ―automatic exposure control.‖ 103. For AEC, the ionization chamber is placed between the ____________ and _____________. a. tube; patient b. patient; image receptor c. grid; image receptor d. image receptor; wall ANS: B The AEC ionization chamber is placed between the patient and image receptor to measure the radiation exiting the patient.

104. Assuming all other factors are accurately adjusted, the exposure time with AEC is based on a. the procedure being done b. the anatomy placed over the ionization chamber c. the age of the patient d. the desired quality of the image ANS: B With all factors set appropriately, the length of exposure will be based on the density of the anatomy covering the detector.

105. The device that adjusts the incoming power supplied to the x-ray circuit to maintain 220 V is the a. circuit breaker b. autotransformer c. timer circuit d. line compensator

ANS: D The line compensator adjusts the incoming voltage to maintain a constant 220 V. 106. The _____________________ converts volts to kilovolts. a. step-up transformer b. step-down transformer c. autotransformer d. none of these ANS: A The step-up transformer converts volts to kilovolts.

107. AEC controls the ________________ of radiation reaching the image receptor. a. type b. quantity c. energy level d. all of these ANS: B AEC controls only the quantity of radiation reaching the image receptor.

108. The devices that change alternating to direct current are the a. transformers b. filaments c. rectifiers d. rheostats ANS: C Rectifiers convert alternating to direct current.

109. The solid-state rectifier has a(n) _______ crystal and a(n) _________ crystal. a. a-type; b-type b. x-type; z-type c. a-type; c-type d. p-type; n-type ANS: D A solid-state rectifier includes both n-type and p-type crystals.

110. The n-type crystal from a solid-state rectifier has an abundance of a. electrons b. electron traps c. protons d. proton traps ANS: A The n-type crystal has many extra freely moving electrons.

111. When a negative charge is placed on the p-type crystal and a positive charge on the n-type crystal, the result is a. current flows through the diode

b. current does not flow through the diode c. voltage decreases d. voltage increases ANS: B When a negative charge is placed on the p-type crystal and a positive charge on the n-type crystal, the traps and electrons remain at the ends of the crystals and do not allow electrons (current) to be conducted.

112. Half-wave rectification a. suppresses half of the AC cycle b. uses two rectifiers c. is seldom used in modern x-ray equipment d. all of these ANS: D Half-wave rectification uses two rectifiers to suppress half of the AC cycle. In that only half of the electrical current would allow production of x-rays, this is seldom used.

113. Ripple refers to a. the fluctuation in amperage in the electricity b. the fluctuation in voltage in the electricity c. the fluctuation in resistance in the electricity d. all of these ANS: B Ripple is the fluctuation of voltage in the electrical signal.

114. Four rectifiers will produce a. alternating current b. half-wave rectified current c. full-wave rectified current d. indirect current ANS: C Four rectifiers produce full-wave rectified electricity.

115. Single-phase full-wave rectified current has a. 10% ripple b. 25% ripple c. 50% ripple d. 100% ripple ANS: D Single-phase full-wave rectified current has voltage that varies from 0 to the maximum or 100%.

116. Three-phase power a. uses three AC waveforms at the same time b. results in three times the ripple c. is half-wave rectified

ANS: A Three-phase power uses three AC waveforms that are 120 degrees out of phase with each other.

117. The less the ripple a. the more consistent the energy of the x-rays produced b. the less consistent the energy of the x-rays produced c. the greater the amount of radiation needed for a specific exposure d. none of these ANS: A As the amount of ripple decreases, a smaller range of voltages will be used to produce the x-rays, resulting in the energy of the x-rays being more consistent.

118. A pulsed DC is produced by a. half-wave rectification b. full-wave rectification c. three-phase power d. high-frequency generators ANS: D High-frequency generators produce pulsed direct current.

119. The least amount of ripple is the result of a. half-wave rectification b. full-wave rectification c. three-phase power d. high-frequency generators ANS: D High-frequency generators produce current with less than 1% ripple, the least of all methods.

120. What must be adjusted to change the temperature of the filament? a. Step-up transformer. b. Rheostat. c. Autotransformer. d. Solid-state rectifier.

ANS: B Adjusting the rheostat changes the amount of current flowing through the filament, thereby adjusting the temperature of the filament.

121. The rate of flow of electrons passing through the x-ray tube during an exposure is a. kVp b. tube current c. filament current d. Hz ANS: B

The tube current, also referred to as mA, is the rate of flow of electrons passing through the x-ray tube during an exposure.

122. Adjusting the rheostat changes the _______________ in the ______________ circuit. a. resistance; primary b. voltage; primary c. resistance; filament d. voltage; filament ANS: C Changing the rheostat results in a change in the resistance in the filament circuit.

123. Which of the following requires the greatest amount of resistance in the filament circuit? a. 50 mA. b. 100 mA. c. 600 mA. d. 1000 mA. ANS: A The higher the resistance, the lower the resulting amperage.

124. The purpose of the filament circuit is to a. produce high voltage b. provide direct current to the x-ray tube c. boil electrons off the filament d. reduce the current reaching the filament ANS: C The goal of the filament circuit is to heat the filament until electrons are boiled off.

125. How much current is needed to heat the filament enough to produce a 100–1000 mA tube current? a. 1 A to 10 A. b. 5 A to 7 A. c. 1 mA to 10 mA. d. 5 mA to 7 mA. ANS: B 5–7 A of current must pass through the filament to produce a tube current from 100 mA to 1000 mA. 126. mAs is a. another name for tube current b. related to the amount of radiation exposing the patient c. the product of mA and exposure time d. the product of mA and exposure time, and related to the amount of radiation exposing the patient ANS: D mAs is the product of mA and exposure time and describes the amount of radiation leaving the x-ray tube and exposing the patient.

127. The primary purpose of the step-down transformer is to a. decrease the voltage going to the filaments b. decrease the amperage going to the filaments c. increase the voltage going to the filaments d. increase the amperage going to the filaments ANS: D Although the step-down transformer does reduce the voltage, its primary purpose is to increase the amperage to the filaments, allowing the thermionic emissions of electrons.

128. A general purpose radiographic tube typically has ____ filament(s). a. one b. two c. three d. four ANS: B A typical tube has two filaments; it is sometimes referred to as a dual-focus tube.

129. Determining which filament will be used during an exposure is done at the operating console by the selection of a. small or large focal spot b. small or large filament c. high or low exposure time d. high or low kVp ANS: A When the large or small focal spot is selected at the operating console, the associated filament will be used.

130. Place the following events leading to the production of x-rays in order, from first to last I. alternating current is converted to direct current II. voltage is adjusted at the autotransformer III. kilovoltage leaves the step-up transformer IV. voltage goes to primary side of step-up transformer a. b. c. d.

I, II, III, IV II, I, IV, III IV, III, II, I II, IV, III, I

ANS: D Leaving the autotransformer, the selected voltage goes through the primary and secondary sides of the step-up transformer and is then rectified.

131. Place the following events leading to the production of x-rays in order, from first to last I. kilovoltage is rectified II. voltage is adjusted at the autotransformer

III. anode and cathode develop large positive and negative charges IV. voltage changes to kilovoltage a. IV, III, II, I b. II, IV, I, III c. III, I, II, IV d. I, III, IV, II ANS: B Leaving the autotransformer, the selected voltage goes to the step-up transformer to change to kilovoltage and, after being rectified, results in the anode having a large positive charge and the cathode a large negative.

132. Place the following events leading to the production of x-rays in order, from first to last I. resistance is adjusted at the rheostat II. filament is heated III. filament current is increased IV. electrons are boiled off filament a. II, IV, III, I b. III, II, IV, I c. I, III, II, IV d. I, IV, III, II ANS: C After the rheostat adjusts the current, it goes to the step-down transformer where the current is increased. This increased current goes to the filament where it heats up and electrons are boiled off.

133. Place the following events leading to the production of x-rays in order, from first to last I. electrons are boiled off filament II. electricity leaves the autotransformer III. electricity goes to primary side of step-down transformer IV. filament is heated a. I, II, III, IV b. III, IV, II, I c. II, IV, III, I d. II, III, IV, I ANS: D The autotransformer serves as the source of electricity for the filament circuit. After the autotransformer, the current will flow to the step-down transformer, heat the filament, and boil off electrons.

134. Which of the following occurs first in the sequence of events leading to x-ray production? a. AC is converted to DC. b. Voltage becomes kilovoltage. c. Anode and cathode become positively and negatively charged. d. Electrons flow from cathode to anode.

ANS: B

Before the kilovoltage is applied to the tube and the electrons travel across, the voltage must first become kilovoltage (step-up transformer) and then be rectified.

135. Which of the following occurs first in the sequence of events leading to x-ray production? a. Voltage is stepped down and amperage increases. b. Electron cloud is produced. c. Current is adjusted by rheostat. d. Electrons flow from cathode to anode.

ANS: C After the rheostat adjusts the current, it is increased by the step-down transformer, heats the filament, and thus produces a cloud of electrons. The electrons then flow from cathode to anode.

136. The selection of the large or small filament in the cathode depends on the choice of focal spot size at the operating console and the a. exposure time selected b. kVp selected c. mA station selected d. all of these ANS: C The choice of mA station determines the size of the filament—small filament for small mA stations and large filament for large mA stations.

TRUE/FALSE 1. Electric charges are found only on the outside of a conductor.

ANS: T Electric charges are found only on the external surface of a conductor.

2. Only electrons are free to move in solid conductors.

ANS: T Only negative charges (electrons) can move in solid conductors.

3. Water is a good conductor.

ANS: T Water is a good conductor because it allows electrons to flow freely.

4. Lines of flux in the same direction attract each other.

ANS: F Lines of flux in the same direction repel each other.

5. Lines of flux in the opposite directions attract each other.

ANS: T Magnetic lines of flux in the opposite directions attract each other.

6. Nonmagnetic material, when interacting with a magnet, distorts the magnetic field.

ANS: F Nonmagnetic materials will not affect the magnetic field.

7. Every magnet has a north, south, east, and west pole.

ANS: F Every magnet has a north and south pole; they do not have east or west poles.

8. With alternating current, the north and south poles of the magnetic field surrounding the wire change each time the electrons change direction.

ANS: T With alternating current, each time the electrons change direction there is a corresponding switch in the north and south poles of the magnetic field, resulting in a changing or moving magnetic field.

9. The autotransformer, using only one coil of wire, does not have both a primary and secondary side.

ANS: F Even though the autotransformer uses only one wire, it still has both a primary and secondary side.

10. The synchronous timer is the most widely used timer in today’s x-ray equipment.

ANS: F The electronic type of timer is the most widely used in modern x-ray equipment.

11. A capacitor and rheostat are used in electronic timers.

ANS: T Electronic timers use a capacitor and variable resistor, or rheostat.

12. During a radiographic procedure using AEC, the exposure time will be shorter when the spine covers the ionization chamber, as compared with air covering the chamber.

ANS: F It will take a longer exposure time for enough radiation to pass through the spine as compared with air.

13. Three-phase power has less than 1% ripple.

ANS: F Three-phase power has from 3.5 to 13% ripple.

14. The number of turns on the primary and secondary sides of the step-down transformer can be adjusted to control the current to the filaments.

ANS: F The turns ratio of the step-down transformer is not adjustable. Adjustment of the current is done by the variable resistor, immediately preceding the transformer.

15. The filament is a straight piece of wire in the cathode.

ANS: F The filament is a coiled piece of wire in the cathode.

16. To produce x-rays, a cloud of electrons are necessary and the cathode must have a strong positive charge.

ANS: F To produce x-rays, the cloud of electrons is necessary, as is a very negatively charged cathode and very positively charged anode.

Chapter 05: The X-Ray Tube Johnston/Fauber: Essentials of Radiographic Physics and Imaging, 4th Edition MULTIPLE CHOICE 1. The protective tube housing is lined with a. aluminum b. copper c. lead d. glass ANS: C The protective tube housing is lined with lead so it can absorb radiation.

2. The protective tube housing helps to a. keep the tube cool b. prevent electrical shocks to the radiographer c. limit leakage radiation d. provide solid, stable mechanical support e. all of these ANS: E The protective tube housing does provide solid, stable mechanical support, electrical insulation and thermal cooling, and limits the radiation that can exit the tube other than toward the patient.

3. By regulation, x-ray tube leakage radiation can be no more than a. 100 R per hour at a distance of one meter b. 100 R per minute at a distance of one meter c. 100 mR per hour at a distance of one meter

d. 100 mR per minute at a distance of one meter ANS: C Leakage radiation can measure no more than 100 mR per hour.

4. After lengthy exposures, the x-ray tube on a mobile fluoroscopic unit may a. become very hot b. produce large electric shocks c. exceed leakage radiation limits d. none of these ANS: A After extended usage, the tube may become very hot.

5. Which of the following helps to keep the x-ray tube cool on the inside of the protective housing? a. Cooling fans b. Oil bath c. Target window d. Cooling fans and an oil bath ANS: D The design of the housing incorporates an oil bath and cooling fans to help dissipate heat away from the tube, protecting it from thermal damage.

6. In terms of the x-ray tube, envelope is another name for the a. anode b. cathode c. glass or metal enclosure d. induction motor ANS: C Envelope is another name for the metal or glass enclosure that contains the anode and cathode.

7. The positive end of the x-ray tube is the a. anode b. cathode c. filament d. envelope ANS: A The anode is the positive end of the x-ray tube.

8. The negative end of the x-ray tube is the a. anode b. cathode c. target d. envelope ANS: B The cathode is the negative end of the x-ray tube.

9. The anode includes the a. target b. induction motor c. envelope d. target and induction motor ANS: D The anode includes both the target and the induction motor.

10. The cathode includes the a. envelope b. filaments c. focusing cup d. filaments and the focusing cup ANS: D The cathode includes both the filaments and the focusing cup.

11. The primary purpose of the glass or metal enclosure is to a. insulate against electrical shock b. maintain a vacuum within the tube c. allow the heat to dissipate d. none of these ANS: B The main purpose of the enclosure is to maintain a vacuum in the tube.

12. Glass enclosures are typically made of borosilicate glass (Pyrex) because a. it is an inexpensive material b. it is an expensive material c. it is very heat-resistant d. none of these ANS: C Glass tube enclosures are made of borosilicate glass because it can withstand high levels of heat.

13. ―Sun-tanning‖ in glass envelopes is caused by a. the color the tube gets as it loses its resistance to heat over time b. oil leaking out of the protective housing c. the breaking of the glass envelope into small pieces of glass d. the vaporized tungsten from filament deposits that coats the inside of the glass ANS: D Vaporized tungsten (from the filament) tends to coat the inside of the glass envelopes.

14. Arcing a. is a problem associated with glass envelopes b. is a problem associated with metal envelopes c. is another name for the flow of electrons from cathode to anode

d. is another name for the flow of electrons from anode to cathode ANS: A Arcing is a problem with glass envelopes, resulting in tube damage.

15. The area of the envelope where x-rays should exit the tube is the a. target door b. target window c. x-ray trap d. x-ray gate ANS: B The target window is the area of the glass or metal envelope where x-rays are intended to leave the tube.

16. The target window is usually approximately a. 1 mm2 b. 5 mm2 c. 1 cm2 d. 5 cm2 ANS: D The target window is typically approximately 5 cm2.

17. The target window is usually approximately a. 1 mm2 b. 5 mm2 c. 1 cm2 d. 5 cm2 ANS: D The target window is typically approximately 5 cm2.

18. The anode serves as a(n) a. target for the electron interaction to produce x-ray b. electrical conductor c. heat conductor d. all of these ANS: D The anode is a target for the electrons, an electrical conductor, and a thermal (heat) conductor.

19. Electrons from the cathode that do not produce x-rays will a. be absorbed in the envelope b. return to the cathode filament c. be absorbed in the air in the tube d. continue as current flow through the circuit ANS: D Electrons that do not result in x-ray production continue on through the electrical circuit.

20. The two types of anode designs are stationary and a. traveling b. rotating c. dynamic d. all of these ANS: B Anode designs include stationary and rotating.

21. The stationary anode includes an area of tungsten embedded in a. a copper rod b. a tungsten rod c. a molybdenum disc d. a copper disc ANS: A The stationary anode has a small area of tungsten embedded in a copper rod.

22. The stationary anode includes an area of ___________ embedded in a copper rod. a. copper b. molybdenum c. tungsten d. glass ANS: C The stationary anode has a small area of tungsten embedded in a copper rod.

23. The primary disadvantage to the stationary anode is that it a. is too expensive b. heats up too quickly during x-ray production c. uses tungsten d. all of these ANS: B Because it does move, heat builds up very quickly with a stationary anode.

24. X-ray tubes with a stationary anode design may be found today a. in small x-ray departments b. in medical areas that only need large x-ray exposures c. in medical areas that only need small x-ray exposures d. in medical areas that only image the head and teeth ANS: C Stationary anode tubes can only be used with small x-ray exposures, no matter what anatomy is being imaged.

25. The rotating anode design uses a disc whose core is made of a. copper b. tungsten c. aluminum d. molybdenum

ANS: D The core of a rotating anode is molybdenum.

26. One of the reasons that molybdenum is used in rotating anodes is because it is a. light but strong b. heavy but strong c. dense but strong d. an excellent conductor ANS: A Molybdenum is used, in part, because it is light but strong, making it easier for the anode to rotate.

27. Copper can be found __________________ of the rotating anode. a. in the shaft b. in the target area c. surrounding the tungsten d. in the base ANS: A Copper is part of the shaft of the rotating anode.

28. Copper is used in the rotating anode because a. it is an excellent conductor b. it is an excellent insulator c. it has a high atomic number d. it has a low atomic number ANS: A Copper is used in the shaft of the rotating anode because it is an excellent conductor of heat and electricity.

29. Molybdenum a. is an excellent conductor of electricity b. is a poor conductor of electricity c. is an excellent thermal conductor d. is a poor thermal conductor ANS: D Because molybdenum is a poor thermal conductor it keeps the heat from the bearings of the rotor, minimizing damage.

30. Molybdenum is found in the _______________ of the rotating anode. a. base b. disc c. target area d. base and disc ANS: D Molybdenum is found in both the base and disc of the rotating anode.

31. Tungsten is used to coat the anode disc because it a. has a low melting point b. has a low atomic number c. has a high atomic number d. does not conduct heat well ANS: C Tungsten is used to coat the anode because it has a high atomic number, which improves the efficiency of x-ray production.

32. Tungsten is used to coat the anode disc because it a. has a high melting point b. has a low melting point c. has a low atomic number d. does not conduct heat well ANS: A Tungsten is used to coat the anode because it has a high melting point.

33. Tungsten is used to coat the anode disc because it a. has a low atomic number b. has a low melting point c. conducts heat well d. does not conduct heat well ANS: C Tungsten is used to coat the anode because it is an excellent thermal conductor.

34. The atomic number of tungsten is a. 14 b. 34 c. 74 d. 104 ANS: C The atomic number of tungsten is 74.

35. The melting point of tungsten is a. 400 °C b. 1400 °C c. 2400 °C d. 3400 °C ANS: D The melting point of tungsten is 3400° C. 36. For the covering of the rotating anode disc, _____________ may be added to tungsten to increase thermal capacity and tensile strength. a. molybdenum b. rhenium c. copper

d. none of these ANS: B For the covering of the rotating anode disc, rhenium may be added to tungsten to increase thermal capacity and tensile strength.

37. The portion of the induction motor that is outside the tube enclosure is the a. rotor b. stator c. disc d. bearings ANS: B The stator is the portion of the induction motor that is found outside the tube enclosure.

38. The portion of the induction motor that is made of electromagnets that are energized in opposing pairs and induce an electric current with associated magnetic field is the a. rotor b. stator c. disc d. bearings ANS: B The stator is the portion of the induction motor that is made of electromagnets that are energized in opposing pairs and induce an electric current in the rotor with associated magnetic field.

39. The motor that turns the anode operates through a. the use of a transformer b. the physical connection between rotor and stator c. self-induction d. mutual induction ANS: D Mutual induction is the basis for how the induction motor turns the anode.

40. The induction motor turns the rotor of a general-purpose x-ray tube ___________ revolutions per minute. a. 1400 b. 3400 c. 6000 d. 10,000 ANS: B The induction motor turns the rotor of a general-purpose x-ray tube 3400 revolutions per minute.

41. The induction motor turns the rotor of a specialty x-ray tube ___________ revolutions per minute. a. 1400 b. 3400

c. 6000 d. 10,000 ANS: D Because they tend to produce more x-rays and more heat, the induction motor turns the rotor of a specialty x-ray tube 10,000 revolutions per minute.

42. The rotor is found a. near the tube window b. outside the envelope c. inside the envelope d. towards the cathode end of the tube ANS: C The rotor is found within the glass or metal envelope.

43. The rotor is made of a. an iron core b. a molybdenum core c. tungsten d. rhenium ANS: A The rotor is made of an iron core surrounded by coils.

44. With a rotating anode, the focal spot becomes a focal a. path b. stripe c. track d. circle ANS: C Because the target is constantly moving during the exposure, the focal spot becomes a focal track.

45. The primary advantage of a rotating anode is that it allows a. greater heat capacity at the anode b. less heat capacity at the anode c. greater heat capacity at the cathode d. less heat capacity at the cathode ANS: A By having a focal track, the rotating anode allows greater heat capacity at the anode, allowing larger x-ray exposures.

46. The effective focal spot is the a. same as the actual focal spot b. actual area where the electrons hit the target c. actual size of the origin of the x-ray beam d. origin of the x-ray beam as seen from below the tube ANS: D

The effective focal spot is the origin of the x-ray beam as seen from below the tube, or from the patient’s perspective.

47. The size of the actual focal spot depends on a. the size of the anode b. the size of the cathode filament being used c. the size of the effective focal spot d. the size of the tube ANS: B The size of the area of the target that is bombarded with electrons depends on the size of the electron stream, which is determined by the size of the filament.

48. The advantage to using a larger actual focal spot is a. the image is sharper b. there is less wear and tear on the tube c. the image is less sharp d. higher exposures can be used ANS: D A large actual focal spot spreads out the heat that is produced over a larger area, allowing greater x-ray exposures.

49. The advantage to using a small effective focal spot is a. the image is sharper b. the image is less sharp c. less exposure can be used d. it is safer for the patient ANS: A A small effective focal spot results in a sharper image.

50. The best situation for less heat production and a quality image is to have a: 1. small actual focal spot 2. large actual focal spot 3. small effective focal spot a. small actual focal spot and large actual focal spot only b. small actual focal spot and small effective focal spot only c. large actual focal spot and small effective focal spot ANS: C The best situation is to have a large actual focal spot (greater heat capacity) and a small effective focal spot size (sharper image).

51. Most x-ray tube target angles range between a. 2 and 5 degrees b. 7 and 18 degrees c. 20 and 28 degrees d. 35 and 45 degrees

ANS: B Most x-ray tubes have a target angle between 7 and 18 degrees.

52. The most common x-ray tube target angle is a. 2 degrees b. 8 degrees c. 12 degrees d. 18 degrees ANS: C The most common target angle is 12 degrees.

53. The relationship between the actual focal spot size, effective focal spot size, and anode target angle is called a. the anode heel effect b. the line-focus principle c. Roentgen’s principle d. all of these ANS: B The relationship between the actual focal spot size, effective focal spot size, and anode target angle is called the line-focus principle.

54. The smaller the anode angle, the _________ the effective focal spot. a. smaller b. larger c. wider d. longer ANS: A The smaller the anode angle, the smaller the effective focal spot.

55. If the anode angle becomes too small, which of the following may result? a. X-ray beam size may start to be limited. b. The intensity of the beam near the anode may be reduced. c. The anode heel effect may be seen. d. All of these.

ANS: D As the anode angle becomes very small, the beam area will be restricted and the intensity of the beam toward the anode reduced (the anode heel effect).

56. Based on the anode heel effect, an image taken at 40 inches source-to-image receptor distance (SID) may demonstrate a. decreased exposure toward the anode end of the beam b. decreased exposure toward the cathode end of the beam c. decreased exposure toward the left side of the beam d. decreased exposure toward the right side of the beam ANS: A The anode heel effect results in decreased exposure toward the anode end of the beam.

57. Part of the useful x-ray beam gets absorbed in the anode when a. the anode angle is too large b. the anode angle is too small c. the actual focal spot is too large d. the actual focal spot is too small ANS: B The anode heel effect, when part of the useful x-ray beam gets absorbed in the anode, occurs when the anode angle gets very small.

58. An x-ray tube with two filaments is called a a. two-focus tube b. dual-focus tube c. trifocal tube d. highly specialized tube ANS: B The dual-focus tube, with two filaments, is the most common x-ray tube.

59. The cathode a. includes a focusing cup b. provides electrons for x-ray production c. includes filaments d. all of these ANS: D The cathode includes filaments and focusing cup, and produces the electrons for x-ray production.

60. The focusing cup is part of the a. filament circuit b. primary circuit c. secondary circuit d. none of these ANS: C The focusing cup is part of the secondary circuit.

61. The filaments are connected to the a. filament circuit b. primary circuit c. secondary circuit d. none of these ANS: A The filaments are connected to the filament circuit.

62. Each cathode filament is approximately a. 1–3 mm long b. 5–6 mm long

c. 7–15 mm long d. 20–35 mm long ANS: C Each cathode filament is approximately 7–15 mm long.

63. The filament is primarily made of a. rhenium b. copper c. molybdenum d. tungsten ANS: D The filament is primarily made of tungsten.

64. There is about 1–2% of ______________ added to the filament to increase thermionic emission. a. thorium b. rhenium c. tungsten d. copper ANS: A About 1–2% of thorium is added to the tungsten in the filament to increase thermionic emission.

65. The purpose of the focusing cup is to a. produce thermionic emission from the filament b. prevent thermionic emission from the filament c. allow the electrons to spread out d. keep the electrons together ANS: D The purpose of the focusing cup is to keep the electrons from spreading apart.

66. To keep the electrons together, the focusing cup a. has a negative charge b. has a positive charge c. physically surrounds the electron cloud d. physically surrounds the electron cloud and has a negative charge ANS: D The focusing cup does physically surround the electron cloud and it is its negative charge that keeps the electrons together.

67. The focusing cup is able to keep the electrons together because a. opposite charges repel b. opposite charges attract c. like charges repel d. like charges attract ANS: C

The focusing cup is able to keep the electrons together because the negative charge of the focusing cup repels the negative charges of the electrons, pushing them together.

68. The focusing cup is made of a. tungsten b. nickel c. thorium d. molybdenum ANS: B The focusing cup is made of nickel.

69. When the exposure switch at the operating console is first depressed: a. high voltage creates a large potential difference between the anode and cathode b. the anode starts rotating c. thermionic emission is achieved d. the anode starts rotating and thermionic emission is achieved ANS: D When the exposure switch at the operating console is first depressed the anode gets up to rotational speed and thermionic emission occurs.

70. Which of the following occurs first during x-ray production? a. Thermionic emission. b. Voltage is transformed to kilovoltage. c. Current passes through the rheostat. d. Kilovoltage is applied to anode and cathode.

ANS: C The correct level of current (adjusted through the rheostat) must occur first.

71. The cloud of electrons produced through thermionic emission is more accurately called a(n) a. space cloud b. electron cloud c. electron charge d. space charge ANS: D Space charge is the more accurate name for the cloud of electrons.

72. Which of the following occurs first during x-ray production? a. Voltage is transformed to kilovoltage. b. Kilovoltage is applied to anode and cathode. c. Alternating current is changed to direct current. d. Voltage level is adjusted at autotransformer.

ANS: D Prior to the alternating electrical voltage being transformed to kilovoltage and changed to direct current, the correct voltage level must be adjusted at the autotransformer.

73. Which of the following occurs last during x-ray production?

a. b. c. d.

Kilovoltage is applied to anode and cathode.

Thermionic emission.

AC is converted to DC. The rotor starts turning.

ANS: A The last step in x-ray production is applying the kilovoltage, creating the great potential difference and forcing the electrons across the tube.

74. Thermionic emission is a. heating the filament b. heating the filament until electrons are boiled off c. using the focusing cup to create a space charge d. the space charge that travels from cathode to anode ANS: B Thermionic emission is the boiling off of electrons from the filament as a result of heating the filament.

75. The space charge effect describes a. how the space charge is created b. the effect the cloud of electrons has on the voltage across the tube c. the effect the cloud of electrons has on limiting additional electrons being emitted d. the effect of the heat of the filament on the size of the space charge ANS: C The space charge effect describes how the space charge is self-limiting, making the emission of additional electrons difficult.

76. The electrons in the tube current travel a. at the speed of light b. approximately half the speed of light c. approximately one-fourth the speed of light d. none of these ANS: B Although their speed varies based on the amount of potential difference between the anode and cathode, the electrons in the tube current travel approximately at half the speed of light.

77. Heat from x-ray production is removed through radiation by the heat traveling from a. the tube into the room by cooling fans b. the tube to heat-tolerant materials c. the tube to the oil bath in the protective housing d. none of these ANS: C Heat is removed from the tube by radiation from the tube to the oil bath in the protective housing.

78. Heat from x-ray production is removed through convection by the heat traveling from a. the tube into the room by cooling fans

b. the tube to heat tolerant materials c. the tube to the oil bath d. none of these ANS: A Heat is removed from the tube by convection by the cooling fans into the room.

79. Heat from x-ray production is removed through conduction by the heat traveling from a. the tube into the room by cooling fans b. the tube to heat tolerant materials c. the tube to the oil bath d. none of these ANS: B Heat is removed from the tube by conduction to heat-tolerant materials used in tube construction.

80. Of all the energy involved in x-ray production, 99% is converted to a. x-rays b. electrons c. heat d. none of these ANS: C 99% of the energy used in x-ray production is converted to heat.

81. Of all the energy involved in x-ray production, _________ is converted to x-ray energy. a. 1% b. 10% c. 50% d. 99% ANS: A Only approximately 1% of the energy used to produce x-rays actually results in x-radiation, a very inefficient process.

82. Prior to the introduction of protective circuits, ______________ were used to determine if a particular combination of exposure factors was safe or unsafe for a given x-ray tube. a. tube cooling charts b. tube rating charts c. technique charts d. QC charts ANS: B Prior to the introduction of protective circuits, tube rating charts were used to determine if a particular combination of exposure factors was safe or unsafe for a given x-ray tube.

83. The tube rating chart plots mA, kVp, and a. amperage b. voltage c. exposure time

d. resistance ANS: C The tube rating chart plots mA, kVp, and exposure time.

84. Using Figure 5-11, A, in the textbook, 75 kVp, 600 mA, and 0.1 sec is a. a safe exposure b. an unsafe exposure c. it is impossible to tell ANS: A 75 kVp, 600 mA, and 0.1 sec is a safe exposure because that point falls below the 75-kVp line.

85. Using Figure 5-11, B, in the textbook, 90 kVp, 100 mA, and 0.1 sec is a. a safe exposure b. an unsafe exposure c. it is impossible to tell ANS: A 90 kVp, 100 mA, and 0.1 sec is a safe exposure because that point falls below the 90-kVp line.

86. The _______________ is used to calculate the length of time needed between exposures so that the tube’s heat loading capacity is not exceeded. a. tube rating chart b. exposure factor chart c. cooling chart d. heating chart ANS: C The cooling chart is used to calculate the length of time needed between exposures so that the tube’s heat loading capacity is not exceeded.

87. Cooling charts are available for the a. tube housing b. anode c. cathode d. tube housing and anode ANS: D Both anode and tube housing cooling charts are available.

88. The formula kVp  mA  s  c is used to calculate a. heat units b. cooling time c. heating time d. Hounsfield units ANS: A The formula kVp  mA  s  c is used to calculate heat units (HU).

89. How many heat units are produced with an exposure using a single phase x-ray unit, 75 kVp, and 50 mAs? a. 3,750 HU. b. 5,063 HU. c. 5,288 HU. d. 5,438 HU. ANS: A Because mAs is the product of mA and time, the heat units for this single-phase unit (correction factor of 1) is 1  75  50, or 3,750 HU. 90. How many heat units are produced with five consecutive exposures using a three-phase, 12-pulse x-ray unit; 85 kVp; and 20 mAs? a. 1,700 HU. b. 2,295 HU. c. 2,397 HU. d. 11,985 HU. ANS: D This requires multiplying the total number of heat units (85  20  1.41—the correction factor) by 5 because there are five consecutive exposures = 11985 HU. 91. Which of the following practices extend tube life? a. Press the prep and exposure switches at almost the same time. b. Warm up the tube just before turning the equipment off. c. Use high exposures. d. None of these.

ANS: A Pressing the prep and exposure switches at almost the same time limits heating of the filament to only what is needed, extending tube life.

92. Which of the following practices extend tube life? a. Warm up the tube as appropriate. b. Press the prep and exposure switches at almost the same time. c. Avoid using very high exposures or long exposure times. d. All of these.

ANS: D To extend tube life, the tube should be warmed up properly, the prep and exposure switch pressed almost simultaneously, and high or long exposures avoided.

93. Pitting of the anode is the result of a. very long exposure times b. high exposures c. failure to warm up the tube d. all of these ANS: B Consistently using very high exposures may result in pitting of the anode.

94. Excessive heat may be transferred to the rotor bearings, resulting in a. the anode rotating faster than it should b. the anode rotating slower than it should c. the anode rotating unevenly d. all of these ANS: C When the rotor bearings heat up they may prevent the anode from rotating evenly.

TRUE/FALSE 1. The oil surrounding the x-ray tube serves only to provide electrical insulation.

ANS: F The oil surrounding the tube helps to dissipate the heat produced during x-ray production.

2. When x-rays are produced inside the tube they travel in all directions.

ANS: T When x-rays are produced, they do travel in all directions, which is why it is so important that the tube housing be lined with lead.

3. Modern x-ray equipment is so safe that it is all right to use the high-voltage cables as handles.

ANS: F It is never a good idea to touch the high-voltage cables entering the tube housing because of the potential for a significant electric shock.

4. In addition to the oil surrounding the tube, mechanical fans help to eliminate the heat produced during x-ray production.

ANS: T Fans inside the tube housing do help to dissipate the heat produced during x-ray production.

5. When the electrons strike the anode target, they all produce x-rays.

ANS: F Only some of the electrons interact with the target to produce x-rays.

6. The stator is made up of electromagnets.

ANS: T The stator is the part of the induction motor that is made of electromagnets.

7. The rotor is supplied with electric current through the x-ray circuit.

ANS: F The rotor is not supplied with any electric current.

8. To get the rotor to move, each pair of stators are energized sequentially.

ANS: T The rotor turns because all the pairs of stators are energized in a row, creating magnetic fields that attract the rotor.

9. The physical area of the target where electrons strike is the effective focal spot.

ANS: F The physical area that is bombarded with electrons is the actual focal spot.

10. The effective focal spot is smaller than the actual focal spot because the face of the anode is angled.

ANS: T By angling the face of the anode, the actual focal spot appears smaller when viewed from below.

11. To take advantage of the anode heel effect, the thicker part of the anatomy being imaged should be placed under the anode end of the tube.

ANS: F In that the intensity of radiation is less on the anode end, the thicker part should be placed beneath the cathode end of the tube.

12. To take advantage of the anode heel effect, the thinner part of the anatomy being imaged should be placed under the cathode end of the tube.

ANS: F In that the intensity of radiation is less on the anode end, the thinner part should be placed beneath the anode end of the tube.

13. Tungsten is used in the cathode filament because it is able to vaporize very easily.

ANS: F Tungsten is used in the cathode filament because it does not vaporize easily.

14. Using very long exposures can lead to tube damage or failure.

ANS: T It is true that using very long exposures can lead to tube damage or failure.

15. Prepping the rotor but not making an exposure on a regular basis is related to tube damage or failure.

ANS: T Regular failure to make an exposure after prepping the rotor results in excessive heat in the filament and may cause tube problems.

16. Many newer x-ray units do not need the radiographer to go through a warm-up procedure.

ANS: T Many of the newer units have automatic built-in tube warm-ups so the radiographer does not need to go through the warm-up procedure.

Chapter 06: X-Ray Production Johnston/Fauber: Essentials of Radiographic Physics and Imaging, 4th Edition MULTIPLE CHOICE 1. During x-ray production the electrons penetrate the target approximately a. 0.2 mm b. 0.5 mm c. 2 mm d. 5 mm ANS: B Electrons travel approximately 0.5 mm into the target during x-ray production.

2. X-rays are produced by a. characteristic interactions b. uncharacteristic interactions c. bremsstrahlung interactions d. characteristic and bremsstrahlung interactions ANS: D X-rays are produced by both characteristic and bremsstrahlung interactions between the filament electron and the tungsten atom.

3. When filament electrons enter the anode target, they interact with a. outer shell electrons of tungsten atoms b. inner shell electrons of tungsten atoms c. outer shell electrons of copper atoms d. inner shell electrons of copper atoms ANS: A When filament electrons enter the anode target, they interact with the outer shell electrons of tungsten atoms.

4. At the anode target, how much of the energy from filament electrons is lost as heat and how much will result in x-ray production. a. 1% energy lost as heat ; 99% result in x-rays b. 99% energy lost as heat; 1% result in x-rays c. 0% energy lost as heat; 100% result in x-rays

d. 100% energy lost as heat; 0% result in x-rays ANS: B At the anode target, 99% of the energy from filament electrons is lost as heat and 1% will result in x-ray production.

5. When a filament electron knocks out a K shell electron from the tungsten atom, it leads to a. a brems x-ray photon b. a 39.5 keV energy x-ray photon c. a characteristic x-ray photon d. none of these ANS: C Characteristic x-ray photons are the result of a filament electron knocking out a K shell electron.

6. Characteristic x-ray photons result a. when an outer-shell electron is knocked out b. when an inner-shell electron is knocked out c. when outer-shell electrons fill the vacancy in an inner shell d. when an outer-shell electron is knocked out and when outer-shell electrons fill the vacancy in an inner shell e. when an inner-shell electron is knocked out and when outer-shell electrons fill the vacancy in an inner shell ANS: E To produce characteristic radiation, outer-shell electrons must drop into an inner-shell vacancy created by the filament electron knocking out the orbital electron.

7. The energy of the characteristic x-ray photon depends on a. the binding energy of the inner-shell electron b. the energy level of the filament electron c. the shell of the electron that is dropping into the vacancy d. all of these ANS: D Characteristic radiation depends on the energy level of the incoming electron, the binding energy of the electron that is knocked out and the shell of the orbital electron that drops into the vacancy.

8. A filament electron removes a K shell electron and an M shell electron fills the vacancy.

The K shell binding energy is 69.5 keV and the M shell binding energy is 2.8 keV. What is the energy of the K-characteristic photon produced? a. 2.8 keV. b. 57.4 keV. c. 66.7 keV. d. 69.5 keV. ANS: C The energy of the photon produced is determined by subtracting the binding energy of the outer shell electron from the binding energy of the inner shell electron (69.5 – 2.8 = 66.7).

9. The process of a series of outer-shell electrons filling inner-shell vacancies right after the other is called a. characteristic tumble b. characteristic x-rays c. characteristic sequence d. characteristic cascade ANS: D The process of a series of outer-shell electrons filling inner-shell vacancies right after the other is called characteristic cascade.

10. Bremsstrahlung means a. electron b. x-ray production c. braking radiation d. all of these ANS: C Bremsstrahlung is the German word for ―braking‖ or ―slowing down‖ radiation.

11. During the bremsstrahlung interaction, the filament electron a. knocks out an inner-shell electron b. knocks out an outer-shell electron c. is absorbed by the nucleus d. is attracted to the nucleus but not absorbed ANS: D During the bremsstrahlung interaction, the filament electron is attracted to the nucleus, causing it to slow down and change direction.

12. The energy of the brems photon depends on a. the original energy of the filament electron b. the strength of the attraction between the electron and the nucleus c. the energy of the filament electron as it leaves the tungsten atom d. all of these ANS: D The energy of the brems photon depends on the energy of the filament electron as it enters and exits the tungsten atom. The energy level as it exits depends on the strength of the attraction between the electron and the nucleus.

13. If a filament electron enters the tungsten atom with 80 keV of energy and leaves the atom with 75 keV of energy: a. the brems photon will be 75 keV b. the brems photon will be 80 keV c. the filament electron traveled very close to the nucleus d. the filament electron traveled very far from the nucleus ANS: D

In that the filament electron lost only 5 keV of energy (which is a 5-keV brems photon) this means that it has lost almost no energy, and it must have traveled far from the nucleus.

14. If a filament electron leaves the tungsten atom with 15 keV of energy and the brems photon produced was 65 keV, how much energy did the incoming filament electron have? a. 15 keV. b. 50 keV. c. 65 keV. d. 80 keV. ANS: D The incoming electron had 80 keV of energy because it only had 15 keV when it left the atom and the brems photon was 65 keV. 15. The average energy of a brems photon is ______ of the kVp selected at the control panel. a. 1/3 b. 1/2 c. 2/3 d. the same as that ANS: A The average energy of a brems photon is one third of the kVp selected at the control panel.

16. The majority of the x-ray photons produced are a. characteristic photons b. brems photons c. photons with energy higher than 70 keV d. photons with energy higher than 80 keV ANS: B The majority of the x-ray photons produced are the result of brems interactions.

17. The total number of x-ray photons in a beam is referred to as the a. x-ray quality b. x-ray number c. x-ray quantity d. x-ray beam ANS: C X-ray quantity refers to the total number of photons in the beam.

18. Which of the following factors affect beam quantity? a. kVp. b. Distance. c. Filtration. d. All of these.

ANS: D Distance, tube filtration, and kVp affect the quantity of the x-ray beam.

19. Which of the following is the primary factor controlling quantity?

a. b. c. d.

kVp. mAs. Distance.

Filtration.

ANS: B mAs are the primary controlling factor for beam quantity.

20. To double the beam quantity a. halve the mAs b. leave the mAs as is and reduce the kVp c. double the mAs d. increase the mAs by a factor of 4 (22) ANS: C To double the beam quantity, double the mAs. 21. If kVp is doubled, the quantity of radiation increases by a factor of a. 2 b. 3 c. 4 d. 8 ANS: C Doubling the kVp increases beam quantity by a factor of 4 (the square of the ratio of the change).

22. If the kVp is changed from 30 kVp to 90 kVp, the quantity of radiation in the beam increases by a factor of a. 2 b. 3 c. 6 d. 9 ANS: D Tripling the kVp increases beam quantity by a factor of 9 (the square of the ratio of the change).

23. To increase the quantity of radiation by adjusting the kVp to the same level as doubling the mAs, increase kVp by a factor of a. 2% b. 5% c. 15% d. 25% ANS: C Increasing the kVp by 15% is equivalent to doubling the mAs, which results in doubling the quantity of radiation.

24. Which of the following results in the equivalent of doubling the mAs if the original kVp is 80?

a. b. c. d.

85 kVp. 92 kVp. 120 kVp. 160 kVp.

ANS: B Because 15% of 80 is 12, changing from 80 kVp to 92 kVp results in the equivalent of doubling the mAs. 25. It is recommended that kVp not be used to control beam quantity because a. it affects scatter production b. it affects penetrability of the beam c. it is less predictable in how changing the kVp affects the image d. all of these ANS: D kVp is not recommended as the primary method to control beam quantity because it affects so many other factors in image production.

26. As the distance increases, the beam quantity reaching a specific area a. increases b. decreases c. stays the same d. doubles ANS: B Increasing the distance results in the number of photons in a specific area being decreased.

27. If the distance from the source is doubled, the quantity of radiation reaching a specific area is _________ the original. a. double b. quadruple c. half d. one fourth ANS: D Doubling the distance from the source reduces the beam intensity to one-fourth the original (half squared = one fourth), as specified in the inverse square law.

28. The inverse square law describes the relationship between ____________ and beam intensity. a. kVp b. distance c. exposure time d. kVp ANS: B The inverse square law describes the relationship between distance and beam intensity.

29. If the distance from the source is changed from 72 inches to 36 inches and the original beam intensity was 200 mR, what is the new intensity?

a. b. c. d.

50 mR. 100 mR. 400 mR. 800 mR.

ANS: D Based on the inverse square law, halving the distance from the source results in four times the intensity. I1/I2 = d22/d21 ; 200/I2 = 362/722 ; 1296 I2 = 1036800 ; I2 = 800 mR. 30. If the intensity of the beam is 900 mR at a distance of 21 inches, what does the distance need to be for the intensity to measure 100 mR? a. 7 inches. b. 10.5 inches. c. 42 inches. d. 63 inches.

ANS: D Based on the inverse square law, because the intensity is reduced by a factor of 9, the distance must be three times the original. I1/I2 = d22/d21; 900/100 = d22/212; 100 d22 = 396900; d22 = 3969; d2 = 63 inches.

31. Filtration placed in the path of the x-ray beam a. absorbs low-energy photons b. absorbs high-energy photons c. increases patient dose d. reduces the quantity of radiation by a factor of 2 ANS: A Beam filtration reduces beam quantity by absorbing low-energy photons.

32. The purpose of beam filtration is to a. control beam quantity b. reduce patient dose c. reduce the wear and tear on the tube d. all of these ANS: B Although it does reduce beam quantity, filtration serves only to reduce patient dose by absorbing low-energy photons that do not contribute to image formation.

33. To produce a radiographic image, which of the following conditions should NOT occur? a. some photons penetrate the body b. some photons do not penetrate the body c. all photons penetrate the body d. some photons, but not all, penetrate the body ANS: C To produce a radiographic image, there must be x-ray photons that do and do not penetrate the body. If all the x-rays penetrated, the image would be black.

34. Beam quality refers to

a. b. c. d.

the energy level of the radiation the amount of radiation how useful the radiation is all of these

ANS: A Beam quality refers to the energy level or penetrability of the x-ray beam.

35. The primary controlling factor for beam quality is a. mAs b. kVp c. distance d. filtration ANS: B Although affected by filtration, kVp is the controlling factor for beam quality.

36. As the kVp increases a. beam energy decreases b. beam energy increases c. beam penetrability increases d. beam energy and beam penetrability increases ANS: D As the kVp increases, the beam energy and penetrability increases.

37. A higher energy beam is said to be a ______ beam. a. soft b. hard c. weak d. powerful ANS: B A higher energy beam is said to be a hard beam.

38. Placing filtration in the path of the beam results in a. a harder beam b. a softer beam c. more x-ray photons d. a sharper image ANS: A Adding filtration removes lower energy photons, resulting in a higher energy, or harder beam.

39. Half-value layer (HVL) is used to measure a. beam intensity b. beam quantity c. beam quality d. all of these ANS: C

HVL is a measure of beam quality.

40. One HVL reduces the intensity of the x-ray beam to _______ of its original. a. one fourth b. one third c. one half d. two thirds ANS: C One HVL reduces the beam intensity to one half of the original.

41. Normal HVL of diagnostic x-ray beams is a. 1–2 mm Al b. 3–5 mm Al c. 6–10 mm Al d. 15–25 mm Al ANS: C The normal HVL of diagnostic beams is 3–5 mm Al. 42. How many HVLs are needed to reduce beam intensity from 600 mR to 300 mR? a. One. b. Two. c. Three. d. Four.

ANS: A One HVL reduces the intensity by half.

43. The beam that is found leaving the collimator and exposes the patient is called the a. collimator beam b. remnant beam c. primary beam d. transmitted beam ANS: C The primary beam is found leaving the collimator and exposes the patient.

44. The x-ray beam that leaves the patient to expose the IR is called the a. remnant beam b. primary beam c. patient beam d. none of these ANS: A The remnant beam is that portion of the primary beam that exits the patient to expose the IR. 45. The remnant radiation that did not interact with any anatomic structures is a. primary radiation b. secondary radiation c. transmitted radiation

d. scattered radiation ANS: C Radiation that passes through the patient without any interactions is called transmitted radiation.

46. The remnant radiation that have an interaction with an anatomic structure is a. primary radiation b. filtered radiation c. transmitted radiation d. scattered radiation ANS: D Scattered radiation is the part of the remnant radiation that interacts with the part.

47. A discrete emission spectrum is a graphic representation of a. characteristic radiation b. bremsstrahlung radiation c. remnant radiation d. characteristic radiation and bremsstrahlung radiation e. bremsstrahlung radiation and remnant radiation ANS: A A discrete emission spectrum illustrates characteristic radiation.

48. A continuous emission spectrum is a graphic representation of a. characteristic radiation b. bremsstrahlung radiation c. remnant radiation d. characteristic radiation and bremsstrahlung radiation e. bremsstrahlung radiation and remnant radiation ANS: B A continuous emission spectrum illustrates bremsstrahlung radiation.

49. The x-ray emission spectrum is a graphic illustration of a. characteristic radiation b. bremsstrahlung radiation c. remnant radiation d. characteristic radiation and bremsstrahlung radiation e. bremsstrahlung radiation and remnant radiation ANS: D The x-ray emission spectrum combines both the discrete and continuous spectra.

50. The x-axis for all emission spectra represents the a. energy level of the photons b. number of photons c. HVL of the photons d. amount of transmitted radiation ANS: A

The x-axis measures the energy level of the x-ray photons.

51. The y-axis for all emission spectra represents the a. energy level of the photons b. number of photons c. HVL of the photons d. penetrability of the radiation ANS: B The y-axis provides information regarding the number of photons.

52. The discrete emission spectrum typically displays a. K-characteristic photon energies b. L-characteristic photon energies c. both of these d. neither of these ANS: C The discrete emission spectrum typically displays K- and L-characteristic photon energies.

53. L-characteristic and higher photon energies are not usually displayed on a discrete emission spectrum because a. there are no photons produced at those levels b. the energies of the photons produced are too high for image production c. the energies of the photons produced are too low for image formation d. none of these; these photon energies are displayed ANS: C The L-characteristic and higher photons produced have such low energies that they do not contribute to image formation.

54. With a 75-kVp exposure, the energy of the photons displayed on the continuous emission spectrum can range from just above 0 to a. 25 keV b. 50 keV c. 75 keV d. 150 keV ANS: C Brems radiation photon energy can range from just above 0 to the kVp set at the control panel up to 75 keV. 55. With a 75-kVp exposure, the peak of the curve on the continuous emission spectrum is approximately a. 25 keV b. 50 keV c. 75 keV d. 150 keV ANS: A

The highest number (peak of the curve) of brems photons produced has an energy level one third of the kVp set at the control panel (1/3 of 75 = 25 keV).

56. With the x-ray emission spectrum, the discrete line is the highest energy ___________________ bar. a. J-characteristic b. K-characteristic c. L-characteristic d. M-characteristic ANS: B Combining both the discrete and continuous emission spectra, the only part of the discrete spectra displayed is the highest-energy K-characteristic photons.

57. For a tungsten target, the discrete line on the x-ray emission spectrum is approximately a. 59 keV b. 69 keV c. 79 keV d. 89 keV ANS: B The discrete line is at 69 keV, based on a tungsten target.

58. Changes along the x-axis of the x-ray emission spectrum reflect changes in the _________________ of the x-ray beam. a. quality b. velocity c. quantity d. size ANS: A Changes along the x-axis of the x-ray emission spectrum reflect changes in the quality of the x-ray beam 59. If all other factors remain constant, a decrease in the mA results in a. a shift to the right along the x-axis b. a shift to the left along the x-axis c. an increase along the y-axis d. a decrease along the y-axis ANS: D Decreasing the mA results in fewer photons, decreasing the levels along the y-axis.

60. Increasing the kVp results in a. an increase along the y-axis b. a shift to the right along the x-axis c. a shift to the left along the x-axis d. an increase along the y-axis and a shift to the right along the x-axis ANS: D

Increasing the kVp affects both the quantity (y-axis) and quality (x-axis) of the spectrum.

The spectrum shifts to the right because higher-energy photons are produced.

61. Adding filtration in the path of the beam results in the x-ray emission spectrum a. changing along the x-axis b. changing along the y-axis c. both of these d. neither of these ANS: C Increasing filtration will result in fewer photons (y-axis) and higher-energy photons (x-axis).

62. Changing from a high-frequency generator to a single-phase generator results in the x-ray emission spectrum a. changing along the x-axis b. changing along the y-axis c. both of these d. neither of these ANS: C Changing from a high-frequency to a single-phase unit results in fewer x-ray photons (y-axis) and lower-energy photons (x-axis).

63. Changing the ___________________ results in changes to the x-axis, y-axis, and location of the discrete line of the x-ray emission spectrum. a. kVp b. mAs c. target material d. distance ANS: C The only variable that changes the discrete line of the spectrum along with the quantity and quality of the beam is changing target material.

TRUE/FALSE 1. As filament electrons enter the anode target, most interact with inner shell electrons of the tungsten atoms.

ANS: F As filament electrons enter the anode target, most interact with outer shell electrons of the tungsten atoms.

2. At the anode target, 1% of the energy from filament electrons is lost as heat and 99% will result in x-ray production.

ANS: F At the anode target, 99% of the energy from filament electrons is lost as heat and 1% will result in x-ray production.

3. A 65-keV filament electron is not able to produce characteristic radiation.

ANS: T To produce characteristic radiation, the electron’s energy must be greater than the orbital electron’s binding energy.

4. When 65 kVp is set on the operating console, no K-characteristic radiation is produced.

ANS: T To knock out a K shell electron, a minimum of 69.5 keV is required.

5. The stronger the attraction between the filament electron and the nucleus, the less energy the brems photon has.

ANS: F The stronger the attraction between the filament electron and the nucleus, the more energy the filament electron loses, producing a higher energy brems photon.

6. Generally speaking, an increase in the quantity of radiation results in decreased patient dose.

ANS: F An increase in x-ray quantity results in an increase in patient dose.

7. HVL is described as a certain amount of tungsten or its equivalent that will reduce beam intensity by one half.

ANS: F HVL is described in terms of the amount of aluminum or its equivalent.

8. The discrete emission spectrum is limited to a few specific values.

ANS: T The discrete emission spectrum is limited to a few specific values based on the different energy levels of the shells of the tungsten atom.

9. When the kVp is changed from 60 to 120, the discrete line on the x-ray emission spectrum shifts to the right.

ANS: F The discrete line does not move because it is determined by the target material.

Chapter 07: X-Ray Interactions with Matter Johnston/Fauber: Essentials of Radiographic Physics and Imaging, 4th Edition MULTIPLE CHOICE 1. Classical interactions are also known as a. coherent scattering b. Compton scattering

c. Thomson scattering d. Thomson and coherent scattering ANS: D Classical interactions are also known as Thomson or coherent scattering.

2. Coherent scattering involves a. an x-ray photon with high energy b. an x-ray photon with low energy c. ionization d. all of these ANS: B A low-energy x-ray photon may produce coherent scattering.

3. The x-ray photon produced during classical scattering a. is lower energy than the incident photon b. is higher energy than the incident photon c. is the same energy as the incident photon d. continues traveling in the same direction as the incident photon ANS: C Classical scatter photons have the same energy level as the incident photon.

4. The x-ray photon produced during classical scattering a. is lower energy than the incident photon b. is higher energy than the incident photon c. travels in a different direction than the incoming photon d. travels in the same direction than the incoming photon ANS: C The x-ray photon produced during classical scattering changes direction when compared with the incoming photon.

5. Most classical scatter photons a. are the result of the removal of an orbital electron b. are absorbed in the body c. are transmitted through the body d. become remnant radiation ANS: B Most classical scatter photons are absorbed in the body through other interactions.

6. In reference to patient radiation dose, classical scatter interactions a. slightly increase dose b. significantly increase dose c. have no effect on dose d. slightly decrease dose ANS: A Classical scatter interactions contribute only slightly to patient radiation dose.

7. The effect of classical scatter interactions on image quality a. is significant b. is minimal c. is that quality is improved d. is minimal, and that quality is improved ANS: B In that most classical scatter photons never leave the body, the effect on image quality is minimal.

8. Compton scattering involves the x-ray photon a. being absorbed in the atom b. removing an inner-shell electron c. removing a middle- or outer-shell electron d. passing through the atom without any change ANS: C Compton scattering occurs when the x-ray photon removes a middle- or outer-shell electron.

9. Compton scattering typically occurs with x-ray photons in the energy range of a. 5–10 keV b. 20–40 keV c. 60–90 keV d. 100–135 keV ANS: B Most x-ray photons that produce Compton interactions range from 20 to 40 keV. 10. Compton scattering typically occurs with a. low-energy photons b. high-energy photons c. moderate-energy photons d. all of these ANS: C Although Compton scattering can occur with photons with energy that spans the diagnostic range, the majority have moderate energy.

11. With Compton scattering, the incident photon a. loses none of its energy b. loses up to one third of its energy c. loses up to one half of its energy d. does not lose any energy ANS: B After a Compton interaction, the photon loses up to one third of its original energy.

12. Compton scattering results in a. a Compton electron b. a secondary electron c. a Compton scatter photon

d. all of these ANS: D Scattering results in a Compton electron (also known as a secondary electron) that was removed from orbit along with a Compton scatter photon.

13. If a middle-shell electron is removed during a Compton interaction a. secondary photons are produced b. a secondary electron is ejected c. a Compton scatter photon is produced d. all of these ANS: D Compton scattering involves the incident photon removing an electron (now called a Compton or secondary electron) and the photon becoming a scatter photon. In addition, a characteristic cascade occurs when a middle-shell electron is removed, resulting in secondary photons.

14. The x-ray photons that are produced as a result of a characteristic cascade during a Compton interaction a. are characteristic x-ray photons b. are brems x-ray photons c. are Compton x-ray photons d. are ―braking‖ x-ray photons ANS: A The x-ray photons produced when the outer-shell electrons move into the vacancies produced when a middle-shell electron is removed are characteristic photons.

15. Secondary photons produced during a Compton interaction a. are typically high energy b. usually exit the patient and interact with the image receptor c. contribute to patient dose d. none of these ANS: C In that the secondary photons produced as a result of the characteristic cascade during a Compton interaction are low energy, they are unable to exit the body and instead add to patient dose.

16. The Compton electron a. typically exits the patient as part of the remnant radiation b. does not have enough energy to produce further interactions c. has enough energy to produce further interactions d. typically exits the patient as part of the remnant radiation and does not have enough energy to produce further interactions ANS: C The electron ejected during the Compton interaction often has enough energy to produce further interactions.

17. The Compton scatter photon a. may exit the patient as part of the remnant radiation b. does not have enough energy to produce further interactions c. has enough energy to produce further interactions d. may exit the patient as part of the remnant radiation and has enough energy to produce further interactions ANS: D The Compton scatter photon may produce other interactions within the body or exit the body as part of the remnant radiation.

18. When it does interact with the image receptor, the Compton scatter photon contributes no useful information because a. its energy is too low b. its energy is too high c. it has changed direction d. none of these ANS: C Because the Compton scatter photon has changed direction, the interaction with the image receptor is in the wrong location, providing no useful information.

19. The fog seen on the radiographic image is the result of a. coherent scattering b. classical scattering c. photoelectric interactions d. Compton interactions ANS: D Compton interactions produce most of the scatter that fogs the image.

20. Reducing the amount of Compton scattering a. is impossible b. is extremely difficult to accomplish c. is very important in producing quality images d. all of these ANS: C To produce quality images it is important to minimize Compton scattering.

21. The greater the angle of deflection of a Compton scatter photon: a. the more likely it will interact with the image receptor b. the lower the energy of the photon c. the higher the energy of the photon d. the more likely it will become part of the remnant radiation ANS: B The closer the angle of deflection gets to 180 degrees, the lower the energy of the scattered photon.

22. No matter which direction the Compton scatter photon goes, it retains approximately

a. b. c. d.

one fourth of its energy one third of its energy one half of its energy two thirds of its energy

ANS: D The Compton scatter photon retains approximately two thirds of its energy.

23. The major source of radiation exposure to technologists is due to a. coherent scattering b. classical scattering c. photoelectric interactions d. Compton interactions ANS: D Compton scattering is the major source of occupational exposure.

24. The primary source of radiation exposure to the radiographer who is in the room during an exposure is a. from scatter from the table b. from scatter from the patient c. from scatter from the image receptor d. hard to determine; it depends on the procedure being done ANS: B Scatter radiation produced in the patient is the primary source of exposure to the radiographer.

25. To minimize occupational exposure, the technologist should a. never go in an x-ray room during a procedure b. wear protective apparel (lead aprons, gloves, etc.) when in the room during imaging c. stand close to the patient’s head d. wear protective apparel (lead aprons, gloves, etc.) when in the room during imaging and stand close to the patient’s head ANS: B To limit occupational exposure, the technologist should wear protective apparel when appropriate and stand away from the patient.

26. For photoelectric interactions to occur, the energy of the incident photon a. may be greater than the binding energy of an inner-shell electron b. may be the same as the binding energy of an inner-shell electron c. may be less than the binding energy of an inner-shell electron d. may be the same or greater than the binding energy of an inner-shell electron ANS: D For a photoelectric interaction to occur, the energy of the incident photon must be greater than or equal to the binding energy of an inner-shell electron.

27. After a photoelectric interaction, the incident photon

a. b. c. d.

loses some of its energy and changes direction loses most of its energy and changes direction loses all of its energy and no longer exists none of these

ANS: C The photoelectric interaction results in the incident photon giving up all its energy and being totally absorbed by the atom.

28. During a photoelectric interaction, a. an inner-shell electron is ejected b. an outer-shell electron is ejected c. no electrons are ejected d. electrons for all levels are ejected ANS: A During a photoelectric interaction, an inner-shell electron is knocked out.

29. The end product(s) of a photoelectric interaction is (are) a. a high-energy x-ray photon b. a photoelectron c. an ionized atom d. a photoelectron and an ionized atom ANS: D A photoelectric interaction produces no x-ray photons, only a photoelectron and the remaining ionized atom.

30. The energy of the photoelectron is equal to a. the binding energy of the orbital electron minus the energy of the incident photon b. the energy of the incident photon minus the binding energy of the orbital electron c. the binding energy of the orbital electron plus the energy of the incident photon d. none of these ANS: B The energy of the photoelectron is equal to the energy of the incident photon minus the binding energy of the orbital electron 31. The energy of the photoelectron is the least as a result of a photoelectric interaction in a. bone b. soft tissue c. fat d. the energy level is the same for all ANS: A Because it will use up most of its energy ejecting the tightly bound inner-shell electron of bone, the photoelectron will be left with very little energy.

32. Photoelectric interactions a. contribute significantly to patient dose b. should be reduced as close to 0 events as possible

c. negatively affect radiographic image quality d. all of these ANS: A Photoelectric interactions contributes significantly to patient dose accrued with each diagnostic image.

33. Secondary x-ray photons are the result of a. Compton interactions b. Thompson scattering c. photoelectric interactions d. Compton interactions and photoelectric interactions ANS: D Both Compton and photoelectric interactions produce secondary photons.

34. Photoelectric interactions are more likely to occur in a. air b. soft tissue c. bone d. they occur equally in all types of tissues ANS: C Photoelectric interactions are more likely to occur in bone because of the higher atomic number of the atoms that make it up.

35. Protective apparel is often made of lead because it a. is very heavy b. can be made very thick c. is inexpensive d. has a high atomic number ANS: D Lead is often used in protective apparel because its high atomic number results in most x-ray photons being absorbed through the photoelectric interaction.

36. Barium sulfate is used as a contrast agent to visualize soft tissue structures such as the stomach because it a. is very heavy b. can be made very thick c. is inexpensive d. has a high atomic number ANS: D Barium sulfate is used as a contrast agent because its high atomic number results in most x-ray photons being absorbed through the photoelectric interaction, making it visible on the radiographic image.

37. Pair production a. occurs very often during radiographic procedures b. seldom occurs during radiographic procedures

c. never occurs during radiographic procedures d. it depends on the procedure ANS: C Pair production never occurs during radiographic procedures because it requires a photon with more energy than is used in diagnostic radiography.

38. Pair production requires an x-ray photon with an energy of at least a. 0.51 MeV b. 1.02 MeV c. 1.51 MeV d. 2.04 MeV ANS: B Pair production requires an x-ray photon with an energy of at least 1.02 MeV. 39. Pair production occurs when the incident photon interacts with a. an inner-shell electron b. an outer-shell electron c. inner- and outer-shell electrons d. the nucleus of the atom ANS: D During pair production the incident photon interacts with the atom’s nucleus.

40. What is produced as a result of pair production, a(n) 1. positron 2. electron 3. proton a. positron and electron only b. positron and proton only c. electron and proton only d. positron, electron, and proton ANS: A Pair production results in an electron and positron leaving the nucleus.

41. A positron is a. the same as an electron b. a positively charged electron c. the same as a proton d. a positively charged neutron ANS: B A positron is a positively charged electron.

42. After ejection from the atom’s nucleus, when the positron interacts with an electron a. an annihilation event occurs b. the electron and positron are both destroyed c. two x-ray photons are produced

d. all of these ANS: D Interaction of a positron and electron results in an annihilation event in which both particles are destroyed and their energy converted into two x-ray photons.

43. For photodisintegration to occur, the incident photon must have an energy level of at least a. 100 keV b. 1.02 MeV c. 10 MeV d. 100 MeV ANS: C For photodisintegration to occur, the incident photon must have an energy level of at least 10 MeV. 44. Photodisintegration occurs when the incident photon interacts with a. an inner-shell electron b. an outer-shell electron c. inner- and outer-shell electrons d. the nucleus of the atom ANS: D Photodisintegration occurs when the incident photon interacts with the nucleus of the atom.

45. Following absorption of the incident photon in the atom’s nucleus, a photodisintegration interaction may result in a. positrons being ejected from the nucleus b. neutrons being ejected from the nucleus c. protons being ejected from the nucleus d. neutrons and protons being ejected from the nucleus ANS: D A photodisintegration interaction results in ejection of a proton, neutron, or alpha particle from the atom’s nucleus.

46. Photodisintegration interactions a. occur very often during radiographic procedures b. seldom occur during radiographic procedures c. never occur during radiographic procedures d. it depends on the procedure ANS: C Photodisintegration never occurs during radiographic procedures because it requires an incident photon with more energy than is used in diagnostic radiography.

47. To produce a radiographic image that represents anatomy, there must be a. photoelectric interactions b. transmitted photons c. differential absorption d. all of these

ANS: D To demonstrate anatomy, differential absorption is necessary. Some of the primary beam must be absorbed (photoelectric interaction) and some must be transmitted.

48. Radiation interacting with bone is more likely to be absorbed, resulting in that area of the image being a. a light shade of gray b. a dark shade of gray c. black d. all of these ANS: A Because few photons are transmitted through the bone, the result is that this area of the image is lighter than the rest.

49. Radiation interacting with air is more likely to be _____________, resulting in that area of the image being darker. a. absorbed b. transmitted c. attenuated d. scattered ANS: B Air absorbs very few, if any, x-ray photons, resulting in most of the photons being transmitted.

50. Dense material, like bone, is considered: a. radiopaque b. easy to penetrate c. radiolucent d. less likely to absorb radiation ANS: A Bone is radiopaque because it readily absorbs radiation and prevents it from reaching the image receptor.

51. Air is considered: a. radiopaque b. hard to penetrate c. radiolucent d. more likely to absorb radiation ANS: C Air is radiolucent because it readily transmits radiation, allowing it to reach the image receptor.

52. The greater the absorption of radiation a. the greater the differential absorption b. the better the image quality c. the greater the patient dose d. all of these

ANS: C Although absorption is necessary to produce an anatomic image, too much absorption results in excessive patient dose.

53. A breakage of the major structure and framework of the macromolecule as a result of an x-ray interaction is called a. main-chain scission b. cross-linking c. point lesions d. all of these ANS: A Main-chain scission refers to a breakage of the major structure, the framework if you will, of the macromolecule itself in response to an x-ray interaction.

54. The most radiosensitive molecule in the body is a. DNA b. RNA c. nerve cells d. gonadal cells ANS: A Damage may occur in DNA as a result of radiation exposure and manifest as a range of responses from minor damage that is reversible to malignant response and permanent damage.

55. Excessive absorption is the result of a. using a mAs that is too low for the examination b. using a kVp that is too low for the examination c. using a kVp that is too high for the examination d. all of these ANS: B Insufficient kVp results in excessive absorption of radiation, increasing patient dose.

TRUE/FALSE 1. Potential biological damage can occur as a result of a Compton interaction because of the ionization of atoms.

ANS: T In that the atom is ionized during the Compton interaction (an electron is removed), the atom is unstable and may cause biological damage.

2. The probability of Compton scattering depends on the atomic number of the atom.

ANS: F The probability of Compton scattering does not depend on the atomic number of the atom.

Compton scattering can occur in any type of tissue.

3. The probability of Compton scattering depends on the energy level of the incident x-ray photon.

ANS: T The probability of Compton scattering does depend on the energy level of the incident x-ray photon.

4. All Compton scatter photons travel in the direction of the image receptor.

ANS: F Compton scatter photons can travel in any direction.

5. The photoelectric interaction and Compton interaction can both result in a characteristic cascade, producing characteristic radiation.

ANS: T In that both interactions result in a missing orbital electron, the characteristic cascade can occur in both.

6. The probability of a photoelectric interaction depends on the atomic number of the atom.

ANS: T The probability of a photoelectric interaction depends on the atomic number of the atom.

7. The probability of a photoelectric interaction depends on the energy level of the incident x-ray photon.

ANS: T The incident photon must be the same as or greater than the binding energy of the inner-shell electron.

8. The more the energy of the incident x-ray photon exceeds the binding energy of the inner-shell electron, the greater the probability of a photoelectric interaction.

ANS: F As the photon’s energy increases beyond the binding energy, it is more likely that the photon will be transmitted through the atom.

9. The higher the atomic number of an atom, the greater the probability of a photoelectric interaction.

ANS: T Atoms with higher atomic numbers will be more likely to have photoelectric interactions.

10. Without differential absorption, the image would be a single shade of gray.

ANS: T

If all tissues absorbed radiation the same, the resulting image would be a single shade of gray and it would be impossible to differentiate between anatomic structures.

11. Irradiation of water can create harmful free radicals that then indirectly damage molecules and cells.

ANS: T Because the human body is about 80% water, irradiation of water (interactions between x-ray photons and water molecules) can create harmful free radicals that then indirectly damage molecules and cells.

12. Deoxyribonucleic acid (DNA) is the most sensitive human cell molecule.

ANS: T The most sensitive of human cell molecules is DNA. Chapter 08: Image Production Johnston/Fauber: Essentials of Radiographic Physics and Imaging, 4th Edition MULTIPLE CHOICE 1. The device that receives the radiation after it exits the patient is the a. Remnant beam b. Primary beam c. Image receptor d. None of these ANS: C The image receptor is the device that receives the radiation leaves the patient.

2. Which of the following is an example of an image receptor? a. Film-screen. b. Digital. c. Analog digital converter. d. Film-screen and digital.

ANS: D Film-screen and digital are types of image receptors.

3. In producing a radiographic image, which of the following occurs first? a. Differential absorption. b. Quality and quantity of x-ray beam exit the x-ray tube. c. Radiation interacts with image receptor. d. Radiation exits the patient with different energies.

ANS: B The quantity and quality of radiation must exit the x-ray tube before the beam can enter the patient.

4. In producing a radiographic image, which of the following occurs last?

a. b. c. d.

Differential absorption.

Quality and quantity of x-ray beam exit the x-ray tube.

Radiation interacts with image receptor.

Radiation exits the patient with different energies.

ANS: C The last step in image production is the image receptor receiving the radiation that exits the patient.

5. The latent image that is formed on the image receptor is the a. visible image b. invisible image c. manifest image d. none of these ANS: B The latent image is the invisible image.

6. Differential absorption occurs because different tissues have varying amounts of a. photoelectric interactions b. Compton scattering c. coherent scattering d. pair production ANS: A Differential absorption occurs because different tissues absorb (photoelectric interaction) different amounts of radiation.

7. Attenuation of the x-ray beam in the diagnostic range is due to a. absorption b. scattering c. transmission d. absorption and scattering ANS: D Attenuation is due to absorption and scattering.

8. Complete absorption of the diagnostic range x-ray photon is the interaction known as a. Compton scattering b. photoelectric interaction c. Coherent scattering d. photodisintegration ANS: B The photoelectric interaction results in complete absorption of the x-ray photon.

9. The photoelectric interaction involves removal of a. an inner-shell electron b. an outer-shell electron c. both inner- and outer-shell electrons d. no electrons

ANS: A The photoelectric interaction involves removal of an inner-shell electron.

10. The ability to remove electrons from an atom is a. absorption b. transmission c. ionization d. none of these ANS: C Ionization is the ability to remove electrons from an atom.

11. With the photoelectric interaction, after the incident photon ejects an inner-shell electron, the energy of the photon is a. greater than the original energy b. the same as the original energy c. slightly less than the original energy d. reduced to zero ANS: D After the photoelectric interaction the photon energy is reduced to 0, the photon has been totally absorbed.

12. The secondary photon that results from the photoelectric interaction a. has the same energy as the incident photon b. has high energy c. has low energy d. typically leaves the patient as scatter radiation ANS: C The secondary photon that is the result of the characteristic cascade following the photoelectric interaction has very low energy and is unlikely to exit the patient.

13. Scattering in the diagnostic range is due to a. photoelectric interactions b. absorption c. the Compton effect d. all of these ANS: C Scattering in the diagnostic range is due to the Compton effect or interaction.

14. The probability of a Compton interaction is _______________ in bone than soft tissue. a. less b. the same c. higher d. none of these ANS: B The probability of a Compton interaction occurring does not depend on the type of tissue, specifically its atomic number.

15. The probability of a Compton interaction occurring depends on a. the type of tissue being imaged b. the binding energy of the atom’s inner-shell electron c. the energy of the incoming x-ray photon d. all of these ANS: C The probability of a Compton interaction occurring depends on the energy of the incident photon.

16. At higher kilovoltages within the diagnostic range a. the percentage of Compton interactions increases b. the percentage of Compton interactions decreases c. the percentage of Compton interactions remains the same d. none of these ANS: A The percentage of Compton interactions increases at higher kilovoltages.

17. At higher kilovoltages within the diagnostic range a. the percentage of photoelectric interactions increases b. the percentage of photoelectric interactions decreases c. the percentage of photoelectric interactions remains the same d. none of these ANS: B The percentage of photoelectric interactions decreases at higher kilovoltages.

18. Coherent scattering a. is a significant interaction during radiographic imaging b. has minimal effect on the imaging process c. never affects the diagnostic image d. none of these ANS: B Coherent scattering seldom affects the image but, if it does, it has a minimal effect.

19. Scatter radiation may a. expose people near the patient b. increase patient exposure c. contributes no useful information if it strikes the image receptor d. all of these ANS: D Scatter radiation increases patient exposure and adds no useful information to the radiographic image. In addition, people near the patient may be exposed by the scatter from the patient.

20. Transmission refers to a. all x-rays in the beam before entering the patient

b. x-rays leaving the patient that did not interact with atoms c. all x-rays leaving the patient d. the process of sending the x-ray beam through the patient ANS: B Transmission refers to x-rays leaving the patient that did not interact with atoms as it passed through the patient.

21. Increasing the thickness of the part being imaged results in a. increased attenuation b. decreased attenuation c. no change in attenuation d. decreased absorption ANS: A Increasing the thickness of the part being imaged results in increased attenuation.

22. Decreased attenuation results in a. more x-ray photons reaching the image receptor b. fewer x-ray photons reaching the image receptor c. no change in the number of photons reaching the image receptor d. increased absorption ANS: A Decreased attenuation means fewer photons were absorbed, leaving more photons to exit the patient and reach the image receptor.

23. For every 4–5 cm of tissue, the x-ray beam quantity is reduced by approximately a. 10% b. 25% c. 50% d. 75% ANS: C The x-ray beam is reduced in quantity to half its original value each time it travels through 4-5 cm of tissue.

24. Imaging a thicker part results in a. more x-ray photons reaching the image receptor b. fewer x-ray photons reaching the image receptor c. no change in the number of photons reaching the image receptor d. none of these ANS: B Imaging a thicker part results in greater attenuation and fewer photons reaching the image receptor.

25. Imaging tissues with lower atomic numbers results in a. increased attenuation b. decreased attenuation c. no change in attenuation

d. increased absorption ANS: B Tissues with lower atomic numbers attenuate fewer photons.

26. Tissues with higher atomic numbers appear __________ on a digital image. a. darker b. grayer c. brighter d. less bright ANS: C Tissues with higher atomic numbers attenuate more x-rays, resulting in fewer interacting with the image receptor, and a brighter area on the digital image.

27. Matter per unit volume defines a. tissue thickness b. tissue density c. tissue composition d. tissue attenuation ANS: B Tissue density is matter per unit volume.

28. Tissue that is more dense results in a. increased attenuation b. decreased attenuation c. no change in attenuation d. decreased absorption ANS: A The more dense the tissue, the greater the attenuation.

29. Imaging which of the following results in the least attenuation? a. Bone. b. Air in lungs. c. Fat. d. Muscle.

ANS: B Air in the lungs absorbs or attenuates very few photons.

30. Imaging which of the following results in the most attenuation? a. Bone. b. Air in lungs. c. Fat. d. Muscle.

ANS: A Bone attenuates many or most x-ray photons.

31. A high-energy beam results in

a. b. c. d.

increased attenuation decreased attenuation no change in attenuation increased absorption

ANS: B A high-energy beam is more likely to penetrate the tissue, minimizing its attenuation.

32. A low-kilovoltage x-ray beam results in a. increased attenuation b. decreased attenuation c. no change in attenuation d. decreased absorption ANS: A A low kVp produces a low-energy beam that is more likely to be absorbed or scattered, increasing attenuation.

33. Another name for remnant radiation is a. transmitted radiation b. scatter radiation c. exit radiation d. absorbed radiation ANS: C Exit radiation and remnant radiation are the same entities.

34. Scatter radiation that reaches the image receptor creates a. unwanted exposure b. useful information c. fog d. unwanted exposure and fog ANS: D Scatter radiation that reaches the image receptor creates fog, or unwanted exposure.

35. The various shades of gray or brightness created on the radiographic image a. are present because of differential absorption b. makes the anatomic structures visible c. represent the radiation that was absorbed or transmitted d. all of these ANS: D The radiographic image that is a variety of shades of gray demonstrates different anatomic structures caused by differential absorption (the degree to which the radiation was absorbed or transmitted).

36. Less than __ of the primary beam entering the part reaches the image receptor. a. 1% b. 5% c. 18%

d. 99% ANS: B Less than 5% of the primary beam entering the anatomic part reaches the image receptor.

37. Before processing, the image receptor has a(n) __________ a. latent b. visible c. invisible d. latent and invisible

image.

ANS: D Before processing, the image receptor has a latent or invisible image.

38. After processing, there is a a. visible b. manifest c. latent d. manifest and visible

image.

ANS: D After processing, there is a manifest or visible image.

39. Intensifying screens are associated with a. film-screen imaging b. digital imaging c. film-screen and digital imaging d. dynamic imaging ANS: A Intensifying screens are used in film-screen imaging to convert the remnant radiation to light energy.

40. The computer is associated with a. film-screen imaging b. digital imaging c. film-screen and digital imaging d. none of these ANS: B The computer is associated with digital imaging only, not with film-screen imaging.

41. Chemical processing is used to create the manifest image in a. film-screen imaging b. digital imaging c. film-screen and digital imaging d. dynamic imaging ANS: A Film-screen imaging requires chemical processing to convert the film’s latent image to a manifest image.

42. Production of a static or still image is associated with a. film-screen imaging b. digital imaging c. dynamic imaging d. film-screen and digital only ANS: C Both film-screen and digital imaging systems produce radiographic or static images.

43. Place the following film-screen imaging steps in the correct order, from beginning to end: I remnant radiation is absorbed by intensifying screens II chemical processing creates a permanent image III light from intensifying screens expose film emulsion IV remnant radiation reaches the intensifying screens a. I, II, IV, III b. IV, III, I, II c. III, II, IV, I d. IV, I, III, II ANS: D After the remnant radiation reaches the intensifying screens, it is absorbed and then emits visible light. The light exposes the film, which is then placed in chemicals to process the image.

44. Place the following digital imaging steps in the correct order, from beginning to end: I remnant radiation reaches the digital image receptor II the manifest image is displayed on a monitor III the manifest image can be adjusted because it is composed of digital data IV the latent image is processed by the computer a. III, IV, I, II b. I, IV, II, III c. I, II, III, IV d. IV, III, II, I ANS: B After the remnant radiation reaches the digital image receptor, it is processed by the computer. The digital image is displayed on a monitor, where it can also be adjusted.

45. What refers to the range of exposure intensities an image receptor can accurately detect? a. ionization b. attenuation c. dynamic range d. dynamic imaging ANS: C Dynamic range refers to the range of exposure intensities an image receptor can accurately detect.

46. A digital image is composed of a combination of rows and columns recorded as a:

a. b. c. d.

dynamic image latent image pixel matrix

ANS: D A digital image is recorded as a matrix, or combination of rows and columns (array).

47. Each ___________ is recorded as a single numerical value representing a single brightness level displayed on a monitor. a. pixel b. crystal c. matrix d. none of these ANS: A Each pixel is recorded as a single numerical value representing a single brightness level displayed on a monitor.

48. A pixel’s ______________ determines the amount the shades of gray that can be displayed in a digital image. a. size b. location c. height d. bit depth ANS: D A pixel’s bit depth determines the amount the shades of gray that can be displayed in a digital image.

49. Fluoroscopy produces dynamic imaging by using __________________ equipment. a. image-intensified b. flat panel detector c. film-screen d. image-intensified and flat panel detector ANS: D Fluoroscopy produces dynamic imaging by using either image-intensified or flat panel detector equipment.

TRUE/FALSE 1. Reduction in the energy or amount of radiation in the primary beam is attenuation.

ANS: T Attenuation is the reduction in the energy or amount of radiation in the primary beam.

2. It is not possible for a photoelectric interaction to occur between a 50-keV x-ray photon and an atom with inner-shell electrons that have a binding energy of 60 keV or more.

ANS: T For the photoelectric interaction to take place, the incident photon must have the same or more energy as the inner-shell electron’s binding energy.

3. Photoelectric interactions occur only at the lower end of the diagnostic range of photon energies.

ANS: F Photoelectric interactions can occur throughout the diagnostic range.

4. After the x-ray beam enters the patient and has many photoelectric interactions, the quantity of the beam increases.

ANS: F After many photoelectric interactions (absorption of the photons), the beam quantity is reduced.

5. Scattered and secondary radiation must be eliminated during radiographic imaging.

ANS: F Although it is impossible to eliminate secondary and scatter radiation, they must be controlled as much as possible.

6. A digital imaging system that displays a greater number of shades of gray has better contrast resolution.

ANS: T The greater the pixel bit depth, the greater the number of shades of gray displayed which increases contrast resolution.

7. All of the radiation that reaches the image receptor is used to create the radiographic image.

ANS: F Only a portion of the radiation that exits the patient and reaches the image receptor is used to create the image.

8. During fluoroscopy, the x-ray tube is usually positioned underneath the table.

ANS: T During fluoroscopy, the x-ray tube is usually positioned underneath the table and the x-rays pass through the patient to interact with a device that converts the x-rays into either light intensities (image-intensified) or digital data (flat panel detector).

9. The process of differential absorption for image formation is different depending on the type of imaging system.

ANS: F The process of differential absorption for image formation is remains the same regardless of the type of imaging system: digital, film, or fluoroscopic.

Chapter 09: Image Quality and Characteristics Johnston/Fauber: Essentials of Radiographic Physics and Imaging, 4th Edition MULTIPLE CHOICE 1. The overall quality of the radiographic image is determined by a. sharpness b. visibility of anatomic structures c. accuracy of structural lines d. all of these ANS: D The visibility of the anatomic structures and the accuracy of their structural lines recorded (sharpness) determine the overall quality of the radiographic image.

2. Visibility of the anatomic structures is accomplished by balancing 1. contrast 2. distortion 3. brightness a. contrast and distortion only b. contrast and brightness only c. distortion and brightness only d. contrast, distortion, and brightness ANS: B Visibility of the anatomic tissues is achieved by the proper balance of image brightness and contrast.

3. The amount of light emitted from the display monitor is a. brightness b. contrast resolution c. spatial resolution d. noise ANS: A Brightness is the amount of luminance (light emission) from the display monitor.

4. A radiographic image that is too light has ________________ to visualize the anatomic structures. a. excessive brightness b. excessive contrast c. insufficient brightness d. sufficient contrast ANS: A A radiographic image that is too light has excessive brightness to visualize the anatomic structures.

5. Who evaluates the overall brightness on the image to determine if it is sufficient to visualize the anatomic area of interest? a. Patient b. Radiologist c. Supervisor d. Radiographer ANS: D The radiographer must evaluate the overall brightness on the image to determine whether it is sufficient to visualize the anatomic area of interest. He or she then decides whether the radiograph is diagnostic or unacceptable.

6. What may happen when the digital image receptor is extremely over exposed? a. excessive density b. increased distortion c. excessive contrast d. saturation ANS: D When the image receptor is extremely over exposed, saturation may occur where the image cannot be properly processed and the quality is severely degraded.

7. What important feature provides a numeric value indicating the level of radiation exposure to the digital image receptor? a. optical density b. detective quantum efficiency c. modulation transfer function d. exposure indicator ANS: D The exposure indicator provides a numeric value indicating the level of radiation exposure to the digital image receptor.

8. The midpoint of the range of brightness levels visible on the digital image is determined by a. window level b. window width c. window pane d. all of these ANS: A The window level or center sets the midpoint of the range of brightness levels visible on the digital image.

9. Which of the following is correct regarding the display feature windowing? a. changing the window level will change the brightness level b. increasing the window width will increase contrast c. decreasing the window width will decrease contrast d. changing the window level has no effect on the brightness level ANS: A

Changing the window level on the display monitor allows the image brightness to be increased or decreased throughout the entire range of brightness levels.

10. The ability of the imaging system to distinguish between small objects that attenuate the x-ray beam similarly is known as: a. grayscale b. spatial frequency c. contrast resolution d. brightness ANS: C Contrast resolution is a term associated with digital imaging and is used to describe the ability of the imaging system to distinguish between small objects that attenuate the x-ray beam similarly.

11. Which of the following refers to the absorption characteristics of the anatomic tissues radiographed and the quality of the x-ray beam? a. brightness levels b. subject contrast c. grayscale d. differential absorption ANS: B Subject contrast refers to the absorption characteristics of the anatomic tissues radiographed and the quality of the x-ray beam.

12. If there were no contrast, the radiographic image would include a. only the same shade of gray b. black and white c. light gray and dark gray d. black and light gray ANS: A No contrast means that all brightness levels in the image are the same.

13. Radiographic contrast a. produces sharper structural lines in the image b. improves the magnification of the image c. allows visibility of anatomic structures d. produces sharper structural lines in the image and allows visibility of anatomic structures ANS: C Radiographic contrast allows visualization of the anatomic tissues.

14. Which of the following contribute to subject contrast? a. Differences in tissue thickness. b. Differences in tissue density. c. Effective atomic number. d. All of these.

ANS: D Differences in tissue thickness, tissue density, and effective atomic number contribute to subject contrast.

15. A radiographic image that has brightness levels that are very different from each other a. has high contrast b. has low contrast c. has long-scale contrast d. has low contrast and has long-scale contrast ANS: A High, or short-scale, contrast has brightness levels that are very different from each other.

16. A radiographic image that has brightness levels that are very similar to each other a. has high contrast b. has low contrast c. has long-scale contrast d. has low contrast and has long-scale contrast ANS: D An image with very similar brightness levels has low or long-scale contrast.

17. What digital characteristics affects the number of shades of gray available for image display? a. pixel size b. pixel density c. pixel pitch d. pixel bit depth ANS: D The pixel bit depth or number of bits affects the number of shades of gray available for image display.

18. The control that adjusts the contrast by adjusting the visible range of brightness levels on the image is the a. window level b. window width c. window pane d. all of these ANS: B The window width is the control that adjusts the contrast by adjusting the visible range of brightness levels on the image.

19. Making the digital image appear with low contrast, many shades of gray, is done by a. raising the window level b. lowering the window level c. increasing the window width (wide WW) d. decreasing the window width (narrow WW) ANS: C

Increasing window width includes more shades of gray in the image, creating a low-contrast appearance.

20. Making the digital image appear with high contrast, more black and white, is done by a. raising the window level b. lowering the window level c. increasing the window width (wide WW) d. decreasing the window width (narrow WW) ANS: D Decreasing window width includes fewer shades of gray in the image, making it more black and white.

21. __________________ is a term used to evaluate accuracy of the anatomic structural lines recorded. a. Contrast resolution b. Spatial frequency c. Spatial resolution d. Detective quantum efficiency ANS: C Spatial resolution is a term used to evaluate accuracy of the anatomic structural lines recorded.

22. The number of pixels per unit area is a. grayscale b. pixel pitch c. pixel density d. contrast resolution ANS: C Pixel density describes how many pixels can fit into a specific area.

23. The distance from the center of one pixel to the center of the one next to it is a. grayscale b. pixel pitch c. pixel density d. contrast resolution ANS: B Pixel pitch is the measurement of the distance between the centers of two adjacent pixels.

24. Spatial resolution can be increased by: 1. Increasing the pixel pitch 2. Increasing the pixel density 3. Decreasing the pixel pitch a. Increasing the pixel pitch and increasing the pixel density only b. Increasing the pixel pitch and decreasing the pixel pitch only c. Increasing the pixel density and decreasing the pixel pitch only d. Increasing the pixel pitch, increasing the pixel density, and decreasing the pixel

pitch ANS: C Increasing the pixel density and decreasing the pixel pitch increases spatial resolution.

25. Increasing the matrix size for a fixed field of view (FOV) results in: a. smaller pixel sizes b. larger pixel sizes c. increased spatial resolution d. smaller pixel sizes and increased spatial resolution ANS: D Increasing the matrix size for a FOV will decrease the size of the pixels and increase spatial resolution.

26. Resolution is typically measured in a. line pairs per millimeter b. line pairs per meter c. line pairs per inch d. line pairs per foot ANS: A Resolution is typically measured in line pairs per millimeter (Lp/mm).

27. To measure resolution, a ______________ is necessary. a. sensitometer b. densitometer c. resolution test pattern d. resolution camera ANS: C A resolution test pattern is needed to measure resolution.

28. A line pair is made up of a. a line b. two lines c. a space d. a line and a space ANS: D A line pair consists of a line and a space.

29. An imaging system that can resolve a greater number of line pairs within 1 mm is said to have ___________________. a. improved spatial resolution b. decreased spatial resolution c. improved brightness d. decreased brightness ANS: A An imaging system that can resolve a greater number of line pairs within 1 mm (e.g., 8–10 Lp/mm) is said to have improved spatial resolution.

30. Which of the following is correct regarding spatial frequency.

1. small objects have lower spatial frequency 2. small objects have higher spatial frequency 3. large objects have lower spatial frequency a. small objects have lower spatial frequency and small objects have higher spatial frequency only b. small objects have lower spatial frequency and large objects have lower spatial frequency only c. small objects have higher spatial frequency and large objects have lower spatial frequency only d. small objects have lower spatial frequency, small objects have higher spatial frequency, and large objects have lower spatial frequency ANS: C General feedback; Small objects have higher spatial frequency and large objects have lower spatial frequency.

31. What is a measure of the imaging system’s ability to display the contrast of anatomic objects varying in size? a. detective quantum efficiency b. spatial resolution c. spatial frequency d. modulation transfer function ANS: D Modulation transfer function (MTF) is a measure of the imaging system’s ability to display the contrast of anatomic objects varying in size, and the value will be between 0 and 1.0.

32. The radiographic misrepresentation of either the size (magnification) or the shape of the anatomic part is called a. brightness b. contrast c. sharpness d. distortion ANS: D Distortion results from the radiographic misrepresentation of either the size (magnification) or the shape of the anatomic part.

33. An increase in the image size of an object compared with its true, or actual, size, is called a. size distortion b. shape distortion c. sharpness distortion d. size distortion and sharpness distortion ANS: A An increase in the image size of an object compared with its true, or actual, size, is called size distortion.

34. Size distortion is also called a. elongation b. magnification c. shape distortion d. foreshortening ANS: B Magnification and size distortion are one and the same.

35. Which of the following combinations would result in an image with the least amount of magnification? a. Decreased OID and decreased source-to-image receptor distance (SID). b. Decreased OID and increased SID. c. Increased OID and decreased SID. d. Increased OID and increased SID. ANS: B Decreased OID and increased SID will produce an image with the least amount of magnification.

36. Objects that are being imaged can be misrepresented radiographically by distortion of their shape. This is called a. size distortion b. shape distortion c. sharpness distortion d. size distortion and sharpness distortion ANS: B Shape distortion is when objects that are being imaged are misrepresented radiographically by distortion of their shape.

37. When the image of a structure appears longer than the actual structure, there is a. size distortion b. foreshortening c. magnification d. elongation ANS: D When the image of a structure is longer than the actual object, there is elongation.

38. When the image of a structure appears shorter than the actual structure there is a. size distortion b. elongation c. magnification d. shape distortion ANS: D When the image of an object is shorter than the actual item, there is shape distortion, specifically foreshortening.

39. Shape distortion is caused by

a. b. c. d.

increasing the OID angling the x-ray tube angling the patient angling the x-ray tube and angling the patient

ANS: D Angling the x-ray beam or the part will result in shape distortion.

40. In order to minimize shape distortion in the image, what three factors should be properly aligned?

1. x-ray tube.

2. image receptor.

3. object-to-image distance.

4. anatomic part. a. x-ray tube, image receptor, and object-to-image distance. b. x-ray tube, object-to-image distance, and anatomic part. c. image receptor, object-to-image distance, and anatomic part. d. x-ray tube, image receptor, and anatomic part.

ANS: D Any misalignment of the CR among these three factors—tube, part, or image receptor—alters the shape of the part recorded in the image.

41. Scatter radiation a. is the result of photoelectric interactions b. decreases the visibility of anatomic structures c. results in higher radiographic contrast d. is the result of photoelectric interactions and decreases the visibility of anatomic structures ANS: B Scatter radiation is the result of Compton interactions, results in lower contrast, and decreases the visibility of the structures being imaged.

42. Brightness fluctuations on the image is called a. brightness b. contrast resolution c. spatial resolution d. quantum noise ANS: D Quantum noise is seen as brightness fluctuations on the image.

43. Quantum noise is a result of a. too few x-ray photons reaching the image receptor b. too many x-ray photons reaching the image receptor c. x-rays with energy that is too high reaching the image receptor d. too few x-ray photons reaching the image receptor and x-rays with energy that is too high reaching the image receptor

ANS: A Quantum noise is the result of too few x-ray photons reaching the IR. 44. Which of the following is a method of describing the strength of the radiation exposure compared with the amount of noise apparent in a digital image? a. contrast-to-noise ratio b. detective quantum efficiency c. modulation transfer function d. signal-to-noise ratio ANS: D Signal-to-noise ratio (SNR) is a method of describing the strength of the radiation exposure compared with the amount of noise apparent in a digital image.

45. Which of the following will increase the ability to visualize anatomic tissues?

1. Increasing the contrast-to-noise ratio 2. Increasing the signal-to-noise ratio 3. A higher modulation transfer function a. Increasing the contrast-to-noise ratio and increasing the signal-to-noise ratio only b. Increasing the contrast-to-noise ratio and a higher modulation transfer function only c. Increasing the signal-to-noise ratio and a higher modulation transfer function only d. Increasing the contrast-to-noise ratio, increasing the signal-to-noise ratio, and a higher modulation transfer function ANS: D Increasing the contrast-to-noise ratio and signal-to-noise ratio, and a higher modulation transfer function will all increase the ability to visualize anatomic tissues.

46. An artifact that is imaged within the patient’s body is a a. anatomic anomaly b. foreign body c. double exposure d. none of these ANS: B Foreign bodies are a classification of artifacts imaged within the patient’s body.

TRUE/FALSE 1. It is impossible to produce a radiographic image without unsharpness.

ANS: T There will always be some level of unsharpness in a radiographic image.

2. Distortion is misrepresentation of the size or shape of the anatomy being imaged.

ANS: T Distortion is misrepresentation of the size or shape of the anatomy being imaged.

3. When the image is distorted, spatial resolution is also decreased.

ANS: T When the image is distorted, spatial resolution is also decreased.

4. When the foot is placed directly on the image receptor, there is no OID. ANS: F Even when the part is placed directly on the image receptor, there is always some distance between the anatomy being imaged and the image-recording material, even if very small.

5. Because it reduces spatial resolution, it is always best to minimize shape distortion in radiographic imaging.

ANS: F Even though shape distortion does reduce spatial resolution, there are times that it can be used as an advantage, such as angling the tube to eliminate superimposition of parts.

6. With digital imaging, the intensity of radiation reaching the image receptor determines the brightness of the image.

ANS: F With digital imaging, the intensity of radiation reaching the IR does not control image brightness.

7. The resolution of the monitor used to display the digital image has an enormous effect on spatial resolution.

ANS: T A high-resolution monitor displays significantly better spatial resolution than a low-resolution monitor.

8. The radiographer should evaluate the exposure indicator value along with the quality of the digital image before determining whether a repeat image is warranted.

ANS: T Because digital image processing can compensate for exposure errors, the image may display the appropriate level of brightness and may not need to be repeated.

9. Scatter radiation is not a significant concern with digital imaging because the computer can adjust the image brightness and contrast.

ANS: F Even though the computer can adjust the image, scatter radiation is an important issue with digital imaging because these image receptors are sensitive to low-energy x-ray photons (such as scattered photons).

10. Artifacts can make diagnosis of pathologic conditions difficult or impossible.

ANS: T Artifacts can create significant problems when they interfere with imaging the anatomy of interest.

11. Decreasing the contrast-to-noise ratio will increase image quality.

ANS: F Decreasing the contrast-to-noise ratio will decrease the visibility of anatomic tissues.

12. The higher the detective quantum efficiency (DQE) of a system, the lower the radiation exposure required to produce a quality image.

ANS: T The higher the detective quantum efficiency (DQE) of a system, the lower the radiation exposure required to produce a quality image, thereby decreasing patient exposure.

13. Tissues that attenuate the x-ray beam very differently are said to be high subject contrast.

ANS: T Tissues that absorb and transmit the x-ray beam very differently are said to be high subject contrast.

Chapter 10: Digital Image Receptors Johnston/Fauber: Essentials of Radiographic Physics and Imaging, 4th Edition MULTIPLE CHOICE 1. Which of the following statements is true concerning computed radiography (CR) systems? a. Use storage phosphors to temporarily store energy representing the image signal. b. Require that the storage phosphor undergoes a process to extract the latent image. c. Have detectors that directly capture and read out an electronic image signal. d. Use storage phosphors to temporarily store energy representing the image signal and require that the storage phosphor undergoes a process to extract the latent image.

ANS: D Computed radiography (CR) systems are those that use storage phosphors to temporarily store energy representing the image signal. The phosphor then undergoes a process to extract the latent image.

2. Which of the following statements is true concerning direct radiography (DR) systems? a. Use storage phosphors to temporarily store energy representing the image signal. b. Require that the storage phosphor undergoes a process to extract the latent image. c. Have detectors that directly capture and read out an electronic image signal. d. Use storage phosphors to temporarily store energy representing the image signal and require that the storage phosphor undergoes a process to extract the latent image.

ANS: C

Direct radiography (DR) systems are those that have detectors that directly capture and read out an electronic image signal.

3. Which of the following imaging systems uses a cassette, a photostimulable phosphor plate, a plate reader, and a computer workstation? a. Computed radiography. b. Computed tomography. c. Direct radiography. d. Computed radiography and computed tomography.

ANS: A The primary parts of a CR system are the cassette, photostimulable phosphor (PSP) plate that goes inside the cassette, the plate reader, and a computer workstation.

4. The type of phosphor found in the PSP plate for computed radiography is a. plastic b. amorphous selenium c. barium fluorohalide d. cesium iodide ANS: C Barium fluorohalide is the active material in the phosphor layer of the PSP plate.

5. This part of the plate reflects light released during the reading phase toward the photodetector a. protective layer b. reflective layer c. phosphor layer d. conductive layer ANS: B The reflective layer is a layer that reflects light released during the reading phase toward the photodetector.

6. Where is the PSP plate located? a. On the inside the cassette. b. On the outside the cassette. c. Computer workstation. d. On the detector array.

ANS: A The cassette is simply a container for the PSP plate, which is found on the inside of the cassette.

7. The part of a PSP plate that reduces and carries away static electricity is the a. conductive layer b. structured phosphor layer c. turbid phosphor layer d. support layer ANS: A

The conductive layer of the PSP plate reduces and carries away static electricity.

8. The part of a PSP plate that gives some rigidity to the plate is the a. conductive layer b. structured phosphor layer c. turbid phosphor layer d. support layer ANS: D The support layer provides some rigidity to the plate to keep it from being too floppy.

9. The part of a PSP plate that includes a random distribution of crystals is the a. conductive layer b. structured phosphor layer c. turbid phosphor layer d. support layer ANS: C The turbid phosphor layer includes phosphor crystals that are randomly distributed within the active layer.

10. The part of a PSP plate that includes columnar crystals, resembling needles on end, is the a. conductive layer b. structured phosphor layer c. turbid phosphor layer d. support layer ANS: B The structured phosphor layer includes phosphor crystals that resemble a bundle of needles standing on end.

11. Europium serves as a. an activator for the phosphor b. the protective layer c. the reflective part of the PSP plate d. the conductive layer ANS: A Europium is a silvery rare earth metal that captures some of the energy as the phosphors respond to the x-ray photons.

12. Approximately what percentage of the removed electrons are ―trapped‖ in the conduction band? a. 1%. b. 25%. c. 50%. d. 99%.

ANS: C Approximately half of the removed electrons are ―trapped‖ in the conduction band.

13. The trapped electrons in the conduction band of the PSP form the a. latent image b. manifest image c. visible image d. latent image and visible image ANS: A The trapped electrons in the conduction band of the PSP form the latent or invisible image.

14. At the time of processing, the energy of the trapped electrons is released by exposure to a laser in a process called a. photodetector b. scintillation c. phosphorescence d. photostimulable luminescence ANS: D Photostimulable luminescence is the term for light (luminescence) produced as a result of stimulation by light (in this case, laser light).

15. When the PSP plate is exposed to the laser of the reader, the energy is released and converted to a digital signal, becoming a a. latent image b. analog image c. primary image d. manifest image ANS: D When the PSP plate is exposed to the laser of the reader, the energy is released and converted to a digital signal, becoming a manifest image.

16. The part of the reader that senses the light released from the PSP plate is the a. photodetector b. optical system c. drive mechanism d. photostimulable luminescence ANS: A The photodetector is the part of the reader that detects the light released from the plate.

17. The part of the reader that consists of a laser, optical filters, light-collecting optics, and beam-shaping devices that is designed to project and guide a precisely controlled laser beam back and forth across the plate as the plate moves through the scan area is the a. photodetector b. optical system c. drive mechanism d. photostimulable luminescence ANS: B

An optical system that is made up of a laser, beam-shaping optics, light-collecting optics, and optical filters and is designed to project and guide a precisely controlled laser beam back and forth across the plate as the plate moves through the scan area.

18. This part of the reader moves the PSP plate through the reader a. Photodetector b. Optical system c. Drive mechanism d. Photostimulable luminescence ANS: C The drive mechanism moves the plate through the reader and scanning process.

19. Following the detection of the light released from the phosphor layer, the amplified signal is sent to the _____________ to convert it to a digital electronic signal for the display computer a. digital-to-analog converter (DAC) b. analog-to-digital converter (ADC) c. light-to-computer device (LTC) d. none of these ANS: B The signal from the reader is sent to the ADC to convert the analog data to digital data, which is usable by the computer.

20. Which of the following digital detectors continues to improve and can be used in mammography and dental radiography? a. photostimulable phosphor (PSP) b. complementary metal oxide semiconductor (CMOS) c. light-to-computer device (LTC) d. charge-coupled device (CCD) ANS: B Recent advances in complementary metal oxide semiconductor (CMOS) technology, particularly the creation of crystal light tubes that prevent light spread and methods for increasing their size, make them future possibilities. They are currently options for mammography and dental radiography machines and their quality and performance continues to improve.

21. Part of a DR system with indirect capture, the light-sensitive device that is commonly found in digital cameras is the a. charge-coupled device (CCD) b. photoconductor c. thin film transistor (TFT) d. scintillator ANS: A The charge-coupled device (CCD) is one of the indirect methods for capturing the image with a direct radiography.

22. A material that absorbs x-ray energy and emits visible light in response is a a. charge-coupled device (CCD) b. photoconductor c. thin film transistor (TFT) d. scintillator ANS: D A scintillator is a material that absorbs x-ray energy and emits visible light in response.

23. The scintillator for the indirect capture DR system that utilizes a CCD is a. cesium iodide b. barium fluorohalide c. amorphous selenium d. crystalline silicon ANS: A The scintillator for the indirect capture DR system that utilizes a CCD is made of cesium iodide.

24. The placement of several CCD detectors close together to form a larger detector is called a. roofing b. tiling c. butting d. none of these ANS: B Tiling is the placement of several CCD detectors close together to form a larger detector.

25. Put the following sequence of events in this indirect capture system in order 1. The CCD converts the energy to an electronic signal 2. The light energy is then transmitted to the CCD 3. X-rays are absorbed by the scintillator and converted to light 4. Electronic signal is sent to the computer work station for processing and display a. The CCD converts the energy to an electronic signal, The light energy is then transmitted to the CCD, X-rays are absorbed by the scintillator and converted to light, Electronic signal is sent to the computer work station for processing and display b. The light energy is then transmitted to the CCD, The CCD converts the energy to an electronic signal, X-rays are absorbed by the scintillator and converted to light, Electronic signal is sent to the computer work station for processing and display c. X-rays are absorbed by the scintillator and converted to light, The CCD converts the energy to an electronic signal, The light energy is then transmitted to the CCD, Electronic signal is sent to the computer work station for processing and display d. Electronic signal is sent to the computer work station for processing and display, X-rays are absorbed by the scintillator and converted to light, The CCD converts the energy to an electronic signal, The light energy is then transmitted to the CCD ANS: C

With this indirect capture, x-rays are absorbed by the scintillator and converted to light. This light energy is then transmitted to the CCD where it is converted to an electronic signal and sent to the computer work station for processing and display.

26. Which of the following statements is false concerning Complimentary Metal Oxide Semiconductors (CMOSs) a. Each detector element has its own amplifier, photodiode, storage capacitor, and are surrounded by transistors b. They have light sensitivity and resolution that is equal to or better than CCDs c. They are very inexpensive to manufacture d. They are currently options for mammography and dental radiography machines ANS: B CMOSs do not have quite the light sensitivity or resolution of CCDs. 27. The two methods for indirect capture in DR include a. charge-coupled devices b. photoconductors and TFT arrays c. scintillator and TFT array d. charge-coupled devices and scintillator and TFT array ANS: D Two methods for indirect capture in DR systems include using CCDs and devices using a scintillator and TFT array.

28. The scintillator for the indirect capture DR system that utilizes a CMOS is a. cesium iodide b. barium fluorohalide c. amorphous selenium d. crystalline silicon ANS: D The scintillator for the indirect capture DR system that utilizes a CMOS is made of crystalline silicon.

29. Amorphous silicon is used as the a. photoconductor for indirect capture DR imaging b. photodetector for indirect capture DR imaging c. photoconductor for direct capture DR imaging d. photodetector for direct capture DR imaging ANS: B Amorphous silicon is used as the photodetector for indirect capture DR imaging.

30. In an indirect capture DR system, the electronic components that is configured in a network of detector elements is the a. charge-coupled device (CCD) b. photoconductor c. thin film transistor (TFT) d. scintillator

ANS: C The thin-film transistor the electronic component that is configured in a network of detector elements.

31. Which of the following digital image receptors does not use a network of TFTs? a. Indirect capture with CCD b. Indirect capture with amorphous silicon photodetector c. Direct capture with amorphous selenium photoconductor d. None of these use TFTs ANS: A The use of the charge-coupled device eliminates the need for a network of thin-film transistors.

32. Which of the following describes the extra step, and is therefore a limitation, of indirect-capture methods? a. X-rays are converted to light and then to electrons b. X-rays are converted directly to electrons c. Electrons are converted to light, and then to x-rays d. X-rays are converted to light and then to electrons and X-rays are converted directly to electrons ANS: A One problem with the indirect-capture methods is that there is an extra step during which x-rays are converted to light, and then to electrons, which causes a loss of resolution.

33. Which of the following is true concerning direct capture DR imaging? a. The DR direct-capture method does not use a scintillator b. The DR direct-capture method uses a photoconductor and TFT array c. The DR direct-capture method avoids the loss of resolution caused by indirect-capture methods d. All of these are true ANS: D The DR direct-capture method does not use a scintillator. Rather, it uses a photoconductor and TFT array, thereby avoiding the loss of resolution caused by indirect capture.

34. Part of the direct-capture DR system, the __________ absorbs x-rays and produces an electric signal a. CCD b. photoconductor c. TFT d. scintillator ANS: B The photoconductor absorbs x-rays and produces an electric signal.

35. Amorphous selenium is used as the a. photoconductor for indirect capture DR imaging b. photodetector for indirect capture DR imaging c. photoconductor for direct capture DR imaging

d. photodetector for direct capture DR imaging ANS: C Amorphous selenium is used as the photoconductor for direct capture DR imaging.

36. The histogram is a. the digital radiographic image b. the electronic signal from the ADC c. a graphic representation of a data set d. the digital radiographic image and a graphic representation of a data set ANS: C A histogram is a display of data in a graphical format.

37. During image acquisition, the computer analyzes the histogram using: a. algorithms b. look-up-tables (LUT) c. analog-digital-converter (ADC) d. detective quantum efficiency (DQE) ANS: A During image acquisition, the computer analyzes the histogram using processing algorithms.

38. Histogram analysis compares the histogram with pre-established histograms specific to a. the x-ray equipment b. the computer monitor c. the patient data d. the anatomic part being imaged ANS: D Histogram analysis compares the histogram with pre-established histograms specific to the anatomic part being imaged.

39. The values of interest (VOI) a. is the graphic representation of a data set b. determine what section of the histogram data set should be included in the displayed image c. determine what section of the histogram data set should be included in the latent image d. is the useful area of the digital image receptor ANS: B The VOI determines what section of the histogram data set should be included in the displayed image 40. When processing the PSP plate, the analog signal is digitized and divided into a matrix of squares. Each square is a a. detector element (DEL) b. field of view (FOV) c. picture element (pixel) d. volume element (voxel)

ANS: C Each square in a matrix is a picture element, or pixel.

41. The useful imaging area of the digital receptor is the a. detector element (DEL) b. field of view (FOV) c. thin-film transistor (TFT) d. values of interest (VOI) ANS: B The FOV is the useful area of the digital image receptor.

42. Which of the following is the correct sequence for extracting and processing the image with the PSP plate? a. Signal sent to ADC; scanned by laser light; sampling of signal; converted to electronic signal b. Converted to electronic signal; sampling of signal; signal sent to ADC; scanned by laser light c. Scanned by laser light; converted to electronic signal; sampling of signal; signal sent to ADC d. None of these ANS: C After being scanned by the laser light, the released light energy is converted to an electronic signal by the photodetector. The electronic signal is then sampled and sent to the ADC to be converted to digital data.

43. As the matrix size increases, the pixel size a. increases b. decreases c. remains the same ANS: B The larger the matrix (the more rows and columns), the smaller the pixel size.

44. As the matrix size decreases, the spatial resolution a. increases b. decreases c. remains the same ANS: B The smaller the matrix (the fewer rows and columns), the larger the pixel and the poorer the spatial resolution.

45. As the pixel size decreases, the spatial resolution a. increases b. decreases c. remains the same ANS: A The smaller the pixel size the better the spatial resolution.

46. The rate at which a data sample is acquired from the detector is the a. sampling rate b. sampling frequency c. sampling pitch d. bit depth ANS: B The sampling frequency is the rate at which a data sample is acquired from the detector.

47. Sampling pitch is a. the distance between laser beam positions during processing of the plate b. the distance between adjacent detector elements c. how digital detectors sample the x-ray exposure d. all of these ANS: D Sampling pitch describes how digital detectors sample the x-ray exposure. For CR digital systems, sampling pitch is the distance between laser beam positions during sampling, whereas for DR systems sampling pitch is the distance between adjacent detector elements.

48. Bit depth: a. is the available grayscale b. refers to the number of shades of gray that can be displayed within a pixel c. is measured in microns d. is the available grayscale and refers to the number of shades of gray that can be displayed within a pixel ANS: D Bit depth is the available grayscale and refers to the number of shades of gray that can be displayed within a pixel.

49. The range of exposures that produce acceptable information to be used in image formation is a. exposure range b. dynamic range c. exposure latitude d. bit depth ANS: C Exposure latitude is the range of exposures to the IR that produce densities or data that can be used in a diagnostic image.

50. The range of exposure intensities that an IR can respond to and acquire image data is a. dynamic range b. exposure range c. brightness range d. All of these ANS: A

Dynamic range is the range of exposure intensities that an IR can respond to and acquire image data.

51. As compared with the H&D curve of film with the toe, straight-line portion, and shoulder, the response of a digital IR to exposure is a. more curved b. a straight line c. a line and a curve d. a curve but less than film ANS: B The digital IR’s response to exposure forms a straight line, very different from the film’s H&D curve.

52. The dynamic range of a digital imaging system is ____________ than that of a film-screen system a. much less b. much greater c. approximately the same d. a little bit greater ANS: B The digital imaging system can respond to a very wide range of exposures as compared with film, which is quite limited.

53. Primary workstation displays are used for a. viewing the image for the first time b. official image interpretation c. viewing the digital image at the radiographer’s console d. All of these ANS: B The display at a primary workstation has high quality standards because it is where official image interpretation takes place.

54. Digital imaging preprocessing includes a. rescaling b. histogram analysis c. rescaling and histogram analysis d. none of these ANS: C Histogram analysis and automatic rescaling are common preprocessing functions.

55. Postprocessing functions a. allow changes to be made in the appearance of the digital image b. are done through computer software operations c. include window width and window level adjustments d. All of these ANS: D

There are many postprocessing functions, all of which involve computer software and make the digital image more useful, including adjustments in window width and level.

56. The human eye can appreciate approximately _______ shades of gray at a particular level. a. 5 b. 10 c. 30 d. 300 ANS: C The human eye can appreciate approximately 30 shades of gray at a particular level.

57. Digital radiography uses a _____ dynamic range a. 8-bit b. 10-bit c. 12-bit d. 14-bit ANS: D Digital radiography uses a 14-bit dynamic range, which equates to 16,384 distinct shades of gray.

58. In CR, this represents the exposure level to the PSP plate a. exposure indicator b. dose-area product (DAP) c. detective quantum efficiency (DQE) d. dynamic range ANS: A In CR systems, the exposure indicator represents how much exposure the imaging plate received.

59. Which manufacturers use exposure indicators that are inversely related to the exposure to the digital detector? a. Carestream b. Fuji c. Konica d. Fuji and Konica ANS: D Fuji and Konica use sensitivity (S) numbers, and the value is inversely related to the exposure to the digital detector.

60. For CR systems, the exposure index is based on a. how much exposure leaves the x-ray tube b. how much mAs was set at the operators console c. the histogram analysis d. the kVp being used ANS: C For CR systems, the EI is based on the histogram analysis.

61. The value that indicates exposure to the IR with a DR imaging system is the a. exposure indicator b. dose-area product (DAP) c. detective quantum efficiency (DQE) d. dynamic range ANS: B DAP is a measure of exposure in air, followed by computation to estimate absorbed dose to the patient.

62. Related to the DR detector,______________ is an expression of the radiation exposure level that is required to produce an optimal image. a. exposure indicator b. dose-area product (DAP) c. detective quantum efficiency (DQE) d. target exposure index ANS: C DQE is a measurement of the efficiency of an image receptor in converting the x-ray exposure it receives to a quality radiographic image.

63. Having a higher DQE indicates a potentially a. lower patient dose b. higher patient dose c. similar patient dose ANS: A A higher DQE (within limits) indicates that a lower exposure can be used to produce an optimal image, potentially reducing patient dose.

64. If the DQE is too high a. the image will be very noisy/grainy b. the image will be very dark c. the image will be very light d. the patient dose will be too high as well ANS: A Having a DQE that is too high results in using very low mAs, which can then appear as quantum noise, or as a grainy appearance.

65. DQE is a. evaluated by comparing the image noise of a detector with that of an ―ideal‖ detector with the same signal-response characteristics b. a measure of how well the signal-to-noise ratio (SNR) is preserved in an image c. the exposure necessary to produce a desired level of brightness d. evaluated by comparing the image noise of a detector with that of an ―ideal‖ detector with the same signal-response characteristics and a measure of how well the signal-to-noise ratio (SNR) is preserved in an image ANS: D

DQE is evaluated by comparing the image noise of a detector with that of an ―ideal‖ detector with the same signal-response characteristics. DQE is a measure of how well the signal-to-noise ratio (SNR) is preserved in an image.

66. What would the DQE measure if the imaging system has 100% conversion efficiency? a. 0 b. 0.5 c. 1.0 d. 2.0 ANS: C If an image receptor system is able to convert x-ray exposure into a quality image with 100% efficiency (meaning no information loss), the DQE would measure 100% or 1.0.

67. Undesirable fluctuations in the brightness of an image is a. contrast b. noise c. DQE d. DAP ANS: B Noise is the undesirable fluctuation in the brightness in a digital image.

68. The ideal expression of digital detector image resolution is the a. Nyquist frequency b. modulation transfer function (MTF) c. limiting spatial resolution (LSR) d. detective quantum efficiency (DQE) ANS: B Modulation transfer function is the ideal expression of digital detector image resolution.

69. The ability of a detector to resolve small structures is the a. Nyquist frequency b. modulation transfer function (MTF) c. limiting spatial resolution (LSR) d. detective quantum efficiency (DQE) ANS: C The limiting spatial resolution (LSR) is the ability of a detector to resolve small structures.

70. The highest spatial frequency that a digital detector can record is the a. Nyquist frequency b. modulation transfer function (MTF) c. limiting spatial resolution (LSR) d. detective quantum efficiency (DQE) ANS: A The Nyquist frequency is the highest spatial frequency that a digital detector can record.

71. One half of the Nyquist frequency is equal to

a. b. c. d.

the spatial resolution of a digital system the contrast resolution of a digital system the amount of noise in a digital system the detective quantum efficiency (DQE)

ANS: A The spatial resolution of a digital system is equal to one half of the Nyquist frequency.

72. The role of mAs in digital imaging includes a. being the controlling factor of density b. determining the quantity of radiation and patient dose c. selecting optimum mAs to provide sufficient quanta to expose the receptor and avoid excessive noise d. determining the quantity of radiation and patient dose and selecting optimum mAs to provide sufficient quanta to expose the receptor and avoid excessive noise ANS: D The mAs no longer controls density of the image, but still determines the quantity of radiation exposing the patient. An optimum mAs should be selected to provide sufficient quanta to expose the receptor and avoid excessive noise.

73. The primary factor influencing contrast with digital imaging is the a. values of interest (VOI) b. modulation transfer function (MTF) c. lookup table (LUT) d. detective quantum efficiency (DQE) ANS: C The primary factor influencing contrast with digital is the lookup table (LUT).

74. Because digital systems are more sensitive to low-energy radiation, how can scatter radiation be effectively controlled? a. Pre-exposure collimation b. Post-processing masks c. Use of a grid d. Pre-exposure collimation and use of a grid ANS: D There are two ways to effectively control scatter radiation’s effect on the image: collimation and grid use. The use of post-processing masks and cropping tools to give the appearance of collimation of the finished radiograph does not serve the same purpose and is unethical.

75. Digital imaging quality control focuses on a. the display monitors b. the viewing environment c. the radiologist’s comfort d. the display monitors and the viewing environment ANS: D Digital imaging quality control focuses on display monitors and the viewing environment.

76. Display device performance is determined by use of a. a special imaging monitor b. various test patterns c. a variety of different types of image receptors d. none of these ANS: B There are many different test patterns that can be used to assess the display image quality.

77. This test assesses the system’s ability to display images of different parts of an image with high fidelity a. reflection b. luminance response c. luminance dependencies d. resolution ANS: D The resolution test assesses the system’s ability to display images of different parts of an image with high fidelity.

78. This test assesses the displayed luminance values versus the input values from the display system a. reflection b. luminance response c. luminance dependencies d. resolution ANS: B The luminance response test assesses the displayed luminance values versus the input values from the display system.

79. If they have not been used for _______ hours, PSP plates should be erased a. 4 b. 12 c. 24 d. 48 ANS: D If they have not been used for 48 hours, PSP plates should be erased.

80. __________ allows for the acquisition, display, and storage of digital images a. PACS b. DICOM c. teleradiology d. RIS ANS: A Picture Archiving and Communication System (PACS) allows for the acquisition, display and storage of digital images.

81. The average size of a digital radiographic study is

a. b. c. d.

35 kB 35 MB 35 TB none of these

ANS: B The average size of a digital radiographic study is 35 megabytes.

82. __________ is the standard format for digital medical images a. PACS b. DICOM c. Teleradiology d. RIS ANS: B Digital Imaging and Communication in Medicine (DICOM) provides the standard format for digital medical images, allowing images to be easily shared.

83. Electronically sending radiologic images over a distance is a. PACS b. DICOM c. teleradiology d. RIS ANS: C Teleradiology allows image files can be accessed from various workstations located throughout the facility or even by clients outside the facility if they are given access to the system.

84. What size display monitor is required for digital image interpretation by a radiologist? a. 1 Mp b. 2 Mp c. 4 Mp d. 5 Mp ANS: B The radiologist should have a 2 Mp display monitor for interpretation of general radiographic studies, whereas a 5 Mp monitor is needed for digital mammograms.

85. This type of storage involves data stored on hard drives with access times in milliseconds and transfer times in the range of tens and hundreds of megabytes per second a. Online storage b. Offline storage c. Nearline storage ANS: A Online storage has very quick access time (in ms) and can transfer very large amounts of data.

86. Which of the following describes a new method of storing digital studies that makes use of a special recording medium and laser technology to record data throughout the depth of the medium? a. Network attached storage (NAS) b. High capacity flash drives c. Magnetic tape d. Holographic disk device ANS: D Another solution recently introduced is the holographic storage device. This technology makes use of a special recording medium and laser technology to record data throughout the depth of the medium. A holographic disk device stores data through its entire depth rather than just the surface. This approach has the potential of storing 1 TB per cubic centimeter.

Chapter 11: Radiographic Exposure Technique Johnston/Fauber: Essentials of Radiographic Physics and Imaging, 4th Edition MULTIPLE CHOICE 1. Which of the following is not a primary exposure technique factor? a. KVP b. Time of exposure c. SID d. MA ANS: C SID is not a primary exposure technique factor; mAs and kVp are.

2. What type of relationship does mAs have with the quantity of x-rays produced? a. Direct proportional b. Direct but not proportional c. Inverse proportional d. Inverse but not proportional ANS: A There is a direct proportional relationship between mAs and quantity of radiation; if the mAs is doubled, the amount of radiation is doubled.

3. Given 40 mAs, doubling the mA produces the same result as a. doubling the time of exposure b. doubling the mAs c. halving the mAs d. doubling the time of exposure and doubling the mAs e. doubling the time of exposure and halving the mAs ANS: D Doubling the mA doubles the mAs, as does doubling the exposure time.

4. How much mAs is produced when the mA is 200 and the exposure time is 0.5 s? a. 25 mAs. b. 50 mAs.

c. 100 mAs. d. 200 mAs. ANS: C In that mAs = mA  exposure time (in seconds), 200  0.5 = 100 mAs. 5. How much mAs is produced when the mA is 800 and the exposure time is 30 ms? a. 24 mAs. b. 240 mAs. c. 2,400 mAs. d. 24,000 mAs ANS: A In that mAs = mA  exposure time (in seconds), 800  0.03 = 24 mAs. 6. How much mAs is produced when the mA is 50 and the exposure time is 4000 ms? a. 20 mAs. b. 200 mAs. c. 2,000 mAs. d. 20,000 mAs. ANS: B In that mAs = mA  exposure time (in seconds), 50  4 = 200 mAs. 7. If the mA is 600 and exposure time is 10 ms, how can the mAs be doubled? a. Increase the mA to 1200. b. Increase the time to 20 ms. c. Increase the mAs to 12 mAs. d. Any of the above is correct.

ANS: D Doubling the mA, time, or mAs results in the same outcome.

8. Which of the following exposure factors produces 25 mAs? a. 250 mA @ 1 s. b. 125 mA @ 2 s. c. 200 mA @ 0.2 s. d. 250 mA @ 100 ms.

ANS: D 250 mA  100 ms (which is 0.10 s) = 25 mAs. 9. What mA should be selected to produce 32 mAs using a 0.04 exposure time? a. 128 mA. b. 200 mA. c. 400 mA. d. 800 mA. ANS: D If mAs = mA  seconds, mA = mAs/seconds, or 32/.04, which equals 800 mA.

10. In maintaining the same mAs, there is a(n) ________________ relationship between mA and exposure time. a. direct b. inverse c. double d. half ANS: B To maintain the same mAs, if the mA is increased, the exposure time must decrease, resulting in an inverse relationship.

11. 200 mA @ 80 ms (0.08 s) produces 16 mAs. Which of the following exposure factors maintains 16 mAs while using a shorter exposure time? a. 100 mA @ 0.16 s. b. 200 mA @ 0.16 s. c. 400 mA @ 0.04 s. d. 400 mA @ 0.08 s.

ANS: C By doubling the mA and halving the seconds, 400 mA @ 0.04 s equals 16 mAs while shortening the exposure time.

12. Which of the following is true regarding the effect of mAs on the brightness of a digital image. a. doubling the mAs will increase brightness b. halfing the mAs will decrease brightness c. mAs does not affect brightness in a digital image d. doubling the mAs will increase brightness and halfing the mAs will decrease brightness ANS: C The mAs does not have a direct effect on image brightness. During computer processing, image brightness is maintained when the mAs is too low or too high (within reason).

13. Which of the following has an effect on how much mAs is used for a specific examination? a. generator type. b. pathologic conditions present. c. a patient’s age. d. All of these.

ANS: D Many factors influence the amount of mAs selected for an examination, including generator type, pathologic conditions, and the patient’s age.

14. If a radiographic image has to be repeated because of an exposure error, the mAs must be changed by a factor of at least a. 1 b. 2 c. 4 d. 30

ANS: B If a radiographic image has to be repeated because of an exposure error, the mAs must be changed by a factor of at least 2 (either doubled or halved).

15. With digital imaging, if the mAs is too high a. the image brightness will appear low (dark) b. the image brightness will appear appropriate c. the image brightness will appear high (light) d. none of these ANS: B With digital imaging, overexposure will result in an image with appropriate brightness because the computer will make the required adjustments.

16. With digital imaging, if the image receptor is not exposed to enough radiation a. the image brightness will appear low (dark) b. the image brightness will appear appropriate c. the image brightness will appear high (light) d. none of these ANS: B With digital imaging, underexposure will result in an image with appropriate brightness because the computer will make the required adjustments.

17. With digital imaging, the relationship between mAs and image brightness is a. direct b. direct proportional c. inverse d. none of these ANS: D Generally speaking, there is no relationship between mAs and image brightness with a digital image; the computer will adjust image brightness to an appropriate level regardless of the mAs used.

18. Using 20 mAs on a knee image (digital IR) when only 8 mAs was needed to produce sufficient remnant radiation results in a. a dark image b. an image with appropriate brightness c. unnecessary patient radiation dose d. an image with appropriate brightness and unnecessary patient radiation dose ANS: D Using more than two times the mAs needed to produce an excellent digital image results in appropriate brightness (due to computer adjustment) but a patient who has been overexposed to radiation.

19. With digital imaging, overexposure of the IR and excessive dose to the patient can be a. easily determined by looking at the image brightness b. difficult to determine by looking at the image brightness c. determined by checking the exposure indicator value

d. difficult to determine by looking at the image brightness and determined by checking the exposure indicator value ANS: D Overexposure of the digital image is difficult to determine by evaluating image brightness but monitoring exposure indicator values will help the radiographer reduce unnecessary patient exposure.

20. With digital imaging, if the image receptor is not exposed to enough radiation, 1. The image brightness will appear appropriate.

2. The image brightness will appear high (light).

3. Quantum noise will be visible a. The image brightness will appear appropriate and the image brightness will appear high (light) only b. The image brightness will appear appropriate and quantum noise will be visible only c. The image brightness will appear high (light) and quantum noise will be visible only d. The image brightness will appear appropriate, the image brightness will appear high (light), and quantum noise will be visible ANS: B Insufficient remnant radiation results in a digital image that has appropriate brightness (resulting from computer adjustment) but has quantum noise.

21. In digital imaging, the reason(s) mAs does not affect brightness is/are: a. wide dynamic range b. matrix size c. computer processing d. wide dynamic range and computer processing only ANS: D Digital IRs can detect a wider range of radiation intensities exiting the patient and are not dependent on mAs. In addition, during computer processing, image brightness is maintained for exposure errors.

22. Increasing the kVp a. increases the energy of the x-ray photons b. increases the quantity of x-ray photons c. increases the penetrating power of x-ray photons d. all of these ANS: D Increasing kVp produces more photons with greater energy (more penetrating).

23. What type of relationship does kVp have with exposure to the image receptor (IR)? a. direct b. inverse c. inverse proportional d. no relationship

ANS: A When kVp is increased, the amount of remnant radiation increases, thereby a direct relationship.

24. With digital imaging, if the kVp is too high a. the image brightness appears low (dark) b. the image brightness appears appropriate c. the image brightness appears high (light) d. none of these ANS: B With digital imaging, overexposure (whether caused by excessive mAs or kVp) results in an image with appropriate brightness because the computer makes the required adjustments.

25. Too little remnant radiation caused by low kVp results in a digital image in which a. the image brightness appears high (light) b. the image brightness appears appropriate c. quantum noise is visible d. the image brightness appears appropriate and quantum noise is visible ANS: D Insufficient remnant radiation (caused by low mAs or kVp) results in a digital image that has appropriate brightness (because of computer adjustment) but has quantum noise.

26. Kilovoltage is not usually used to adjust exposure to the image receptor because a. it has too great an effect on patient exposure b. it is too difficult to determine how much it should be changed c. it affects contrast d. it affects spatial resolution ANS: C Even though changing kVp has an effect on exposure to the IR, it is not usually used to make exposure adjustments because it also affects the image contrast.

27. The relationship between kVp and the quantity of remnant radiation is known as the a. kVp-brightness rule b. 15% rule c. 25% rule d. kVp-quantity rule ANS: B The 15% rule represents the relationship between kVp and the quantity of radiation reaching the image receptor.

28. Increasing the kVp by 15% has the same effect as a. increasing the mAs by 15% b. doubling the mAs c. decreasing the mAs by 15% d. halving the mAs ANS: B

Increasing the kVp by 15% has the same effect on the amount of radiation reaching the IR as does doubling the mAs. 29. Which of the following exposure factors produces the same amount of remnant radiation as does 20 mAs at 70 kVp? a. 10 mAs @ 70 kVp. b. 10 mAs @ 81 kVp. c. 40 mAs @ 60 kVp. d. 10 mAs @ 81 kVp and 40 mAs @ 60 kVp. ANS: D A 15% increase in 70 kVp = 81 and 20 mAs ÷ 2 = 10. A 15% decrease in 70 kVp = 60 and 20 mAs x 2= 40. Changing the kVp by 15% with the appropriate change in mAs will produce the same amount of remnant radiation.

30. 100 kVp should be changed to __________ kVp to decrease the exposure to the IR by a factor of 2. a. 50 b. 85 c. 115 d. none of these ANS: B To decrease the exposure to the IR by a factor of 2, the kVp must be reduced by 15%; 15% of 100 is 15, so 100 minus 15 is 85 kVp. 31. To maintain the same amount of radiation reaching the IR, if the kVp is increased by 15% the mAs needs to a. remain the same b. be doubled c. be halved d. be decreased by a minimum of 30% ANS: C Increasing the kVp by 15% doubles the radiation reaching the IR, so to maintain the original amount of exit radiation, the mAs needs to be halved.

32. When the intensities of radiation exiting the patient are very similar, the resulting radiographic image has a. low contrast b. high contrast c. short-scale contrast d. low contrast and short-scale contrast ANS: A When the intensities of radiation exiting the patient are very similar, the resulting image has low contrast (many shades of gray).

33. When the intensities of radiation exiting the patient are very different from each other, the resulting radiographic image has a. low contrast

b. high contrast c. short-scale contrast d. high contrast and short-scale contrast ANS: D Using a low kVp results in a greater variation of x-ray intensities exiting the patient, producing a high or short-scale contrast image.

34. Low kVp results in the intensities of radiation exiting the patient being ____________. a. similar b. different c. homogenous d. none of these ANS: B Low kVp results in the intensities of radiation exiting the patient being very different from each other.

35. Increasing the kVp has what affect on wavelength and frequency? a. decreases wavelength and decreases frequency b. increases wavelength and decreases frequency c. increases wavelength and increases frequency d. decreases wavelength and increases frequency ANS: D Increasing the kVp will decrease wavelength and increase frequency.

36. What is the relationship between kVp and subject contrast? a. high kVp creates low subject contrast b. low kVp creates low subject contrast c. high kVp creates high subject contrast d. kVp does not affect subject contrast ANS: A Low kVp produces high subject contrast and high kVp produces low subject contrast.

37. To adjust the kVp to produce the desired contrast level, the kVp must first be a. as low as possible for radiation safety b. as high as possible c. high enough to penetrate the part d. high enough to reduce noise ANS: C To manipulate the kVp to change radiographic contrast, it first must be at a level high enough to penetrate the anatomy.

38. If the original exposure factors are 20 mAs @ 70 kVp, which of the following factors produce a radiographic image with the same IR exposure but higher contrast? a. 10 mAs @ 81 kVp. b. 10 mAs @ 70 kVp. c. 40 mAs @ 70 kVp.

d. 40 mAs @ 60 kVp. ANS: D Based on the 15% rule, decreasing the kVp by 15% (to create more contrast) and doubling the mAs results in the same amount of radiation reaching the IR. 39. Using excessive kVp with a digital IR results in 1. increased IR exposure 2. increased scatter reaching the IR 3. decreased contrast a. increased IR exposure and increased scatter reaching the IR only b. increased IR exposure and decreased contrast only c. increased scatter reaching the IR and decreased contrast only d. increased IR exposure, increased scatter reaching the IR, and decreased contrast ANS: D Using kVp that is too high results in more exposure to the IR, more scatter reaching the IR, as well as decreased contrast.

40. Standard radiographic x-ray tubes have small focal spots that measure approximately a. 0.5–0.6 mm b. 0.6–1 mm c. 1–1.2 mm d. 0.5–1.2 mm ANS: A Standard radiographic x-ray tubes have small focal spots that measure approximately 0.5–0.6 mm.

41. Standard radiographic x-ray tubes have large focal spots that measure approximately a. 0.5–0.6 mm b. 0.6–1 mm c. 1–1.2 mm d. 0.5–1.2 mm ANS: C Standard radiographic x-ray tubes have large focal spots that measure approximately 1–1.2 mm.

42. Focal spot size is determined by a. the amount of kVp b. cathode filament size c. cathode focusing cup size d. the exposure time ANS: B The size of the focal spot depends on the size of the cathode filament selected to be energized.

43. Focal spot size affects only

a. b. c. d.

brightness contrast sharpness noise

ANS: C Focal spot size affects only sharpness (spatial resolution).

44. The line-focus principle is used to produce: 1. A larger actual focal spot 2. A smaller effective focal spot 3. Increased spatial resolution a. a larger actual focal spot and a smaller effective focal spot only b. a larger actual focal spot and increased spatial resolution only c. a smaller effective focal spot and increased spatial resolution only d. a larger actual focal spot, a smaller effective focal spot, and increased spatial resolution ANS: D The largest actual focal spot allows higher heat (and exposure), whereas the smallest effective focal spot produces increased spatial resolution.

45. Changes in the SID will affect: 1. Image receptor exposure 2. Spatial resolution 3. Magnification a. image receptor exposure and spatial resolution only b. image receptor exposure and magnification only c. spatial resolution and magnification only d. image receptor exposure, spatial resolution, and magnification ANS: D Changes in the SID will affect radiation exposure to the IR, spatial resolution, and magnification. Increasing the SID will decrease exposure to the IR, increase spatial resolution, and decrease magnification.

46. What type of relationship does distance have with x-ray beam intensity? a. Direct proportional. b. Direct but not proportional. c. Inverse proportional. d. Inverse but not proportional.

ANS: C Distance and beam intensity have an inverse proportional relationship (as the distance increases the beam intensity decreases by a proportional amount).

47. Which of the following is the correct formula for the inverse square law? a. I1/I2 = D1/D2. b. I1/I2 = D2/D1.

c. I1/I2 = (D1)2/(D2)2. d. I1/I2 = (D2)2/(D1)2.

ANS: D I1/I2 = (D2)2/(D1)2 is the formula for the inverse square law, describing the relationship between the intensity of radiation (I) and distance (D).

48. When the distance is increased from 20 inches to 40 inches, the beam intensity a. increases  2 b. decreases by c. stay the same d. decreased by ANS: B Increasing the distance from 20 to 40 inches results in lowered beam intensity by

.

49. If the intensity of radiation at 36 inches is 480 mR, what is the intensity if the SID is increased to 72 inches? a. 360 mR. b. 240 mR. c. 120 mR. d. 60 mR. ANS: C If the distance is increased by a factor of 2 (doubled), the beam intensity will be one fourth (1/22) of the original.

50. If the intensity of radiation at 36 inches SID is 600 mR, what is the intensity if the distance is increased to 60 inches SID? a. 216 mR. b. 359.9 mR. c. 1000 mR. d. 1666.7 mR. ANS: A Before actually calculating the answer, it is important to think through that increasing the distance results in decreased intensity. Using the inverse square formula, I1/I2 = (D2)2/(D1)2, 600/X = 602/362, X = 216 mR. 51. If the intensity of radiation at 72 inches is 225 mR, what is the intensity if the SID is decreased to 40 inches? a. 69.4 mR. b. 125 mR. c. 405 mR. d. 729 mR. ANS: D Before actually calculating the answer, it is important to think through that decreasing the distance results in increased intensity. Using the inverse square formula, I1/I2 = (D2)2/(D1)2, 225/X = 402/722, X = 729 mR.

52. To maintain the same exposure to the IR, if the SID is increased, the mAs must be a. decreased b. increased c. left the same d. doubled ANS: B If the SID is increased, the beam intensity and radiation reaching the IR is decreased. To maintain the original exposure to the IR, the mAs must be increased.

53. Which of the following is the correct formula for the direct square law formula? a. mAs1/mAs2 = SID1/SID2. b. mAs1/mAs2 = SID2/SID1. c. mAs1/mAs2 = (SID1)2/(SID2)2. d. mAs1/mAs2 = (SID2)2/(SID1)2.

ANS: C mAs1/mAs2 = (SID1)2/(SID2)2 is the formula for the direct square law (exposure maintenance formula), describing the relationship between the mAs and SID. 54. When the SID is decreased from 72 inches SID to 40 inches SID, how is the radiation reaching the IR affected and does mAs need to be adjusted? a. increases radiation to IR; decrease mAs b. decreases radiation to IR; increase mAs c. increases radiation to IR; increase mAs d. decreases radiation to IR; decrease mAs ANS: A When the SID is decreased from 72 inches SID to 40 inches SID, the radiation reaching the IR increases, requiring a decrease in mAs to compensate.

55. If 12 mAs produce appropriate IR exposure at 36 inches SID, how much mAs is needed at 72 inches SID to maintain that amount of exposure to the IR? a. 3 mAs. b. 6 mAs. c. 24 mAs. d. 48 mAs. ANS: D If the SID is increased by a factor of 2 (doubled), the beam intensity will be one fourth (1/22) of the original, requiring a 4´ increase in mAs to maintain IR exposure.

56. If 16 mAs produce appropriate IR exposure at 72 inches SID, how much mAs is needed at 48 inches SID to maintain that amount of IR exposure? a. 7 mAs. b. 11 mAs. c. 24 mAs. d. 36 mAs. ANS: A

Before actually calculating the answer, it is important to think through that decreasing the SID will require a decrease in mAs to maintain IR exposure. Using the direct square law formula, mAs1/mAs2 = (SID1)2/(SID2)2, 16/X = 722/482, X = 7.1 rounded to 7mAs. 57. If 10 mAs produce appropriate IR exposure at 40 inches SID, how much mAs is needed at 48 inches SID to maintain that amount of IR exposure? a. 7 mAs. b. 8 mAs. c. 12 mAs. d. 14 mAs. ANS: D Before actually calculating the answer, it is important to think through that increasing the SID will require an increase in mAs to maintain IR exposure. Using the direct square law formula, mAs1/mAs2 = (SID1)2/(SID2)2, 10/X = 402/482, X = 14.4 rounded to 14 mAs. 58. When a known mAs at 72 inches SID produces appropriate exposure to the IR, one half of the mAs can be used at a. 40 inches SID b. 48 inches SID c. 56 inches SID d. 60 inches SID ANS: C A useful tip is that changing from 72 inches SID to 56 inches SID can be compensated for by reducing the mAs in half.

59. Doubling the mAs produces appropriate exposure to the IR when the SID is changed from 40 inches to a. 36 inches SID b. 48 inches SID c. 56 inches SID d. 60 inches SID ANS: C A useful tip is that changing from 40 inches SID to 56 inches SID requires a doubling of the original mAs to maintain exposure to the IR. 60. Changes in SID affect 1. spatial resolution 2. beam intensity 3. magnification a. spatial resolution and beam intensity only b. spatial resolution and magnification only c. beam intensity and magnification only d. spatial resolution, beam intensity, and magnification ANS: D Changes in SID affect beam intensity (inverse square law) as well as magnification and spatial resolution.

61. Using a higher SID a. increases magnification b. decreases magnification c. increases size distortion d. none of these ANS: B Using a higher SID decreases magnification (size distortion).

62. _________ SID is typically used to image the chest so that the heart is seen with minimal magnification. a. 30 inches b. 40 inches c. 72 inches d. 90 inches ANS: C An SID of 72 inches is typically used with chest imaging to reduce magnification. An SID of 90 inches would further reduce magnification, but it is not a standard SID. 63. The distance between the object being imaged and the image receptor is the a. OID b. SID c. MF d. SOD ANS: A The OID is the distance between the object being imaged and the image receptor.

64. The distance between the x-ray focal spot and the image receptor is the a. OID b. SID c. MF d. SOD ANS: B The SID is the distance between the x-ray focal spot (source of radiation) and the image receptor.

65. The indication of how much magnification is seen on a radiograph is a. OID b. SID c. MF d. SOD ANS: C The indication of how much magnification is seen on a radiograph is the magnification factor.

66. The distance between the x-ray focal spot and the object being imaged is the

a. b. c. d.

OID SID MF SOD

ANS: D The SOD distance between the x-ray focal spot (source of radiation) and the object being imaged.

67. When OID cannot be reduced, it is possible to reduce size distortion by a. increasing mAs b. increasing SID c. using a small focal spot size d. decreasing SID ANS: B An increase in SID can, to some extent, compensate for an increase in OID. 68. The only factor that affects exposure to the image receptor, size distortion, and image contrast is a. SID b. mAs c. focal spot size d. OID ANS: D OID is the only factor that affects exposure to the image receptor, size distortion, and image contrast.

69. Which of the following is the formula to determine magnification factor? a. MF = SID/OID. b. MF = SID/SOD. c. MF = SOD/SID. d. MF = OID/SID. ANS: B The formula for magnification factor (MF) is MF = SOD/SID. 70. The formula to determine the SOD (source to object distance) is a. SOD = SID – OID b. SOD = OID – SID c. SOD = SID + OID d. none of these ANS: A The formula to calculate SOD is SOD = SID – OID. 71. When a patient cannot fully extend his or her leg for a knee image, resulting in the knee being 4 inches away from the image receptor, what is the MF if the SID is 40 inches? a. 0.1. b. 0.9.

c. 1.11. d. 10.

ANS: C The MF equals the SID (40) divided by the SOD (36). It is important to recall that it is impossible to have a MF of less than 1.

72. If the magnification factor is 1.25, the image is ______________ than the object. a. 25% smaller b. 125% smaller c. 25% larger d. 125% larger ANS: C A MF of 1.25 indicates that the image is 25% larger than the object. The image can never be smaller than the object.

73. The formula that can be used to calculate the object size when the image size and MF are known is a. object size = MF/image size b. object size = image size/MF c. object size = image size  MF d. none of these ANS: B The formula object size = image size/MF can be used to calculate image size, object size, or magnification when two of the variables are known.

74. What is the size of the object when the MF is 1.3 and the image size is 3 cm? a. 0.43 cm. b. 1.3 cm. c. 2.3 cm. d. 3.9 cm.

ANS: C Using the formula, object size = image size (3 cm) / MF (1.3).

75. Whenever magnification is increased a. spatial resolution is decreased b. spatial resolution is increased c. spatial resolution stays the same d. exposure to the image receptor increases ANS: A Increased magnification results in decreased spatial resolution.

76. Which of the following results in the greatest amount of magnification.

1. increased SID 2. decreased SID 3. increased OID

a. increased SID and increased OID only b. decreased SID and increased OID only ANS: B Decreased SID and increased OID results in increased magnification.

77. The device used to absorb scatter radiation in the radiation leaving the patient before it reaches the image receptor is the a. cassette b. flat panel detector c. grid d. filter ANS: C The grid is the device used to absorb scatter radiation in the radiation, leaving the patient before it reaches the image receptor.

78. Reducing the amount of scatter radiation reaching the IR results in a. lower radiographic contrast b. higher radiographic contrast c. improved spatial resolution d. reduced spatial resolution ANS: B Less scatter radiation reaching the IR results in higher radiographic contrast.

79. Deciding to use a grid for a radiographic examination requires use of a. increased SID b. decreased SID c. increased mAs d. decreased mAs ANS: C Adding a grid to a procedure, or increasing the grid ratio, requires additional mAs. 80. If a 12:1 ratio grid is used in a procedure that previously used no grid and 4 mAs, how much mAs should be used with the grid? a. 0.33 mAs. b. 1.25 mAs. c. 20 mAs. d. 48 mAs. ANS: C Adding a 12:1 ratio grid requires five times the mAs (based on the grid conversion factor).

Adding a grid or increasing grid ratio always requires more mAs. 81. Which of the following is the correct formula to determine how mAs should be adjusted when different ratio grids are used? a. mAs1/mAs2 = grid conversion factor2/grid conversion factor1. b. mAs1/mAs2 = grid conversion factor1/grid conversion factor2. c. mAs1/mAs2 = (grid conversion factor2)2/(grid conversion factor1)2.

d. mAs1/mAs2 = (grid conversion factor1)2/(grid conversion factor2)2.

ANS: B mAs1/mAs2 = grid conversion factor1/grid conversion factor2 is the formula used to adjust mAs based on grid changes.

82. If 25 mAs produce a diagnostic image when a 6:1 ratio grid is used, how much mAs should be used with a 12:1 ratio grid? a. 13 mAs. b. 15 mAs. c. 42 mAs. d. 50 mAs. ANS: C Using the formula, mAs1(25)/mAs2 = grid conversion factor1(3)/grid conversion factor2(5)X = 41.67 rounded to 42 mAs. 83. If 40 mAs produce a diagnostic image when a 12:1 ratio grid is used, how much mAs should be used with a 5:1 ratio grid? a. 6.7 mAs. b. 16 mAs. c. 96 mAs. d. 100 mAs. ANS: B Using the formula, mAs1(40)/mAs2 = grid conversion factor1(5)/grid conversion factor2(2).

X = 16 mAs. 84. Increased collimation results in 1. less volume of tissue irradiated 2. less exit radiation reaching the IR 3. less scatter reaching the IR a. less volume of tissue irradiated and less exit radiation reaching the IR only b. less volume of tissue irradiated and less scatter reaching the IR only c. less exit radiation reaching the IR and less scatter reaching the IR only d. less volume of tissue irradiated, less exit radiation reaching the IR, and less scatter reaching the IR ANS: D Increasing collimation results in less volume of tissue irradiated, less scatter produced and reaching the IR, and fewer photons reaching the IR. 85. If 8 mAs produces an appropriate exposure to the IR with a single-phase generator, a three-phase generator should a. use higher mAs b. use lower mAs c. keep the mAs the same d. none of these ANS: B

Because a three-phase generator produces radiation more efficiently, less mAs is required.

86. Increasing tube filtration a. increases the beam energy b. decreases radiographic contrast c. increases radiographic contrast d. increases the beam energy and decreases radiographic contrast ANS: D Although the effect is minimal, increased tube filtration results in a beam with higher energy photons, producing a lower-contrast image.

87. A compensating filter a. is used for specific anatomic areas b. produces a more uniform exposure to the IR c. requires an increase in mAs d. all of these ANS: D The purpose of a compensating filter is to produce a more uniform exposure to the IR for specific anatomy that has both thick and thin parts. This requires additional mAs and therefore additional patient exposure.

88. Which type of body habitus is the thickest, requiring higher exposure factors? a. Asthenic. b. Hypersthenic. c. Sthenic. d. Hyposthenic.

ANS: B The hypersthenic patient is the thickest, requiring an increase in exposure factors.

89. Which type of body habitus is the thinnest, requiring a reduction in exposure factors? a. Asthenic. b. Hypersthenic. c. Sthenic. d. Hyposthenic.

ANS: A The asthenic patient is the thinnest, requiring a decrease in exposure factors.

90. The ________________ body habitus accounts for approximately 50% of the adult population. a. asthenic b. hypersthenic c. sthenic d. hyposthenic ANS: C The sthenic body habitus accounts for approximately 50% of the adult population.

91. Typically, every increase in the thickness of the part being imaged of 4–5 cm requires the mAs a. to be halved b. to be doubled c. to be increased by 4–5% d. to be decreased by 4–5% ANS: B A thicker body part requires more radiation to produce the same exit radiation reaching the IR. The mAs should be doubled for every 4–5 cm increase in thickness.

92. If a small adult’s 24-cm thick abdomen requires 20 mAs to produce the appropriate level of radiation reaching the IR, how much mAs should be used with a larger patient whose abdomen measures 32 cm? a. 5 mAs. b. 20 mAs. c. 40 mAs. d. 80 mAs. ANS: D For every 4–5 cm the mAs needs to be doubled, so if the patient measures 28 cm, then 40 mAs is needed. Adding another 4 cm of thickness (to get to 32 cm) requires a doubling of 40 mAs to get to 80 mAs. 93. Which of the following results in increased scatter radiation reaching the IR? 1. Imaging a thicker body part 2. Using a lower kVp 3. Decreasing collimation a. imaging a thicker body part and using a lower kVp only b. imaging a thicker body part and decreasing collimation only c. using a lower kVp and decreasing collimation only d. imaging a thicker body part, using a lower kVp, and decreasing collimation ANS: B Opening up the x-ray beam field size (decreasing collimation) and imaging a thicker body part increases the amount of scatter radiation reaching the IR. 94. Imaging a thicker body part will result in a radiographic image with 1. lower contrast 2. increased magnification 3. more exposure to IR a. lower contrast and increased magnification only b. lower contrast and more exposure to IR only c. increased magnification and more exposure to IR only d. lower contrast, increased magnification, and more exposure to IR ANS: A A thicker body part will absorb more radiation (resulting is less exposure to the IR), produce more scatter radiation (lower contrast), and increase magnification due increased OID.

95. Assuming that all of these produce a similar amount of exit radiation, which of the following exposure factors minimize the radiation dose to the patient? a. 5 mAs @ 80 kVp. b. 10 mAs @ 70 kVp. c. 20 mAs @ 60 kVp. d. None of these.

ANS: A Exposure factors with higher kVp and lower mAs reduce radiation exposure to the patient.

96. Which of the following statements is true regarding practices to minimize patient exposure? a. The beam should be open as large as the image receptor being used. b. Grids should be used as much as possible to improve contrast. c. Use more mAs in digital imaging because the brightness is computer adjusted and it reduces noise. d. None of these.

ANS: D To minimize radiation exposure, the beam should be restricted as much as possible, the grid used only when needed, and use the least amount of mAs to produce a diagnostic image.

TRUE/FALSE 1. Higher kVp results in less absorption and more variation in the intensities exiting the patient.

ANS: F Higher kVp results in less absorption and less variation in the intensities exiting the patient.

2. More scatter radiation exits the patient when using higher kilovoltage.

ANS: T More Compton interactions producing scatter radiation that exits the patient occur when using higher kilovoltage.

3. At higher levels, kVp always decreases radiographic contrast.

ANS: T At higher levels, kVp does always decrease radiographic contrast.

4. Exposure factors that include higher kVp and lower mAs are recommended for digital imaging to reduce patient exposure.

ANS: T Higher kVp and lower mAs exposure factors are primarily recommended for digital imaging because the computer can adjust the contrast.

5. The small focal spot should be used with every examination to produce the best spatial resolution.

ANS: F Although the small focal spot size should be used whenever possible, there are examinations that require exposures so high (lateral lumbar spine for example) that the small cathode filament and focal spot cannot handle the heat and the large filament and focal spot must be used.

6. As the SID increases, the x-ray photons in the beam become less perpendicular to the object being imaged.

ANS: F As the SID increases the x-ray photons in the beam become more perpendicular to the object being imaged, decreasing size distortion.

7. A magnification factor of 1 cannot be achieved with radiographic imaging.

ANS: T A MF of 1 indicates there is absolutely no magnification, a situation that is not possible with radiographic imaging.

8. Shape distortion is only due to misalignment of the tube or part.

ANS: F Shape distortion is due to misalignment of the tube, part, or image receptor.

9. Tube angulation may result in elongation and reduced exposure to the IR. ANS: T Tube angulation may result in elongation (shape distortion) and reduced exposure to the IR if the SID is not adjusted.

10. The higher the grid ratio, the less efficient the grid is at absorbing scatter radiation.

ANS: F The higher the grid ratio, the more efficient the grid is at absorbing scatter radiation.

11. Increasing collimation means decreasing the x-ray field size.

ANS: T Increased collimation can also be said as decreased field size.

12. Increased beam restriction means decreased collimation.

ANS: F Increasing beam restriction is the same as increased collimation (resulting in a smaller field size).

13. A radiographic image of a part having different tissue types (such as bone and air) has lower contrast than that of a part with similar tissue types (such as liver and stomach).

ANS: F Images of anatomy that include different tissue types have higher contrast.

14. With digital imaging, it is impossible to overexpose the patient to radiation because the computer makes adjustments to the image.

ANS: F Overexposure with digital imaging is a real issue; overexposing the IR may result in an image with appropriate brightness (because of computer adjustment), but the patient has received an unnecessary radiation dose.

Chapter 12: Scatter Control Johnston/Fauber: Essentials of Radiographic Physics and Imaging, 4th Edition MULTIPLE CHOICE 1. For radiographic procedures, scatter radiation is primarily the result of a. photoelectric interaction b. pair production c. Compton interaction d. classical scattering ANS: C Scatter radiation is primarily due to Compton interactions between x-ray photons and matter.

2. The probability of a Compton interaction occurring is related to a. the energy of the x-ray photon b. the atomic number of the matter c. the mAs used d. all of these ANS: A The probability of a Compton interaction occurring is related to the energy of the photon.

Atomic number does not affect the probability of Compton interactions.

3. As compared with lower kVp exposures, when using high kVp will result in a scattered photon that is of ________ energy. a. lower b. higher c. the same d. half ANS: B Scattered photons produced at a high kVp have higher energy than those produced at a lower kVp.

4. The two major factors that affect the amount of scatter radiation produced and exiting the patient are a. kVp and mAs b. volume of tissue irradiated and mAs c. volume of tissue irradiated and kVp d. kVp and the use of a grid ANS: C The two major factors that affect the amount of scatter radiation produced and exiting the patient are the volume of tissue irradiated and kVp. Use of a grid does not affect the production of scatter.

5. The volume of tissue irradiated is determined by a. the area of collimation b. patient thickness c. field size d. all of these ANS: D The volume of tissue irradiated is determined by the area of collimation (field size) and patient thickness.

6. Which of the following produces the greatest amount of scatter radiation exiting the patient? a. Low kVp and small field size. b. Low kVp and large field size. c. High kVp and small field size. d. High kVp and large field size.

ANS: D A large field size (increasing the volume of tissue) and high kVp produces the greatest amount of scatter exiting the patient.

7. Restriction of the primary beam to the anatomy of interest a. reduces patient exposure b. is a major cause of repeated images c. reduces the amount of scatter produced d. reduces patient exposure and reduces the amount of scatter produced ANS: D Restricting the x-ray beam to the anatomy of interest reduces the amount of scatter produced (volume of tissue) as well as patient exposure (to the primary beam and scatter).

8. The unrestricted primary x-ray beam is a. round b. oval c. square d. rectangular ANS: A The unrestricted primary x-ray beam is round.

9. Increased collimation means: 1. Decreased field size 2. Increased field size 3. Less volume of tissue irradiated a. decreased field size and increased field size only b. decreased field size and less volume of tissue irradiated only c. increased field size and less volume of tissue irradiated only d. decreased field size, increased field size, and less volume of tissue irradiated ANS: B Increasing collimation means decreasing field size and reducing the volume of tissue irradiated.

10. Increasing collimation (decreasing the field size) results in a. increased scatter production and higher contrast b. increased scatter production and lower contrast c. decreased scatter production and higher contrast d. decreased scatter production and lower contrast ANS: C Increasing collimation results in decreased scatter production and higher contrast.

11. A significant amount of collimation will have what effect on the amount of radiation reaching the IR a. reduced b. increased c. the same d. none of these ANS: A Significant collimation reduces scatter radiation, reducing the overall amount of radiation reaching the IR. 12. When imaging the pelvis, changing from 14  17 collimation to 8  10 collimation requires a. an increase in kVp b. a decrease in kVp c. an increase in mAs d. a decrease in mAs ANS: C To compensate for the reduction of scatter radiation reaching the IR, the mAs must be increased.

13. The simplest type of beam-restricting device is a flat piece of lead with a hole in it called a(n) a. aperture diaphragm b. collimator c. cone or cylinder d. none of these ANS: A

The aperture diaphragm is a flat piece of lead with a hole in it.

14. A flat piece of lead with a hole in it that has a flange directed toward the patient is a(n) a. aperture diaphragm b. collimator c. cone or cylinder d. none of these ANS: C A cone or cylinder is a flat piece of lead with a hole in it that has a flange directed toward the patient.

15. The most effective and often used radiographic beam-restriction device is the a. aperture diaphragm b. collimator c. cone or cylinder d. none of these ANS: B The collimator is the most effective and often used radiographic beam-restriction device.

16. The most easily adjusted radiographic beam-restricting device is the a. aperture diaphragm b. collimator c. cone or cylinder d. none of these ANS: B The most easily adjusted radiographic beam-restricting device is the collimator because it has movable lead shutters.

17. The beam-restriction device that results in images with the most unsharpness around the edge of the image is the a. aperture diaphragm b. collimator c. cone or cylinder d. none of these ANS: A Because it is situated very close to the focal spot, using the aperture diaphragm results in an image with the most edge unsharpness.

18. The field shape produced by a collimator is a. round b. oval c. rectangular d. all of these ANS: C The field shape produced by a collimator is rectangular or square.

19. The beam-restriction device that includes a visible light projecting the x-ray field size onto the patient is the a. aperture diaphragm b. collimator c. cone or cylinder d. none of these ANS: B The collimator includes a light that makes the field size visible.

20. The automatic collimator a. automatically collimates to the anatomy of interest b. is required by law on all new equipment c. is seldom found on modern x-ray equipment d. automatically collimates to the IR size ANS: D When the IR is in the bucky tray, the automatic collimator adjusts the beam size to the size of the IR. 21. The purpose of automatic collimation is a. improved image quality b. limiting patient exposure c. increasing the cost of equipment d. none of these ANS: B By not allowing the x-ray beam to be larger than the IR, automatic collimation limits patient exposure.

22. The grid is located a. just below the x-ray tube window b. between the patient and IR c. just below the IR d. between the x-ray tube and the patient ANS: B The grid is found between the patient and IR. 23. Grids are typically used when: 1. Part measures more than 10 cm 2. At least 60 kVp needed for the exam 3. Part transmits little scatter radiation a. part measures more than 10 cm and at least 60 kVp needed for the exam only b. part measures more than 10 cm and part transmits little scatter radiation only c. at least 60 kVp needed for the exam and part transmits little scatter radiation only d. part measures more than 10 cm, at least 60 kVp needed for the exam, and part transmits little scatter radiation ANS: A

Grids are typically only used when the part measures more than 10 cm and transmits a lot of scatter, and 60 kVp is needed for the exam.

24. Grids absorb scatter radiation because the scattered photons a. have high energy b. have extremely low energy c. travel in the same direction as the transmitted photons d. travel at an angle to the transmitted photons ANS: D Because scattered photons change direction after the Compton interaction, they no longer travel in the same direction as the transmitted photons and will be absorbed by the lead strips in the grid.

25. The material between the lead strips of a grid must be a. radiopaque b. radiolucent c. made of lead d. made of tungsten ANS: B The material between a grid’s lead strips must be radiolucent, allowing transmitted photons to pass through to the IR. 26. The number of lead lines per inch or centimeter is the grid a. number b. ratio c. frequency d. rating ANS: C Grid frequency is the number of lead strips per inch or centimeter.

27. The relationship between the height of the lead strips and the distance between them is the grid a. number b. ratio c. frequency d. rating ANS: B The grid ratio is the relationship between the height of the lead strips and the distance between them.

28. Which of the following is the formula used to determine grid ratio? a. Grid ratio = distance between the lead strips/height of the lead strips. b. Grid ratio = width of the lead strips/height of the lead strips. c. Grid ratio = height of the lead strips/distance between the lead strips. d. Grid ratio = height of the lead strips/width of the lead strips.

ANS: C

Grid ratio = h/D, or height of the lead strips / distance between the lead strips.

29. What is the grid ratio if the lead strips are 2.4 mm high, 0.02 mm wide, and 0.3 mm apart? a. 5:1. b. 61. c. 8:1. d. 12:1.

ANS: C Using the formula, grid ratio = h (2.4) / D (0.3), the grid ratio is 8:1.

30. What is the grid ratio if the grid frequency is 100 lines/inch, and the lead strips are 1.8 mm high and 0.3 mm apart? a. 5:1. b. 6:1. c. 8:1. d. 12:1.

ANS: B Using the formula, grid ratio = h (1.8) /D (0.3), the grid ratio is 6:1.

31. Using a higher the grid ratio will 1. increase spatial resolution 2. increase scatter clean up 3. increase radiographic contrast a. increase spatial resolution and increase scatter clean up only b. increase spatial resolution and increase radiographic contrast only c. increase scatter clean up and increase radiographic contrast only d. increase spatial resolution, increase scatter clean up, and increase radiographic contrast ANS: C The higher the grid ratio, the better the scatter clean up and the higher the radiographic contrast. Spatial resolution is not affected when using a grid.

32. A linear grid a. has lead strips that cross each other b. has lead strips that travel in one direction c. is the most common type of radiographic grid d. has lead strips that travel in one direction and is the most common type of radiographic grid ANS: D The linear grid, with lead strips or lines that go in one direction only, is the most commonly used grid.

33. Linear grids are the most commonly used because a. they are very inexpensive b. the tube can be angled in all directions c. the tube can be angled in the direction of the lead lines d. all of these

ANS: C As compared with crossed grids, which do not allow any tube angulation, linear grids are most commonly used because the tube can be angled in the direction of the lead lines.

34. A grid with lead strips that run parallel to each other is a a. parallel grid b. focused grid c. nonfocused grid d. parallel grid and nonfocused grid ANS: D A grid with strips that run parallel to each other is a parallel or nonfocused grid.

35. A focused grid’s lead strips are angled to match a. the anode angle b. the angle of the x-rays in the primary beam c. the angle of the scattered photons leaving the patient d. the tube angulation ANS: B A focused grid’s lead strips are angled to match the angle of the photons in the primary beam as they diverge.

36. The distance between the grid and the convergent line is the a. convergent line b. focal distance c. convergent point d. focal range ANS: B The focal distance is the distance between the grid and the convergent line.

37. The recommended SIDs that can be used with a focused grid is the a. convergent line b. focal distance c. convergent point d. focal range ANS: D The focal range is a range of SIDs that will work with a specific grid.

38. The location in space where all the lines extended from the lead strips meet is the a. convergent line b. focal distance c. convergent point d. focal range ANS: C The point in space where the imaginary lines projected from the lead strips meet is the convergent point.

39. If all the convergent points were connected along the length of the grid they would form the a. convergent line b. focal distance c. convergent point d. focal range ANS: A The convergent line is an imaginary line connecting the convergent points along the length of the grid.

40. An IR that has a grid permanently attached to its front surface is a a. wafer grid b. bucky c. grid cap d. grid cassette ANS: D An IR that has a grid permanently attached to its front surface is a grid cassette.

41. The type of grid that is usually taped to the front of an IR is the a. wafer grid b. bucky c. grid cap d. grid cassette ANS: A The wafer grid is typically taped onto the front of an IR. 42. The type of grid that is permanently mounted but includes a channel for the IR to slide into is the a. wafer grid b. bucky c. grid cap d. grid cassette ANS: C The grid cap is constructed to allow a cassette to slide into place behind the grid.

43. The grid is part of a device located just below the tabletop that also includes a tray to hold the IR. This device is the a. wafer grid b. bucky c. grid cap d. grid cassette ANS: B The bucky includes a tray to hold the IR and a grid located above the IR. 44. During the exposure, the grid in the bucky a. reciprocates b. moves slightly from side to side

c. moves slightly from top to bottom d. reciprocates and moves slightly from side to side ANS: D The grid in the bucky moves slightly from side to side (reciprocates) during the x-ray exposure.

45. The reason the grid in the bucky reciprocates is to a. absorb more scatter radiation b. allow more transmitted radiation to reach the IR c. blur out the grid lines d. all of these ANS: C The only reason for having the grid move slightly during the exposure is to blur out the grid lines.

46. A short-dimension grid a. is typically used for most examinations b. has lead strips running perpendicular to the long axis of the grid c. has lead strips running parallel to the long axis of the grid d. is typically used for most examinations and has lead strips running perpendicular to the long axis of the grid ANS: B A short dimension grid has lead strips running perpendicular to the long axis of the grid, but is only used with special situations.

47. To compensate for the reduction of radiation reaching the IR when a grid is used, ___________ is typically increased. a. SID b. kVp c. mAs d. OID ANS: C When a grid is added or a change is made to a higher ratio grid, mAs is typically increased to compensate for the reduction in radiation reaching the IR. 48. To determine the amount of mAs to be increased when making grid changes, it is necessary to use the GCF, also known as the a. grid constancy figure b. grid compensatory factor c. general compensation factuals d. grid conversion factor ANS: D The grid conversion factor (GCF) is used to make adjustments in mAs when grid changes are made.

49. The formula for the grid conversion factor is

a. b. c. d.

GCF = mAs without grid/mAs with grid GCF = mAs with grid – mAs without grid GCF = mAs with grid/mAs without grid GCF = mAs without grid – mAs with grid

ANS: C The formula for grid conversion factor is GCF = mAs with grid/mAs without grid.

50. If a tabletop exposure (no grid) of a humerus requires 2.5 mAs, how much mAs is needed if a 12:1 ratio grid is added? a. 0.21 mAs. b. 0.5 mAs. c. 12.5 mAs. d. 30 mAs. ANS: C Based on the grid conversion formula, GCF (5) = mAs with the grid/mAs without the grid (2.5), resulting in 12.5 mAs. 51. If a shoulder examination is done using an 8:1 ratio grid and 12 mAs, how much mAs should be used if the examination must be done tabletop (no grid)? a. 1.5 mAs. b. 3 mAs. c. 48 mAs. d. 96 mAs. ANS: B Based on the grid conversion formula, GCF (4) = mAs with the grid (12)/mAs without the grid, resulting in 3 mAs. 52. The formula to use when changing from one grid to another is a. mAs1/mAs2 = GCF2/GCF1 b. mAs1/mAs2 = (GCF2)2/GCF1)2 c. mAs1/mAs2 = GCF1/GCF2 d. mAs1/mAs2 = (GCF1)2/GCF2)2 ANS: C To change from one grid ratio to another, it is necessary to use the formula mAs1/mAs2 = GCF1/GCF2.

53. If 24 mAs and a 6:1 grid results in an excellent radiographic image, how much mAs is needed with a 5:1 grid? a. 16 mAs. b. 20 mAs. c. 28.8 mAs. d. 36 mAs. ANS: A Using the formula mAs1/mAs2 = GCF1/GCF2, 24/X = 3/2, 3X = 48, x = 16.

54. If 22 mAs and a 5:1 grid results in an excellent radiographic image, how much mAs is needed with a 12:1 grid? a. 8.8 mAs. b. 9.2 mAs. c. 52.8 mAs. d. 55 mAs. ANS: D Using the formula mAs1/mAs2 = GCF1/GCF2, 22/X = 2/5, 2X = 110, X = 55.

55. When the x-ray beam is angled across the lead strips you have _________ cutoff. a. upside-down focused b. off-level c. off-center d. off-focus ANS: B Angling the beam in a direction that is not in line with the direction of the lead strips will produce off-level cutoff.

56. When the x-ray beam is not lined up with the center of the grid you have _________ cutoff. a. upside-down focused b. off-level c. off-center d. off-focus ANS: C Failure to line up with the midline of the grid will result in off-center grid cutoff.

57. When the SID is above or below the focal range you have _________ cutoff. a. upside-down focused b. off-level c. off-center d. off-focus ANS: D Off-focus grid cutoff occurs when the SID is outside of the grid’s focal range.

58. The cutoff that appears as a strip of appropriate exposure in the middle of the image and a significant loss of exposure on both sides is a. upside-down focused b. off-level c. off-center d. off-focus ANS: A A focused grid upside down results in cutoff that appears as a strip of appropriate exposure in the middle of the image and a significant loss of exposure on both sides.

59. Absorption of transmitted photons by a grid caused by misalignment is a. grid focus

b. grid reduction c. grid cutoff d. transmission reduction ANS: C Grid cutoff is the absorption of transmitted photons by a grid caused by a misalignment issue.

60. Grid cutoff results in a. reduced exposure to the IR b. may increased patient exposure c. possible increased in quantum noise d. all of these ANS: D In that grid cutoff means that transmitted radiation is not reaching the IR, the reduced exposure can result in the increased presence of quantum noise caused by insufficient photons producing the image. Patient exposure is not increased unless a repeat image is necessary.

61. The Moiré effect is also known as the _________ pattern. a. cheetah b. giraffe c. lion d. zebra ANS: D The Moiré effect is also known as the zebra pattern.

62. The Moiré effect can be caused by a. using a grid with a frequency similar to the CR laser scanning frequency b. using a focused grid upside down in the bucky c. using a grid cassette in the bucky d. using a grid with a frequency similar to the CR laser scanning frequency and using a grid cassette in the bucky ANS: D The Moiré effect can be caused by using a grid with a frequency similar to the CR laser scanning frequency as well as using a grid cassette in the bucky.

63. Which of the following grids provides excellent scatter clean-up but requires close to perfect alignment to avoid cutoff? a. 5:1 ratio grid. b. 8:1 ratio grid. c. 12:1 ratio grid. d. 16:1 ratio grid.

ANS: D A 16:1 grid provides excellent scatter clean-up, but requires close to perfect alignment to avoid cutoff.

64. The only grid cutoff error that can occur with a parallel grid is a. upside-down focused b. off-level c. off-center d. off-focus ANS: B Because the lead strips are not angled, the only grid cutoff error that affects a parallel grid is being off-level.

65. Which of the following factors need to be considered when determining the grid to be used for a specific examination? a. Patient exposure. b. Level of contrast improvement needed. c. Focal range. d. All of these.

ANS: D Since grid selection affects the level of contrast improvement and the amount of patient exposure, these factors must be considered. A grid with the focal range that includes the SID to be used for an examination is a necessity to avoid cutoff.

66. The air gap technique is based on creating a gap by increasing the ______ a. SID b. SOD c. OID d. mAs ANS: C The air gap technique is based on creating a gap by increasing the OID. 67. The air gap technique reduces the amount of scatter radiation reaching the IR because a. more scatter is absorbed in the air b. less scatter is produced in the patient c. more scatter misses the IR d. less scatter misses the IR ANS: C Because of the air gap (increased OID), more scatter misses the IR because the scattered photons leave the patient at angle.

68. To compensate for some of the loss of spatial resolution due to the large OID used with the air gap technique, the ________ should be increased as well. a. grid ratio b. SID c. focal spot size d. grid ratio and SID ANS: B

To compensate for some of the loss of spatial resolution caused by the large OID used with the air gap technique, the SID should be increased as well. There is no grid when the air gap technique is used.

69. For a lateral lumbar spine, placing a lead shield on the table behind the area of exposure a. reduces the scatter exiting the patient b. reduces the scatter reaching the IR c. increases the scatter reaching the IR d. reduces the scatter exiting the patient and reduces the scatter reaching the IR ANS: B The lead strip behind the patient has no effect on how much scatter was produced and exits the patient. It does absorb some of the scatter produced and keeps it from reaching the IR. 70. Anyone in the room during an exposure must wear a lead apron to absorb the scatter radiation produced in the a. table b. patient c. air d. beam ANS: B The scatter produced in the patient is the primary source of exposure to anyone in the room when the exposure is made.

TRUE/FALSE 1. Scatter radiation is of no value in radiographic imaging.

ANS: T Scatter radiation adds unwanted and useless exposure to the IR and increases patient exposure.

2. Beam-restricting devices are located just above the tube housing.

ANS: F Beam-restricting devices are located just below the tube housing.

3. It is safe to assume that the x-ray beam field size and the light field size are exactly the same.

ANS: F Although the light field and the x-ray field should be very similar, this may not always be true and quality control tests need to be done on a regular basis.

4. When using automatic collimation, collimating to the anatomy of interest is an unnecessary waste of time; the beam is already collimated.

ANS: F

Although it is restricted to IR size, automatic beam collimation does not excuse the radiographer from limiting the field size to the anatomy of interest.

5. The grid limits the amount of scatter radiation produced in the patient.

ANS: F The grid has no effect on the amount of scatter produced; it only limits how much of the scatter reaches the IR. 6. Grids absorb almost all of the scatter radiation and allow all of the transmitted photons to pass through to the IR. ANS: F Grids are not able to be so effective that all the transmitted photons pass through to the IR and almost all scattered photons are absorbed.

7. Focused grids are more effective than parallel grids in absorbing scatter radiation and allowing transmitted radiation to reach the IR. ANS: T Because of the angulation of the lead strips to match the beam divergence, focused grids are more effective than parallel grids in absorbing scatter radiation and allowing transmitted radiation to reach the IR. 8. Adding a grid or increasing grid ratio results in an increase in patient dose.

ANS: T Adding a grid or increasing grid ratio requires an increase in mAs, increasing patient exposure.

9. To overcome the decrease in sharpness caused by the air gap technique, an increase in SID is required.

ANS: T To overcome the decrease in image sharpness caused by the air gap technique, an increase in SID is required.

10. The grid used for the air gap technique should be high frequency.

ANS: F There is no grid used with the air gap technique.

Chapter 13: Exposure Technique Selection Johnston/Fauber: Essentials of Radiographic Physics and Imaging, 4th Edition MULTIPLE CHOICE 1. The primary purpose of AEC is to a. reduce patient exposure

b. control the amount of radiation reaching the IR c. control image contrast d. eliminate the need for the radiographer to think critically ANS: B Although AEC can result in reduced patient exposure, its purpose is to control the amount of radiation reaching the IR. 2. AEC controls the amount of radiation reaching the IR by controlling the a. time of exposure b. mA c. kVp d. none of these ANS: A AEC determines the length of exposure and therefore the mAs used to produce an image.

3. The predetermined level of radiation exposure needed to terminate the exposure when using AEC is set by a. the radiologist b. the technologist c. the equipment service personnel d. the manufacturer ANS: C Although it may initially be set by the manufacturer, the service personnel make the final determination following testing.

4. The two types of detectors used in AEC are a. image intensifiers and phototimers b. phototimers and ionization chambers c. ionization chambers and Geiger counters d. direct and indirect capture ANS: B Phototimers and ionization chambers are the two types of AEC detectors.

5. The original type of AEC detector was the a. ionization chamber b. Geiger counter c. phototimer d. none of these ANS: C The early AEC systems used phototimers.

6. Today’s AEC system is most likely use a(n) a. ionization chamber b. Geiger counter c. phototimer d. none of these

ANS: A Modern AEC systems use ionization chambers.

7. There is typically a configuration of at least ______ AEC detectors in place a. 2 b. 3 c. 4 d. 5 ANS: B Three radiation-measuring detectors are commonly used.

8. The device that converts the light energy to an electrical signal is the a. photodiode b. photocathode c. photomultiplier tube d. photodiode and photomultiplier tube ANS: D The older photomultiplier tube and the new photodiode both convert light energy to an electrical signal.

9. Phototimer AEC detectors are usually exit-type devices because a. the x-rays must exit the patient to get to the detectors b. the x-rays must exit the grid to get to the detectors c. the x-rays must exit the image receptor to get to the detectors d. the x-rays must exit the table to get to the detectors ANS: C Exit-type detectors are located behind the IR, requiring the radiation to exit the IR before interacting with the detectors.

10. An ionization chamber is a hollow cell that contains a. water b. lead c. tungsten d. air ANS: D Ionization chambers typically contain air or some other gas.

11. Ionization chamber AEC detectors are entrance-type devices because the x-rays must a. enter the detector after passing through the IR b. enter the detector before entering the patient c. enter the detector before getting to the IR ANS: C The entrance-type detector is positioned immediately in front of the IR. 12. When the x-rays enter the ionization chamber a. atoms become ions

b. atoms give up electrons c. electrons and positive ions become electricity d. All of these ANS: D As the x-rays expose the air in the ionization chamber, electrons are knocked off of atoms, producing ions. This creates an electric charge.

13. Once the electric charge from an AEC detector reaches a preset point a. the patient can leave b. the kVp will be reduced to 0 c. the exposure will stop d. All of these ANS: C Once a predetermined electric charge has been reached (resulting from a set amount of radiation being absorbed by the detector), the exposure is terminated.

14. The shortest exposure time that a radiographic tube with AEC can have is the a. mAs readout b. back-up time c. minimum response time d. density controls ANS: C Minimum response time is the minimum time for the exposure to start and then end.

15. The maximum length of time that an exposure lasts when using AEC is the a. mAs readout b. back-up time c. minimum response time d. exposure adjustment ANS: B Whether set by the radiographer or inherent in the equipment, the back-up time is the longest time the exposure can go on when using AEC. 16. The actual amount of exposure used for an image when AEC is being used is the a. mAs readout b. back-up time c. minimum response time d. density controls ANS: A It may appear for a brief time, but the mAs readout shows exactly how much mAs was used to produce an image.

17. These can be adjusted to increase or decrease the amount of radiation needed to terminate the exposure using AEC a. mAs readout b. back-up time

c. minimum response time d. exposure adjustment ANS: D Exposure adjustment (density controls) is a mechanism to change the preset exposure time by a specific amount.

18. When using AEC, the kVp a. is set higher than usual b. is set lower than usual c. is set as appropriate for the study d. is determined by the AEC system ANS: C kVp must be set at the level appropriate for the examination.

19. Using a higher kVp during an AEC examination results in a. a longer exposure time b. a shorter exposure time c. increased patient exposure d. a shorter exposure time and increased patient exposure ANS: B Higher kVp allows more of the radiation to penetrate the part and reach the AEC detector, reducing the time of exposure. With higher kVp, patient exposure is reduced.

20. Using a lower mA station during an AEC examination results in a. reduced patient exposure b. increased exposure time c. decreased exposure time d. reduced patient exposure and increased exposure time ANS: B Lowering the mA does not affect the mAs (and patient exposure) but does result in an increase in exposure time so the detector has time to reach its preset level.

21. In imaging a child, a short exposure time is needed for an examination using AEC. The mA setting should a. be low b. be mid-range c. be high d. It doesn’t matter—the AEC device will control the exposure time ANS: C The mA should be set on the high side to that the exposure time is reduced.

22. The AEC minimum response time may pose a problem when a. a long exposure time is needed b. the patient can easily hold still for the exposure c. the patient is unable to hold still for the exposure d. All of these

ANS: C If the minimum response time is too long to produce a sharp image of the patient who is unable to hold still, it may be better to not use the AEC. 23. The reason to have a back-up time is to a. protect the patient from excessive exposure b. make sure the right mAs is used c. keep the tube from excessive heat-loading d. protect the patient from excessive exposure and keep the tube from excessive heat-loading ANS: D The back-up time is a safety mechanism to keep both the patient and the tube from excessive exposures when the AEC is not operating correctly.

24. If the back-up time is set by the equipment, the exposure time should end when the mAs reaches a. 100 b. 400 c. 600 d. 1000 ANS: C 600 mAs is the most a patient should receive during an AEC examination. This is far, far beyond the mAs appropriate for producing a radiograph, but if the AEC is not working properly, it is at least a stopping point.

25. When the radiographer has the opportunity to set the back-up time, it should be approximately _____________ of the expected exposure time a. 50% b. 150% c. 250% d. 600% ANS: B When the radiographer has the opportunity to set the back-up time, it should be approximately 150% of the expected exposure time.

26. If the back-up time is shorter than the actual exposure time needed for a properly exposed image a. the image receptor will be overexposed b. the image receptor will be underexposed c. the exposure will be appropriate d. the x-ray tube will overheat ANS: B If the back-up time is too short, the exposure will terminate prematurely, resulting in the image receptor being underexposed.

27. The detector(s) selected for an image should be a. the center detector because that is where the central ray is going

b. the two or three detectors that are covered by the part being imaged c. the ones that are superimposed by the anatomic structure of interest d. all three so that the system can work properly ANS: C It is very important that the detectors that are selected are directly superimposed by the specific anatomy of interest. For example, if the spine is of interest, the center detector alone should be selected because it is the only one superimposed by the spine.

28. If the radiographer is doing an upright chest using AEC but the table bucky is selected a. the back-up time will be reached b. the patient will be exposed to excessive radiation c. the IR will be overexposed d. All of these ANS: D If the AEC detectors in the table are waiting to get enough radiation to terminate the exposure, they will never get to that point if the tube is directed at the upright bucky. The result will be a very long exposure (possibly until the back-up time is reached) and excessive mAs. 29. Which of the following is the most critical aspect of successfully performing an examination using AEC? a. Centering the anatomy of interest over the detector b. Using the right mA c. Using the right kVp d. Setting the best back-up time ANS: A The AEC detector will terminate the exposure based on the set amount of radiation exiting the anatomy directly in front of it. If that anatomy is not the anatomy of interest, the exposure will be wrong.

30. If the patient for an AP thoracic spine image is centered so that the spine is not over the top of the detector a. the IR exposure will be excessive b. the IR exposure will be insufficient c. the mAs will compensate d. the image will be too light ANS: B If the spine is not over top of the detector, then the lung is, allowing the radiation to easily penetrate and the exposure will be terminated quickly. The result is that the lungs are correctly exposed but the spine is underexposed, however, the brightness will be computer adjusted.

31. If the AEC detectors are not completely covered by the part being imaged and are exposed to primary radiation a. the patient will be overexposed b. the exposure will terminate very quickly c. the IR will be exposed to the appropriate amount of radiation

d. none of these ANS: B When the detector is exposed to primary radiation, the preset amount of exposure needed to terminate the exposure is reached almost immediately, resulting in a very short exposure time and the IR being underexposed.

32. When your patient for an AP abdomen is very heavy, using AEC results in a. underexposure to the IR b. overexposure to the IR c. appropriate exposure to the IR d. increased image contrast ANS: C Assuming all is set up correctly, AEC should not terminate the exposure until the present level is reached, regardless of the patient’s thickness.

33. When there is a significant amount of scatter radiation reaching the detector, the exposure time will a. be shorter than needed b. be appropriate c. be longer than needed d. reach the backup mAs ANS: A In that the detector cannot differentiate scatter from transmitted radiation, the excess scatter will cause the exposure to terminate too soon.

34. When using AEC, destructive pathologic conditions can cause a. overexposure b. decreased absorption c. excessive overexposure d. increased absorption ANS: B Destructive pathologic conditions can cause decreased absorption of the area of radiographic interest.

35. Anatomically programmed technique is a system in which the radiographer selects a. AEC settings b. mAs and kVp settings c. preset settings based on the part being imaged d. preset settings based on the size of the patient ANS: C Anatomically programmed technique presents the radiographer with preset exposure factors based on the anatomy being imaged.

36. AEC devices should be calibrated a. when the unit is first installed b. at regular intervals

c. when there is a change in the type of IR being used d. All of these ANS: D It is critical that AEC devices operate accurately, to ensure consistent quality and protect patients from unnecessary exposure. Initial and routine calibration, along with special situations such as changing IR type, are key to AEC devices working properly.

37. The Quality control test of the AEC device to see that exposures for a given set of factors and selected detector result in mR readings within 5%, is known as: a. reproducibility; b. linearity c. reciprocity d. accuracy ANS: A The testing to see that the AEC device has reproducibility of exposures for a given set of factors and selected detector should result in mR readings within 5%.

38. When imaging a patient’s right hip, which detector(s) should be activated? a. right detector only b. right and left detector c. center detector only d. all three detectors ANS: C The patient’s right hip should be positioned over the center detector.

39. When using AEC with digital imaging, errors resulting in overexposure to the IR result in a. images that have insufficient brightness b. images with appropriate brightness c. overexposed patients d. images with appropriate brightness and overexposed patients ANS: D Typically the computer can compensate for overexposure to the digital IR, displaying an image with appropriate brightness. However, the overexposure to the patient cannot be adjusted and it is up to the radiographer to maintain vigilance in using the AEC system properly.

40. With digital imaging using AEC, if an AP thoracic spine is imaged using an outside instead of center detector, the brightness in the area of interest will be _________________ and the patient will be underexposed. (Assume all other factors are appropriate.) a. gray b. light c. dark d. appropriate ANS: D

Because the detector is terminating the exposure when enough radiation exits the lungs, the exposure is insufficient (patient underexposed in spine); however, because of computer adjustment, the brightness in the spine will be appropriate.

41. With digital imaging using AEC, if an AP thoracic spine is imaged with the central ray passing 3 inches to the right of the spine, the right lung exposure is _________________.

(Assume all other factors are appropriate.) a. excessive b. insufficient c. sufficient d. too long ANS: C If the centering is 3 inches to the right of the spine, the lungs are superimposing the detector.

Because the detector is terminating the exposure when enough radiation exits the lungs, the exposure is sufficient to image the lungs but not the thoracic spine.

42. With digital imaging using AEC, if an AP thoracic spine is imaged with the exposure adjustment changed from 0 to +2, the exposure time is _______________________.

(Assume all other factors are appropriate.) a. longer b. shorter c. unchanged d. insufficient ANS: A The exposure time is longer as the detector takes longer to detect 50% more photons and the patient receives additional dose.

43. With digital imaging using AEC, if an AP thoracic spine is imaged using 110 kVp instead of 80 kVp, the patient exposure is _______________________. (Assume all other factors are appropriate.) a. higher b. unchanged c. lower d. excessive ANS: C The increase in kVp results in lower mAs (because the exposure time is shortened for the preset level to be reached) and therefore reduces patient exposure.

44. With digital imaging using AEC, what factor is increased if an AP thoracic spine is imaged using an outside instead of the center detector? (Assume all other factors are appropriate.) a. subject contrast b. brightness c. spatial resolution d. quantum noise ANS: D

Because the detector is terminating the exposure when enough radiation exits the lungs, the exposure is too short. The brightness will be appropriate as a result of computer rescaling, but there will be additional noise because less mAs was used to create the image.

45. With digital imaging using AEC, what factor is decreased if an AP thoracic spine is imaged using an outside instead of the center detector? (Assume all other factors are appropriate.) a. exposure time b. increased; subject contrast c. spatial resolution d. brightness ANS: A Because the detector is terminating the exposure when enough radiation exits the lungs, the exposure time is decreased.

46. Pre-established guidelines used by the radiographer to select standardized manual or AEC exposure factors for each type of radiographic examination are called a. anatomical programming b. exposure technique charts c. mA reciprocity d. quantum noise detectors ANS: B Exposure technique charts are pre-established guidelines used by the radiographer to select standardized manual or AEC exposure factors for each type of radiographic examination.

47. ___________________ identify optimal kVp values and alter the mAs for variations in part thickness a. Fixed kVp–variable mAs b. Variable kVp-variable mAs c. Fixed kVp-Fixed mAs d. Variable kVp-fixed mAs ANS: A Fixed kVp–variable mAs technique charts identify optimal kVp values and alter the mAs for variations in part thickness.

48. Which of the following patient conditions require a change in technical factors?

1. Plaster casts 2. Obesity 3. Additive disease a. plaster casts and obesity only b. plaster casts and additive disease only c. obesity and additive disease only d. plaster casts, obesity, and additive disease ANS: D A change in technical factors is required for patient conditions, such as additive diseases, plaster casts, or if the patient is obese.

TRUE/FALSE 1. AEC and phototiming are accurate terms that can be used in place of automatic exposure control.

ANS: F Although both terms have been used in place of AEC, phototiming is not an accurate term for the current process.

2. When using AEC, the exit radiation is converted into an electrical signal.

ANS: T The radiation that exits the patient is detected and converted to an electrical signal.

3. Phototimer AEC devices convert the radiation to light energy.

ANS: T Phototimer devices absorb the radiation and produce light energy using a fluorescent screen.

4. The amount of electric charge produced in an AEC detector depends on the amount of radiation absorbed by the detector.

ANS: T The electric charge produced is in proportion to the amount of x-rays absorbed by the detector.

5. The mAs readout is of little value because there is no need to set manual exposure factors.

ANS: F Taking note of the mAs readout can be invaluable in gaining an understanding of appropriate exposure factors, making exposure adjustments, and dealing with situations when AEC is unavailable.

6. Using the plus or minus exposure adjustments should be done frequently to produce quality images with AEC. ANS: F If the AEC system is working properly, there should seldom be a need to adjust the density controls.

7. If an AEC has three detectors, only one of the three can be selected for an exposure.

ANS: F As part of the procedure, the radiographer is responsible for selecting the one, two, or all three detectors appropriate for the image.

8. The AEC system will automatically adjust to changing image receptor types.

ANS: F When changing to a different image receptor type, the AEC detectors must be recalibrated, a time-consuming procedure.

9. It is safe to assume that anatomically programmed techniques have been developed to produce the best quality images with the least amount of exposure.

ANS: F Although most anatomically programmed techniques are appropriate, the radiographer should never assume that what is set is the best; the radiographer should carefully evaluate anatomically programmed techniques to see if they need to be adjusted because of the patient, a pathologic condition, or just good practice.

10. Collimation is not required when AEC is used.

ANS: F The size of the x-ray field is a factor when AEC systems are used because the additional scatter radiation produced by failure to accurately restrict the beam may cause the detector to terminate the exposure prematurely.

11. Different radiographic projections and patient positions of the same anatomic part do not require modification of the exposure factors.

ANS: F Different radiographic projections and patient positions of the same anatomic part often require modification of the exposure factors.

Chapter 14: Film-Screen Imaging Johnston/Fauber: Essentials of Radiographic Physics and Imaging, 4th Edition MULTIPLE CHOICE 1. Factors evaluated for a quality radiographic film image include: 1. Sharpness 2. Contrast 3. Density a. sharpness and contrast only b. sharpness and density only c. constrast and density only d. sharpness, contrast, and density ANS: D The visibility (density and contrast) of the anatomic structures and the accuracy of the structural lines recorded (sharpness) determine the overall quality of the radiographic image.

2. In film-screen imaging, film is the medium for a. processing b. display c. image acquisition

d. All of these ANS: D Film is the medium for image acquisition, processing, and display.

3. The material in the film emulsion that is responsive to radiation or light is the a. blue dye b. emulsion c. gelatin d. silver halide ANS: D Silver halide is the material in the film emulsion that responds to radiation or light.

4. The most important layer for creating the image; the layer of the film that is sensitive to light or radiation is the a. blue dye b. emulsion c. gelatin d. silver halide ANS: B The emulsion is the layer of the film that is sensitive to light or radiation.

5. The material in the film emulsion that supports the silver halide crystals is the a. blue dye b. emulsion c. gelatin d. silver halide ANS: C In the film emulsion, the silver halide crystals are suspended in gelatin.

6. The material in the base that reduces eye strain is the a. blue dye or tint b. emulsion c. gelatin d. silver halide ANS: A Blue dye or tint is added to the base of radiographic film to reduce eye strain when the radiograph is displayed on the viewbox.

7. The latent film image is the image a. before exposure to exit radiation b. after exposure to exit radiation and before chemical processing c. after chemical processing d. in digital imaging only ANS: B The latent film image is the image after exposure to exit radiation and before chemical processing.

8. The manifest image a. is the image after exposure to exit radiation and before chemical processing b. is the image after chemical processing c. is the radiographic image d. is the image after chemical processing and is the radiographic image ANS: D The manifest image, also referred to as the radiographic image, is the image that exists on film after exposure and chemical processing.

9. The Gurney-Mott theory explains the process of a. chemical processing b. x-ray production c. latent image formation d. histogram analysis ANS: C The Gurney–Mott theory explains the process of latent image formation.

10. According to the Gurney–Mott theory, _____________ are attracted to the sensitivity specks in the silver halide crystal after exposure to light or radiation a. electrons b. metallic silver c. gelatin d. silver ions ANS: D According to the Gurney–Mott theory, silver ions are attracted to the sensitivity specks in the silver halide crystal after exposure to light or radiation.

11. The latent image center is a. several sensitivity specks with silver ions attached b. what will become metallic silver after chemical processing c. what will become radiographic density after chemical processing d. All of these ANS: D The latent image center is several sensitivity specks that have silver ions attached. After chemical processing, this will become metallic silver, contributing to the radiographic density.

12. How sensitive the film is to x-rays or light is the a. film speed b. film latitude c. film contrast d. spectral sensitivity ANS: A Film speed is the degree to which the emulsion is sensitive to x-rays or light.

13. What factors affect the film speed?

1. Color of the silver halide crystals 2. Size of the silver halide crystals 3. Number of silver halide crystals a. color of the silver halide crystals and size of the silver halide crystals only b. color of the silver halide crystals and number of silver halide crystals only c. size of the silver halide crystals and number of silver halide crystals only d. color of the silver halide crystals, size of the silver halide crystals, and number of silver halide crystals ANS: C Increasing the size or number of silver halide crystals increases film speed.

14. This is the film characteristic regarding how gray or black and white the image is a. film speed b. film latitude c. film contrast d. spectral sensitivity ANS: C Film contrast refers to the ability of radiographic film to provide a certain level of image contrast (black, white, and shades of gray).

15. The range of radiation exposures that will provide diagnostic optical densities is a. film speed b. film latitude c. film contrast d. spectral sensitivity ANS: B Latitude is the range of radiation exposures that can provide diagnostic optical densities.

16. The color of light to which a particular film most responds to is a. film speed b. film latitude c. film contrast d. spectral sensitivity ANS: D The color of light that a film most responds to (blue, green, etc.) is spectral sensitivity.

17. Orthochromatic film is sensitive to a. blue light b. green light c. red light d. any light ANS: B Orthochromatic film is sensitive to green light.

18. The color of light produced by an intensifying screen is the a. spectral matching b. spectral sensitivity c. spectral emission d. spectral sensitivity and spectral emission ANS: C Spectral emission is the color of light produced by an intensifying screen.

19. Using blue-sensitive film with green light–emitting screens is an example of poor a. spectral matching b. spectral sensitivity c. spectral emission d. spectral sensitivity and spectral emission ANS: A Spectral matching is correctly matching the color of light emitted by screens and the color of light to which the film is sensitive. Appropriate spectral matching is using blue-sensitive film with blue light-emitting screens.

20. Intensifying screens are used to a. reduce patient exposure b. convert light photons to x-ray photons c. convert x-ray photons to light photons d. reduce patient exposure and convert x-ray photons to light photons ANS: D Intensifying screens convert x-rays to light, resulting in an intensification of the x-ray photons. Their primary purpose is to reduce patient exposure.

21. The most important c of the intensifying screen is the a. base b. phosphor layer c. protective layer d. absorbing layer ANS: B The phosphor layer, where the x-ray photons are converted to light photons, is the most important screen component.

22. The rare earth phosphors are used in the intensifying screen to a. protect the active layer b. convert x-rays to light photons c. convert light photons to x-rays d. provide physical support for the active layer ANS: B The rare earth phosphors are the most common phosphors used in the screen’s phosphor (active) layer to convert x-ray photons to light photons.

23. When using two intensifying screens, the front screen is found

a. b. c. d.

outside of and in front of the cassette inside the cassette, away from the tube inside the cassette, facing the tube outside and in back of the cassette

ANS: C When using two intensifying screens, the front screen is found inside the cassette facing the tube.

24. With a film-screen IR, __________ of the total energy exposing the film is light energy from the intensifying screens a. 1–10% b. 15–25% c. 40–60% d. 90–99% ANS: D With a film-screen IR, 90–99% of the total energy exposing the film is light energy from the intensifying screens.

25. The ability of phosphors to emit visible light only while exposed to x-rays is a. fluorescence b. brightness c. density d. spectral emission ANS: A The ability of phosphors to emit visible light only while exposed to x-rays is called fluorescence.

26. The capability of a screen to produce visible light is called a. film speed b. luminescence c. screen speed d. spectral emission ANS: C The capability of a screen to produce visible light is called screen speed.

27. Absorption efficiency is the intensifying screen’s a. ability to attenuate the x-ray photons b. ability to produce light from the x-ray photons absorbed c. speed d. luminescence ANS: A Absorption efficiency is how well the screen absorbs (attenuates) the x-ray photons.

Although it affects the speed, it is not the same as speed.

28. Conversion efficiency is the intensifying screen’s a. ability to attenuate the x-ray photons

b. ability to produce light from the x-ray photons absorbed c. speed d. luminescence ANS: B Conversion efficiency is how well the screen produces light from the x-ray photons that were absorbed. Although it affects the speed, it is not the same as speed.

29. The thicker the phosphor layer of an intensifying screen, the _______ light will be produced, and the ________ the screen speed a. more; slower b. more; faster c. less; slower d. less; faster ANS: B The thicker the phosphor layer of an intensifying screen, the more light will be produced, and the faster the screen speed.

30. The smaller the crystal in the phosphor layer of an intensifying screen, the _______ light will be produced, and the ________ the screen speed a. less; slower b. less; faster c. more; slower d. more; faster ANS: A The smaller the crystal in the phosphor layer of an intensifying screen, the less light will be produced, and the slower the screen speed.

31. Having a reflective layer in an intensifying screen results in the screen having a(n) ________ screen speed a. slower b. faster c. identical d. half the ANS: B A reflective layer reflects the light produced toward the film, making the screen faster.

32. For a _____________ screen speed, the recorded detail is __________________. a. faster; increased b. faster; decreased c. slower; increased d. faster; decreased and slower; increased ANS: D For a faster screen speed, recorded detail is decreased and conversely, for a slower screen speed, recorded detail is increased.

33. The relative speed of a standard or fast screen-film system is approximately

a. b. c. d.

50 100 400 800

ANS: C The relative speed of a standard or faster screen-film system is approximately 400.

34. The relative speed of an extremity or detail screen-film system is approximately a. 50 b. 100 c. 400 d. 800 ANS: B The relative speed of an extremity or detail screen-film system is approximately 100.

35. A slower film-screen system has ___________ recorded detail and needs ________ mAs a. worse; less b. worse; more c. better; less d. better; more ANS: D A slower film-screen system has better recorded detail and needs more mAs. 36. What new mAs will maintain radiographic density when changing from a 200 speed to a 400 fill-screen system using 20 mAs? a. 5 b. 10 c. 30 d. 40 ANS: B 20 = 400 ; 400 X = 4,000; X= 4,000; X= 10 (new mAs) X 200 400 37. Poor film-screen contact results in an area of the radiograph that a. is sharp b. is unsharp c. has high contrast d. is white ANS: B Poor film-screen contact results in an unsharp area on the radiographic image.

38. The test tool to evaluate film-screen contact is the a. densitometer b. sensitometer c. wire mesh d. penetrometer

ANS: C The wire mesh test tool is used to evaluate film-screen contact.

39. The purpose of film processing is to convert the _______________ image to the________________ image a. manifest; manifest b. manifest; latent c. latent; manifest d. latent; latent ANS: C The purpose of film processing is to convert the latent image to the manifest image.

40. Which of the following is the correct sequence for automatic film processing? a. Washing, fixing, developing, drying b. Fixing, washing, developing, drying c. Developing, washing, fixing, drying d. Developing, fixing, washing, drying ANS: D Film processing starts with developing and goes on to fixing, washing, and drying.

41. Reducing agents are found in the a. developer b. dryer c. fixer d. none of these ANS: A The developer has reducing agents.

42. The developer solution a. is acidic b. is alkaline c. removes silver halide from the film d. stops the development process ANS: B The developer solution is alkaline.

43. Phenidone and hydroquinone are a. fixing agents b. washing agents c. developing agents d. drying agents ANS: C Phenidone and hydroquinone are developing agents.

44. The fixing stage a. hardens the emulsion

b. removes unexposed silver halide crystals c. makes the image permanent d. All of these ANS: D The fixer removes unexposed silver halide crystals, makes the image permanent, and hardens the emulsion.

45. The ―hypo‖ is a a. fixing agent b. washing agent c. developing agent d. drying agent ANS: A Hypo is another name for a thiosulfate, which is the fixing agent.

46. The fixer solution a. is acidic b. is alkaline c. makes the latent image visible d. continues the development process ANS: A The fixer solution is acidic.

47. The purpose of washing is to remove a. dirt from the film b. fixer solution from the film c. developer solution from film d. fixer solution from the film and developer solution from film ANS: B Washing, the third stage of processing, removes excess fixer solution from the film.

48. Washing the film uses a process called a. infusion b. diffusion c. synergy d. disfusion ANS: B The process used during the washing stage of processing is diffusion.

49. For film archival quality, how much moisture needs to remain in the film after chemical processing? a. 0% b. 5% c. 10% d. 25% ANS: C

Permanent radiographs must retain moisture of 10% to 15% to maintain archive quality.

50. _______________ is a flat metal surface that helps to align the film as it enters the processor a. Recirculation b. The feed tray c. The standby control d. Replenishment ANS: B When introducing the film into the processor it is critical that it goes in straight. The feed tray makes this happen.

51. When the processor is not being used, _______________ stops the rollers from turning. It needs to be activated when it is time to start processing again a. recirculation b. the feed tray c. the standby control d. replenishment ANS: C The standby control stops the rollers from turning during periods of inactivity and often has to be activated to start them up again.

52. _______________ involves the addition of fresh chemicals to the fixer and developer solutions to maintain their activity. a. Recirculation b. The feed tray c. The standby control d. Replenishment ANS: D Developer and fixer solutions become exhausted over time and the more they are used.

Replenishment pumps new chemistry in to maintain activity levels.

53. _______________ system keeps the chemicals mixed, which helps maintain solution activity, agitation, and maintains solution temperature. a. The Recirculation b. The feed tray c. The standby control d. The Replenishment ANS: A The recirculation pump makes sure that the fixer and developer solutions are constantly being mixed.

54. Developer temperature in most automatic film processors is approximately a. 72–76°F b. 78–83°F c. 93–95°F d. 103–108°F

ANS: C Developer temperature in most automatic film processors is approximately 93–95°F. 55. Radiographic film should be stored a. horizontally or flat b. away from heat and ionizing radiation c. after it is taken out of its original packaging, to save space in the darkroom d. all of these ANS: B Radiographic film should be stored on end, in its original packaging, and away from heat and radiation.

56. Kodak Wratten 6B and Kodak GBX are types of a. films b. processors c. safelight filters d. light bulbs ANS: C Kodak Wratten 6B and Kodak GBX are types of safelight filters.

57. Which of the following is a part of a darkroom and processor quality control program? a. Perform safelight fog test b. Check replenishment rates c. Determine the pH level of the fixer and developer solutions d. All of these are part of a recommended darkroom and processor quality control program ANS: D Quality control tests and checks are critical to the processing and storage of film and include these activities and many more.

58. Over time there is an accumulation of silver in the a. developer solution b. fixer solution c. washing solution d. dryer ANS: B Because the fixer removes unexposed silver halide from the film emulsion, silver accumulates there.

59. Silver recovery is important because: 1. it’s a natural resource 2. it’s a heavy metal 3. it can be toxic to environment a. it’s a natural resource and it’s a heavy metal only b. it’s a natural resource and it can be toxic to environment only

c. it’s a heavy metal and it can be toxic to environment only d. it’s a natural resource, it’s a heavy metal, and it can be toxic to environment ANS: D Silver recovery is a natural resource, is a heavy metal that can be toxic to the environment, and needs to be removed from the used fixer.

60. Which of the following statements are correct when using film-screen IRs. 1. Increasing the mAs will increase radiographic density.

2. Decreasing kVp will decrease radiographic density.

3. Changing the intensity of exposure has no effect on radiographic density. a. increasing the mAs will increase radiographic density and decreasing kVp will decrease radiographic density only b. increasing the mAs will increase radiographic density and changing the intensity of exposure has no effect on radiographic density only c. decreasing kVp will decrease radiographic density and changing the intensity of exposure has no effect on radiographic density only d. changing the intensity of exposure has no effect on radiographic density only ANS: A Unlike digital imaging, the intensity of exposure reaching the film-screen IR has a direct effect on the amount of density produced in a processed image.

61. What device determines the amount of light transmitted through film and calculates optical density? a. densitometer b. sensitometer c. odometer d. optimeter ANS: A The densitometer determines the amount of light transmitted and calculates the optical density.

62. A radiograph with a larger number of densities but little differences among them describes: a. high contrast b. low contrast c. long-scale contrast d. low contrast and long-scale contrast ANS: D A radiograph with a larger number of densities but little differences among them is said to have low contrast. Also described as long-scale contrast.

63. Which of the following explains the relationship between the intensity of radiation exposure to the film and the amount of density produced after processing? a. Histogram analysis b. Gurney Mott theory c. Sensitometry d. Nyquist theorem

ANS: C In radiography, sensitometry is the study of the relationship between the intensity of radiation exposure to the film and the amount of blackness produced after processing (density).

64. In sensitometry, for every 0.3 change in log relative exposure, the intensity of radiation exposure changes by a factor of _______. a. 2 b. 3 c. 4 d. 5 ANS: A Along the x-axis, for every 0.3 change in log relative exposure, the intensity of radiation exposure changes by a factor of 2.

65. What region of a sensitometric curve are diagnostic densities produced in? a. shoulder b. toe c. heel d. straight-line ANS: D The straight-line region is where the diagnostic or most useful range of densities is produced.

66. What film characteristics can be evaluated by comparing sensitometric curves?

1. speed 2. film type 3. exposure latitude a. speed and film type only b. speed and exposure latitude only c. film type and exposure latitude only d. speed, film type, and exposure latitude ANS: B Comparing sensitometric curves provides important film characteristics of speed, contrast, and exposure latitude.

67. Calculating a film’s average gradient determines what characteristic? a. speed b. exposure latitude c. contrast d. film type ANS: C When comparing film, the radiographer typically determines contrast by calculating the sensitometric curve’s average gradient.

68. Which of the following is correct regarding the relationship between exposure latitude and film contrast. a. narrow latitude film has low contrast b. wide latitude film has high contrast c. narrow latitude film has high contrast d. exposure latitude has no relationship with a film’s contrast ANS: C Exposure latitude and film contrast have an inverse relationship. High-contrast radiographic film has narrow latitude, and low-contrast film has wide latitude.

69. What can be said about radiographic film quality when optical densities lie within the straight-line region of a sensitometric curve?

1. maximum contrast 2. optimum density 3. visibility of recorded details is optimal a. maximum contrast and optimum density only b. maximum contrast and visibility of recorded details is optimal only c. optimum density and visibility of recorded details is optimal only d. maximum contrast, optimum density, and visibility of recorded details is optimal ANS: D When optical densities lie in the straight-line region, optimum density and maximum contrast have been achieved and the visibility of recorded details is of optimal radiographic quality.

TRUE/FALSE 1. Intensifying screens do not need cleaning because they never get exposed to dust or dirt.

ANS: F Intensifying screens do need regular cleaning because they are exposed to dust and dirt every time the cassette is opened to insert or remove the film.

2. The humidity in the darkroom should be as low as possible.

ANS: F There needs to be some humidity to avoid static electricity.

3. Film-screens have a wide dynamic range.

ANS: F One major deficiency is the limited dynamic range of film-screen.

4. Sensitometric curves positioned to the left (closer to the y-axis) are slower in speed than films positioned to the right (farther from the y-axis).

ANS: F

A faster-speed film is positioned to the left (closer to the y-axis) of slower speed film.

Chapter 15: Fluoroscopic Imaging Johnston/Fauber: Essentials of Radiographic Physics and Imaging, 4th Edition MULTIPLE CHOICE 1. The commercial fluoroscope was developed by a. Rutherford b. Edison c. Newton d. Einstein ANS: B The commercial fluoroscope was invented by Thomas Edison in 1896.

2. The original fluoroscopic image was very a. dim b. bright c. clear d. fuzzy ANS: A A major problem with the original fluoroscopic image was that it was very dim and hard to see.

3. ___________ imaging is used to view the human body in real time, accomplished with fluoroscopy as opposed to radiography. a. Static b. Still c. Dynamic d. Colorized ANS: C Fluoroscopy allows imaging of anatomy in motion, dynamic imaging.

4. The image intensifier was introduced in the a. 1920s b. 1950s c. 1970s d. 1980s ANS: B The image intensifier was introduced in the 1950s.

5. Before the image intensifier, the operator may have worn _____ goggles to make it easier to see the dim fluoroscopic image. a. blue b. green c. red d. yellow

ANS: C Goggles with red lenses allowed the operator’s eyes to adjust to the dark, making it easier to see the dim image from the fluoroscope.

6. The image intensifier improved fluoroscopy by a. making the image easier to see because it is brighter b. making the image easier to see because it is bigger c. allowing indirect viewing of the fluoroscopic image d. making the image easier to see because it is brighter and allowing indirect viewing of the fluoroscopic image ANS: D The fluoroscopic process was improved when the image intensifier made the image significantly brighter and allowed the operator to view the image without standing in the path of the x-ray beam.

7. The electronic vacuum tube that converts the remnant x-ray beam to light, then to electrons, then back to light, increasing the light intensity in the process is the a. x-ray tube b. image intensifier c. recording system d. mirror optics ANS: B The image intensifier is an electronic vacuum tube that converts the remnant beam to light, then to electrons, then back to light, increasing the light intensity in the process.

8. The image intensifier is located a. next to the x-ray tube used for fluoroscopy b. beneath the x-ray table c. on a shelf just inside the room d. inside the fluoroscopic tower ANS: D The image intensifier is located inside the fluoroscopic tower; whereas the x-ray tube used during fluoroscopy is located beneath the x-ray table.

9. This part of the image intensifier is made of cesium iodide. a. Input phosphor. b. Output phosphor. c. Photocathode. d. Accelerating anode.

ANS: A The input phosphor is made of cesium iodide.

10. This part of the image intensifier is made of silver-activated zinc cadmium sulfide. a. Input phosphor. b. Output phosphor. c. Photocathode.

d. Accelerating anode.

ANS: B The output phosphor is made of zinc cadmium sulfide.

11. This part of the image intensifier is made of cesium and antimony compounds. a. Input phosphor. b. Output phosphor. c. Photocathode. d. Electrostatic focusing lenses.

ANS: C The photocathode is made of cesium and antimony compounds.

12. The __________________ absorbs electrons and emits light. a. input phosphor b. output phosphor c. photocathode d. accelerating anode ANS: B The output phosphor absorbs electrons and emits light.

13. The __________________________ absorbs x-rays and emits light. a. input phosphor b. output phosphor c. photocathode d. accelerating anode ANS: A The input phosphor absorbs x-rays and emits light.

14. The _____________________ is designed to set the electron stream in motion at a constant velocity, and is located very close to the output phosphor. a. input phosphor b. output phosphor c. photocathode d. accelerating anode ANS: D The anode is located very close to the output phosphor.

15. This part of the image intensifier absorbs light and produces electrons. a. Input phosphor. b. Output phosphor. c. Photocathode. d. Electrostatic focusing lenses.

ANS: C The photocathode absorbs light and produces electrons.

16. The _______________________ is bonded directly to the input phosphor.

a. b. c. d.

input phosphor output phosphor photocathode accelerating anode

ANS: C The photocathode is attached directly to the input phosphor.

17. The electrostatic focusing lenses have a ___________ charge. a. negative b. positive c. neutral d. changing ANS: A The electrostatic focusing lenses have a negative charge.

18. Photoemission occurs at the a. input phosphor b. output phosphor c. photocathode d. accelerating anode ANS: C Photoemission, when electrons are emitted in response to the presence of light, occurs at the photocathode.

19. The ________________ is (are) found along the entire length of the image intensifier. a. input phosphor b. output phosphor c. photocathode d. electrostatic focusing lenses ANS: D The electrostatic focusing lenses are located at various locations along the length of the image intensifier tube.

20. The ratio of the number of light photons at the output phosphor as compared with the input phosphor is the a. brightness gain b. flux gain c. minification gain d. conversion factor ANS: B Flux gain is the ratio of the number of light photons at the output phosphor as compared with the input phosphor.

21. The term that means an expression of the degree to which the image is made smaller from input phosphor to output phosphor. is the a. brightness gain

b. flux gain c. minification gain d. conversion factor ANS: C Minification gain is an expression of the degree to which the image is minified (made smaller) from input phosphor to output phosphor. This characteristic makes the image appear brighter because the same number of electrons are being concentrated on a smaller surface area.

22. An expression of the ability of an image intensifier tube to convert x-ray energy into light energy and increase the brightness of the image in the process is the a. brightness gain b. flux gain c. minification gain d. conversion factor ANS: A Brightness gain is an expression of the ability of an image intensifier tube to convert x-ray energy into light energy and increase the brightness of the image in the process.

23. ____________ is measured in cd/m2/mR/s. a. Brightness gain b. Flux gain c. Minification gain d. Conversion factor ANS: D Conversion factor is an expression of the luminance at the output phosphor divided by the input exposure rate, and its unit of measure is the candela per square meter per milliroentgen per second (cd/m2/mR/s).

24. The formula for brightness gain is a. brightness gain = minification gain – flux gain b. brightness gain = minification gain / flux gain c. brightness gain = minification gain  flux gain d. None of these ANS: C Brightness gain = minification gain  flux gain.

25. The formula for minification gain is a. minification gain = input phosphor diameter / output phosphor diameter b. minification gain = input phosphor diameter2 / output phosphor diameter2 c. minification gain = output phosphor diameter / input phosphor diameter d. minification gain = output phosphor diameter2 / input phosphor diameter2 ANS: B Minification gain = input phosphor diameter2 / output phosphor diameter2.

26. The diameter of the input phosphor typically measures between

a. b. c. d.

5 and 10 cm 15 and 30 cm 40 and 65 cm 75 and 100 cm

ANS: B The input phosphor typically measures between 15 and 30 cm, or approximately 6–12 inches across.

27. The diameter of the output phosphor is approximately a. 1 cm b. 2.5 cm c. 5 cm d. 25 cm ANS: B The output phosphor measures approximately 2.5 cm or 1 inch in diameter.

28. If the input phosphor measures 30 cm and the output phosphor measures 2.5 cm, what is the minification gain? a. 8.3 b. 12 c. 69.2 d. 144 ANS: D Minification gain = input phosphor diameter2/output phosphor diameter2. In this case, minification gain = 302/2.52.

29. The function of the fluoroscopic unit that maintains the overall appearance of the fluoroscopic image (contrast and brightness) by automatically adjusting the kilovoltage peak (kVp), milliamperage (mA), or both is a. DQE b. APR c. ABC d. AEC ANS: C Automatic brightness control (ABC) maintains the brightness and contrast of the fluoroscopic image during the study.

30. When operated in magnification mode, the electrostatic lenses have ________ of a negative charge. a. less b. more c. the same amount as usual d. one half ANS: B Magnification of the fluoroscopic image occurs when the electrostatic lenses have more voltage applied, creating a more negative charge.

31. The formula to determine the amount of magnification created when in magnification mode, the formula is a. MF = full size input phosphor / selected input phosphor b. MF = full size input phosphor2 / selected input phosphor2 c. MF = selected input phosphor / full size input phosphor d. MF = selected input phosphor2 / full size input phosphor2 ANS: A The magnification factor during fluoroscopy = full size input phosphor / selected input phosphor.

32. If a 30/23/15 cm image intensifier is operated in the 23-cm mode, the fluoroscopic image will be magnified by a factor of a. 1.3 b. 1.5 c. 1.7 d. 2 ANS: A Based on the formula, the MF = the full size of the input phosphor (30 cm)/the selected input phosphor size (23 cm).

33. Magnification of the fluoroscopic image results in improved a. brightness b. spatial resolution c. contrast d. exposure ANS: B Using the magnification mode improves spatial resolution of the fluoroscopic image.

34. A disadvantage of using magnification mode during fluoroscopy is a. the image has less resolution b. the image brightness is inconsistent c. the patient receives additional dose d. the image has increased noise ANS: C Additional x-ray photons must leave the patient and be absorbed by the image intensifier when magnification mode is used, resulting in increased patient dose.

35. Typical fluoroscopic systems have spatial resolution capabilities in the range of a. 2–3 Lp/mm b. 4–6 Lp/mm c. 7–9 Lp/mm d. 10–12 Lp/mm ANS: B Typical fluoroscopic systems have spatial resolution capabilities in the range of 4–6 Lp/mm.

36. In fluoroscopy, shape distortion is caused by a. angling the x-ray tube b. angling the image intensifier in the fluoroscopic tower c. inaccurate control or focusing of the electrons released at the periphery of the photocathode, and the curved shape of the photocathode d. inaccurate control or focusing of the electrons released at the periphery of the photocathode, and the flat shape of the photocathode ANS: C Shape distortion during fluoroscopy is the result of inaccurate control or focusing of the electrons released at the periphery of the photocathode, and the curved shape of the photocathode.

37. Distortion of the fluoroscopic image that appears as unequal magnification is a. noise b. pincushion appearance c. vignetting d. magnification ANS: B Pincushion appearance is the distortion of the fluoroscopic image that appears as unequal magnification.

38. A loss of brightness around the edge of the fluoroscopic image caused by the curve of the photocathode is a. noise b. pincushion appearance c. vignetting d. magnification ANS: C Vignetting is the loss of brightness around the periphery of the fluoroscopic image.

39. Increasing the mA is the way to correct a fluoroscopic image that has a. noise b. pincushion appearance c. vignetting d. magnification ANS: A When a fluoroscopic image has noise, or is noisy, the quantity of radiation (mA) must be increased because the issue is too few photons being used to create the image.

40. Fluoroscopic imaging systems operate at: ________ a. 2–5 mA b. 15–20 mA c. 200–500 mA d. Unlimited mA ANS: A Fluoroscopic imaging systems operate at 2–5 mA.

41. In image-intensified fluoroscopy units the method for viewing the fluoroscopic image is by a. using a handheld device b. looking directly at the output phosphor c. using a television monitor d. using a mirror-optics system ANS: C The fluoroscopic image is viewed by looking at a television monitor.

42. The _____________ is a diode tube contained in a glass envelope to maintain a vacuum, and is connected to the output phosphor of the image intensifier by either a fiberoptic bundle or an optical lens system. a. television monitor b. Vidicon camera tube c. charge coupled device (CCD) d. liquid crystal display (LCD) ANS: B A vidicon camera tube is a diode tube contained in a glass envelope to maintain a vacuum, and is connected to the output phosphor of the image intensifier by either a fiberoptic bundle or an optical lens system.

43. When the electron beam sweeps the anode and travels back and forth across, from top to bottom, it is moving in a ___________ pattern. a. sweep b. brush c. raster d. rooster ANS: C A raster pattern sweeps the anode in a path from left to right, from the top to the bottom.

44. A face plate, signal plate, and target are all components of the a. anode of the camera tube b. charge coupled device (CCD) c. cathode of the camera tube d. x-ray tube ANS: A The anode end of a camera tube includes a face plate, signal plate, and target.

45. The target layer is a photoconductive layer made of antimony trisulfide in ______ tubes. a. x-ray b. CCD c. vidicon d. plumbicon ANS: C The target layer is a photoconductive layer made of antimony trisulfide in vidicon tubes.

46. When the electron beam is sweeping the anode, if the electron beam and light from the output phosphor are incident on the same place at the same time, _____________ through the target to the signal plate. a. x-rays are produced b. the image is magnified c. electrons are transmitted d. light photons are transmitted ANS: C If the electron beam encounters light as it sweeps across the target plate, electrons (an electric current) will be transmitted.

47. A light-sensitive semiconducting device that generates an electrical charge when stimulated by light and stores this charge in a capacitor is the a. anode of the camera tube b. charge coupled device (CCD) c. cathode of the camera tube d. x-ray tube ANS: B The CCD is a light-sensitive semiconducting device that generates an electrical charge when stimulated by light and stores this charge in a capacitor.

48. Semiconductor capacitors made of metal oxide are components of the a. camera tube b. CCD c. image intensifier d. x-ray tube ANS: B The charge-coupled device (CCD) includes metal oxide semiconductor capacitors.

49. Each capacitor in a CCD represents a(n) a. line b. image c. pixel d. electric charge ANS: C Each capacitor in a CCD briefly stores an electric charge that represents a pixel in the digital image.

50. The CCDs electric charge from the capacitors is sent to the a. output phosphor b. television monitor c. x-ray tube d. image intensifier ANS: B After being briefly stored in the capacitor, the charge travels by wire to the television monitor.

51. The fiberoptic bundle or optical lens system is used to a. deliver the image from the camera to the television monitor b. record the fluoroscopic image c. link the output phosphor and camera or CCD d. none of these ANS: C The fiberoptic bundle or optical lens system is used to link or couple the camera or CCD to the output phosphor of the image intensifier.

52. The purpose of a beam-splitting mirror is to a. view the fluoroscopic image on more than one television b. reduce the intensity of the image from the output phosphor c. allow spot filming to be done during fluoroscopy d. all of these ANS: C The beam-splitting mirror is used during fluoroscopy when a spot image is to be recorded on a photospot or cine camera.

53. The purpose of the from the camera tube or CCD into a visible image. a. image intensifier b. fiberoptic bundle c. television monitor d. anode of the vidicon tube

is to convert the electronic signal

ANS: C The purpose of the television monitor is to convert the electronic signal from the camera tube or CCD back into a visible image.

54. Typical television monitors are a. 300 line systems b. 450 line systems c. 525 line systems d. 875 line systems ANS: C Typical television monitors are 525 line systems.

55. A typical high-resolution monitor has a. 825 line systems b. 1024 line systems c. 525 line systems d. 2034 line systems ANS: B A typical high-resolution monitor has 1024 line systems.

56. An electron gun can be found in the

a. b. c. d.

television monitor CCD vidicon camera television monitor and vidicon camera

ANS: D A camera tube and a television monitor both use electron guns.

57. The television is capable of resolving approximately ___________. a. 0.5–.75 Lp/mm b. 1–2 Lp/mm c. 3 Lp/mm d. 5 Lp/mm ANS: B The television monitor can resolve approximately 1–2 Lp/mm.

58. In terms of resolution, the weakest part of the fluoroscopic system is the a. image intensifier b. television monitor c. camera tube d. CCD ANS: B The television monitor has the least amount of spatial resolution of all the components of the fluoroscopic system.

59. The recording device that ―photographs‖ the image off of the output phosphor is the a. cassette spot filming b. film cameras for spot filming c. videotape d. flat panel detector ANS: B The photospot (film) camera is a static imaging system that is used with an optical lens system incorporating a beam-splitting mirror. When the spot-film exposure switch is pressed, the beam-splitting mirror is moved into place, diverting some of the beam toward the photospot camera and exposing the film.

60. This recording system uses 70-mm or 105-mm film. a. Cassette spot filming b. Film cameras for spot filming c. Videotape d. None of these ANS: B Film cameras use either 105-mm ―chip‖ film or 70-mm roll film to record the fluoroscopic image.

61. This allows the film to be divided into 2, 4, or more images. a. Cassette spot filming with masking shutters

b. Film cameras for spot filming c. Videotape d. Flat panel detector ANS: A Cassette spot filming with masking shutters can be set up to record two images per film (two on one), four on one, or more.

62. Which of the following recording systems require the x-ray-beam to be in radiographic mode? a. Cassette spot filming. b. Film cameras for spot filming. c. Cine filming. d. Flat panel detector.

ANS: A When the spot-film exposure button is pressed, the cassette is moved into position between the patient and image intensifier and the machine shifts from fluoroscopic to radiographic mode and exposes the film.

63. Early versions of digital fluoroscopy used the conventional fluoroscopic chain and added a. a different camera b. an analog-to-digital converter (ADC) c. videotape d. a digital-to-analog converter (DAC) ANS: B Early digital fluoroscopy systems added an analog-to-digital converter between the camera and monitor.

64. The analog-to-digital converter a. takes the video (analog) signal and divides it into binary language that the computer ―understands.‖ b. determines the contrast resolution of the system c. is necessary for the computer to process and display the image d. all of these ANS: D The ADC converts analog data (electrical current) to digital data (1s and 0s). Each ADC has a set number of bits that determine image contrast resolution and is necessary for the computer to process and display the image.

65. Digital fluoroscopy is improved by using a. a vidicon camera b. videotape c. a beam-splitting mirror d. a charge-coupling device (CCD) ANS: D A CCD used for converting the light image to an electronic image greatly improved digital imaging.

66. Which of the following is true about the CCD used in digital fluoroscopy?

I. The CCD is more light sensitive with a higher DQE value.

II. The CCD exhibits less noise and no spatial distortion.

III. The CCD has higher spatial resolution.

IV. The CCD requires more radiation to create an image. a. the CCD is more light sensitive with a higher DQE value and CCD exhibits less noise and no spatial distortion only. b. the CCD is more light sensitive with a higher DQE value and the CCD has higher spatial resolution only. c. the CCD is more light sensitive with a higher DQE value, CCD exhibits less noise and no spatial distortion, and the CCD has higher spatial resolution only. d. the CCD is more light sensitive with a higher DQE value, CCD exhibits less noise and no spatial distortion, and the CCD requires more radiation to create an image only. e. the CCD is more light sensitive with a higher DQE value, CCD exhibits less noise and no spatial distortion, the CCD has higher spatial resolution, and the CCD requires more radiation to create an image.

ANS: C The charge-coupled device is more sensitive to the light from the output phosphor as it has a higher DQE. The CCD exhibits less noise with no spatial distortion. The CCD has higher spatial resolution and requires less radiation to create an image (less dose).

67. More advanced fluoroscopy systems use a flat-panel detector a. to record the spot images during fluoroscopy b. to record the overhead images following fluoroscopy c. in place of the image intensifier d. in addition to the image intensifier ANS: C More advanced fluoroscopy systems use a flat-panel detector in place of the image intensifier.

68. Flat-panel detectors for fluoroscopic imaging can be the a. cesium iodide amorphous silicon indirect-capture detector b. amorphous selenium direct-capture detector c. cesium and antimony direct-capture detector d. cesium iodide amorphous silicon indirect-capture detector and amorphous selenium direct-capture detector e. cesium iodide amorphous silicon indirect-capture detector and cesium and antimony direct-capture detector ANS: D Two forms of flat-panel detectors may be used for fluoroscopic applications: the cesium iodide amorphous silicon indirect-capture detector and the amorphous selenium direct-capture detector.

69. For detectors to produce high-quality fluoroscopic images, they must be able to

a. b. c. d.

respond very quickly maintain a large fill factor have application-specific integrated circuits for noise reduction all of these

ANS: D To produce quality fluoroscopic images with digital detectors, they have to be able to respond quickly to the presence of x-rays, maintain a large fill factor, and have an ASIC in place to reduce noise.

70. Which of the following is required in the indirect-capture detector? a. Amorphous silicon photodetector. b. Thin film transistor (TFT) array. c. A scintillator that uses cesium iodide or gadolinium oxysulfide as the phosphor. d. All of these.

ANS: D The cesium iodide amorphous silicon indirect-capture detector requires a scintillator with cesium iodide or gadolinium oxysulfide as the phosphor, an amorphous silicon photodetector, and a TFT array layered onto a glass substrate that include the readout, charge collector, and light-sensitive elements.

71. Which of the following is an advantage of flat panel detectors used in fluoroscopy? a. Much lighter and more compact. b. Produce a digital signal directly. c. Less electronic noise. d. 60 times larger operational dynamic range because they do not exhibit veiling glare. e. All of these.

ANS: E The flat panel detectors used in fluoroscopy are much lighter and more compact, they produce a digital signal directly (no need for a camera tube or ADC), and because it is a digital system producing a digital signal, absent of the electronic components of the old II system, there is less electronic noise. The flat panel detector has 60 times larger operational dynamic range than image intensifier systems and because of this do not exhibit veiling glare.

72. A large fill factor refers to a. the entire detector being filled with x-rays b. the entire detector needing a large amount of x-rays to produce an image c. a large area of the pixel being sensitive to x-rays d. a small area of the pixel being sensitive to x-rays ANS: C A large fill factor refers to a large area of the pixel being sensitive to x-rays. This is important to flat panel fluoroscopy.

73. Which of the following has a rectangular field of view and a wider dynamic range? a. Conventional image intensifier. b. Flat-panel detector in place of image intensifier.

c. There is no difference between the two.

ANS: B The flat-panel detector has a rectangular field of view and a wider dynamic range.

74. Flat panel detector fluoroscopy units operate at what mA? a. 2- 5 mA b. 10-30 mA c. 50-1200 mA d. 1500-2000 mA ANS: C Flat panel detector fluoroscopy units operate at 50-1200 mA. 75. What fluoroscopic feature allows the position of the collimator plates to be viewed on the monitor without exposing the patient to radiation? a. last image hold b. frame averaging c. digital subtraction d. electronic magnification ANS: A With the newest units, lines will appear on the last image hold (LIH) displayed on the monitor, indicating the position of the collimator plates.

76. The ability to adjust the size of the field of view without exposing the patient to radiation is known as: a. frame averaging b. last image hold c. digital subtraction d. virtual collimation ANS: D Virtual collimation allows the displayed field of view to be further adjusted without exposing the patient to additional radiation.

77. Frame averaging is a. an operation that reduces overall patient dose b. an operation that increases overall patient dose c. an operation that reduces image noise by averaging multiple image frames together d. an operation that reduces overall patient dose and an operation that reduces image noise by averaging multiple image frames together ANS: D Frame averaging is an operation that reduces overall patient dose and image noise by averaging multiple image frames together.

78. What feature automatically adjusts the tube current (mA), voltage (kVp), filtration, and pulse width to maintain radiation exposure to the flat panel detector? a. Pulsed fluoroscopy b. Last image hold

c. Frame averaging d. None of these ANS: D AERC automatically adjusts the tube current (mA), voltage (kVp), filtration, and pulse width to maintain radiation exposure to the flat panel detector.

79. The difference between geometric magnification and electronic magnification is a. geometric magnification is where the patient is placed closer to the x-ray tube to create a larger OID and electronic magnification involves the selection of a smaller field of view b. electronic magnification is where the patient is placed closer to the x-ray tube to create a larger OID and geometric magnification involves the selection of a smaller field of view c. there is no difference between geometric magnification and electronic magnification ANS: A Geometric magnification is where the patient is placed closer to the x-ray tube to create a larger OID and electronic magnification involves the selection of a smaller field of view.

80. A fluoroscopic feature that automatically turns the x-ray beam on and off rapidly during operations is known as: a. Pulsed fluoroscopy b. Frame averaging c. Intermittent fluoroscopy d. Pulse rate ANS: A Pulsed fluoroscopy is simply a design of the unit that rapidly turns the x-ray beam on and off during operation.

81. ______________ refers to how long each exposure lasts. a. Pulsed fluoroscopy b. Pulse rate c. Pulse width d. Dose rates ANS: C Pulse width refers to the length of each pulse. Think of this as how long each exposure lasts.

82. What information during fluoroscopy should be documented in the patient’s record?

1. Total amount of fluoroscopy time 2. Cumulative air kerma 3. Dose-area product a. total amount of fluoroscopy time and cumulative air kerma only b. total amount of fluoroscopy time and dose-area product only c. cumulative air kerma and dose-area product only d. total amount of fluoroscopy time, cumulative air kerma, and dose-area product

ANS: D The DAP/KAP and cumulative air kerma provide radiation exposure data and need to be recorded in the patient’s medical record.

83. In LCD monitors, the liquid crystal layer sandwiched between polarizing layers contains crystals that are in an unorganized and twisted state, but become __________ when electric current is applied. a. unorganized and twisted b. unorganized and untwisted c. organized and twisted d. organized and untwisted ANS: D In LCD monitors, the liquid crystal layer sandwiched between polarizing layers contains crystals that are in an unorganized (twisted) state, but become organized (untwisted) when electric current is applied.

84. In LCD monitors, the number of TFTs is equal to the a. number of crystals in the liquid crystal layer b. number of pixels displayed c. Lp/mm of resolution d. display line system ANS: B The number of TFTs is equal to the number of pixels displayed.

85. Which of the following contains a thin layer of pixels instead of a liquid crystal layer, with each pixel containing three neon and xenon gas-filled cells? a. Cathode ray tube monitor. b. Liquid crystal display (LCD) monitor. c. Plasma monitor. d. Radiographic film.

ANS: C The plasma monitor contains a thin layer of pixels instead of a liquid crystal layer, with each pixel containing three neon and xenon gas-filled cells.

86. Fluoroscopic equipment should be inspected every a. week b. month c. 6 months d. year ANS: C Fluoroscopic equipment should be inspected every 6 months.

87. The radiographer may be responsible for the ________________ inspection of the fluoroscopic equipment for quality control purposes. a. performance b. operational c. supervisory

d. performance and operational ANS: B The radiographer may be responsible for the operational inspection of the fluoroscopic equipment for quality control purposes.

88. Inspecting the protective curtain, Bucky slot cover, and fluoroscopic timer are part of the a. performance quality control testing b. operational quality control testing c. supervisory d. performance quality control testing and operational quality control testing ANS: B Inspecting the protective curtain, Bucky slot cover, and fluoroscopic timer are part of the operational quality control testing.

89. Performance quality control testing should be done by a a. radiographer b. medical physicist c. radiologist d. student radiographer ANS: B Performance quality control testing should be done by a medical physicist.

TRUE/FALSE 1. The input phosphor and photocathode form a flat surface.

ANS: F The input phosphor and photocathode form a curved surface so that the electrons all travel the same distance to the output phosphor.

2. Each electron that interacts with the output phosphor results in significantly fewer light photons being produced in comparison with the number of light photons it took to release that electron.

ANS: F Each electron that interacts with the output phosphor results in significantly more light photons being produced in comparison with the number of light photons it took to release that electron.

3. Age has no effect on an image intensifier because everything takes place in a vacuum.

ANS: F As an image intensifier ages, it deteriorates and requires more radiation to produce the same level of image brightness.

4. The ABC can be slow in responding, seen as a brief delay in the brightness adjustment.

ANS: T The automatic brightness control can often be seen lagging behind changes in the intensity of the x-ray exposure to the image intensifier.

5. Automatic exposure rate control (AERC) is used in digital fluoroscopy and is similar to automatic brightness control (ABC).

ANS: T Like automatic brightness control (ABC) used in older image intensified units, automatic exposure rate control (AERC) serves a similar function in modern fluoroscopy.

6. When using magnification mode, the operator needs to set the exposure factors higher to compensate for the reduced diameter useful input phosphor.

ANS: F The need for increased exposure to the input phosphor when using magnification mode is automatically addressed by the ABC. 7. The purpose of a camera tube or CCD is to record the fluoroscopic image.

ANS: F The purpose of a camera tube or CCD is to convert the output phosphor image to an electronic signal that can then travel to the television monitor.

8. Pulse rate refers to how many pulses occur per second of operation.

ANS: T Pulse rate refers to how many pulses occur per second of operation. Think of this as how many exposures occur per second.

9. When a CCD is exposed to light, it generates and briefly stores light energy.

ANS: F When a CCD is exposed to light, it generates and briefly stores an electrical charge.

10. The electron gun in the television monitor sends out a steady stream of electrons to activate the fluorescent screen.

ANS: F The electron gun in the television monitor sends out a constantly changing stream of electrons to activate the fluorescent screen, allowing display of various levels of brightness.

11. The radiographer is responsible for performance and operational quality control testing.

ANS: F The radiographer is responsible for operational quality control testing. The medical physicist is responsible for performance quality control testing.

Chapter 16: Additional Equipment Johnston/Fauber: Essentials of Radiographic Physics and Imaging, 4th Edition

MULTIPLE CHOICE 1. Which of the following is/are unique features of interventional fluoroscopic units? a. Bi-plane technology b. Cine fluoroscopic images at 60 frames/sec c. Digital subtraction d. All of these ANS: D Bi-plane technology, cine fluoroscopic images at 60 frames/sec, and digital subtraction capabilities are all unique features of interventional flouroscopic units.

2. Why would radiographic or fluoroscopic equipment need to be mobile? a. patient too sick b. patient in surgery c. patient is immobile d. all of these ANS: D Radiographic and fluoroscopic equipment are mobile to image patients at the bedside or in surgery. Patient’s condition could prevent him/her from traveling to the radiology department.

3. Which of the following mobile radiographic units does not need to be plugged into the wall while producing x-rays? a. Battery-powered mobile unit. b. Capacitor discharge mobile unit. c. Direct-power mobile unit. d. High-frequency mobile unit.

ANS: A General Feedback: Battery-operated units do not need to be plugged into the wall, but need to be recharged.

4. Which of the following mobile units produce consistent radiation output, similar to a single-phase generator? a. Battery-powered b. Capacitor discharge c. Direct-power d. High-frequency ANS: B Capacitor discharge units must be plugged into a wall outlet during operation but produce consistent radiation output, similar to that of a single-phase generator.

5. This type of radiographic mobile unit is very lightweight and produces a consistent radiation output. a. Battery-powered b. Capacitor discharge

c. Direct-power d. High-frequency ANS: D High-frequency units produce a consistent radiation output and are lightweight but must be plugged into a wall outlet during operation.

6. Which of the following individuals need to be shielded during mobile radiographic procedures? a. The patient. b. The nurse who must stay by the patient. c. The student radiographer. d. All should be shielded.

ANS: D Just as if the procedure was being done in the radiology department, anyone who remains in the room during the exposure should be shielded.

7. The radiographer should stand at least _________ away from the patient and x-ray tube when making an exposure using mobile radiographic equipment a. 2 feet b. 4 feet c. 6 feet d. 10 feet ANS: C The radiographer should stand at least 6 feet away from the patient and x-ray tube when making an exposure using mobile radiographic equipment.

8. Challenges when performing mobile radiographic imaging include a. having to use SIDs other than 40 inches or 72 inches b. difficulty minimizing off-level grid cutoff c. patient condition d. all of these ANS: D There are many challenges when performing mobile radiographic imaging, including the patient’s condition as well as making adjustments for nonstandard SID and accurately aligning the x-ray beam and grid.

9. A common site for mobile fluoroscopy is a. in the patient’s room b. in the radiology department c. in nursing home d. in the operating room ANS: D Although it may be done elsewhere, mobile fluoroscopy is commonly performed in the operating room.

10. There are generally _____ sets of locks on mobile C-arm fluoroscopic equipment.

a. b. c. d.

1 2 3 6

ANS: C There are generally 3 sets of locks on mobile C-arm fluoroscopic equipment. One set moves the entire ―C‖ toward or away from the base. Another set allows the pivot of the ―C‖ about its axis. The last set allows the ―C‖ to slide along its arc.

11. Unless otherwise required, during C-arm fluoroscopic procedures the x-ray tube should be located. a. beneath the patient b. to the right side of the patient c. above the patient d. to the left side of the patient ANS: A To reduce radiation exposure to the operator, the C-arm should be set up so the tube is beneath the patient.

12. Which of the following provides a lower fluoroscopic dose rate and greater flexibility while imaging the extremities? a. Bi-plane b. DR mobile imaging c. O-arm imaging d. Mini C-arm imaging ANS: D The mini C-arm provides a lower dose rate and greater equipment flexibility while providing quality images of extremities 13. Which of the following provides both static and dynamic images along with 2D and 3D during surgical procedures. a. Bi-plane b. DR mobile imaging c. O-arm imaging d. Mini C-arm imaging ANS: C The O-arm system can provide both static and dynamic images along with 2D and 3D. Fluoroscopic and radiographic images are obtained with flat panel detector technology which allows the surgeon to assess the outcome of the procedure before closing the patient.

14. When operating mobile C-arm fluoroscopic equipment, it is the _____________ responsibility to monitor and apply radiation safety measures. a. surgeon’s b. radiographer’s c. radiologist’s d. anesthesiologist’s

ANS: B When operating mobile C-arm fluoroscopic equipment, it is the radiographer’s responsibility to monitor and apply radiation safety measures.

15. The dedicated unit designed to image the thorax in the upright position is the a. chest b. panorex c. mammography d. bone densitometry ANS: A The chest unit is very helpful in efficient performance of chest radiography, the most commonly performed radiographic studies.

16. This type of dedicated unit is typically used to image the mandible and teeth a. Chest b. Panorex c. Mammography d. Bone densitometry ANS: B The panoramic x-ray (Panorex) unit is designed to image curved surfaces, typically the mandible and teeth.

17. When a panorex examination is being performed, the x-ray tube and image receptor move a. from side to side b. around the patient c. up and down d. very little ANS: B The x-ray tube and IR move around the stationary patient during a panorex procedure.

18. This dedicated unit is used to determine if a patient’s bone mineral density or mass is normal or low (osteoporosis). a. Chest b. Panorex c. Mammography d. Bone densitometry ANS: D Bone densitometry units evaluate a patient’s bone density to determine whether osteoporosis is an issue.

19. Bone densitometry uses x-rays to determine the level of bone density because a. different density tissues absorb radiation differently b. x-rays are very good at imaging the bony anatomy c. A and B d. none of these ANS: A

X-rays are used in bone densitometry because dense bones absorb more radiation than porous (osteoporotic) bones.

20. DXA stands for a. double x-ray absorption b. density x-ray attenuation c. dual energy x-ray absorptiometry d. double density x-ray attenuation ANS: C DXA stands for dual energy x-ray absorptiometry.

21. The most common areas scanned for bone densitometry are the a. hip and cervical spine b. lumbar spine and skull c. shoulder and thoracic spine d. hip and lumbar spine ANS: D The most common areas scanned for bone densitometry are the hip and lumbar spine.

22. DXA uses an x-ray tube and a(n) a. scintillation detector b. photostimulable image receptor c. image intensifier d. flat panel detector ANS: A Once the x-rays pass through the part, they interact with a scintillation detector and produce light energy, which is then converted to an electronic signal.

23. The T-score from the DXA data analysis indicates a. how large the patient is compared with young, healthy adults b. how large the patient is compared with adults of the same age and gender c. fracture risk resulting from osteoporosis compared with young, healthy adults d. fracture risk resulting from osteoporosis compared with adults the same age and gender ANS: C The T-score from the DXA data analysis indicates fracture risk resulting from osteoporosis compared with young, healthy adults.

24. Breast imaging is done on this type of dedicated radiographic unit a. Chest b. Panorex c. Mammography d. Bone densitometry ANS: C Mammographic units are designed for breast imaging.

Incorrect Answer Reply: 25. Mammography uses a ___________ exposure. a. low kVp b. mid-range kVp c. high kVp d. low mAs ANS: A Mammography uses a low kVp exposure, such as 24–34 kVp. 26. The kVp is very low because of the breast’s a. size b. low subject contrast c. high subject contrast d. density ANS: B Because the breast is made of very similar tissues, this low subject contrast requires a low kVp to make the anatomy visible.

27. To produce the low kVp photons needed for mammography, the tube anode target is made of a. molybdenum b. beryllium c. rhodium d. a and c ANS: D To produce the low kVp photons needed for mammography, the tube anode target is typically made of molybdenum or rhodium.

28. The focal spot size of a dedicated mammography unit measures________________ a. 0.1–0.3 mm b. 1–3 mm c. 0.1–0.3 mm d. 1–3 mm ANS: A The focal spot size of a dedicated mammography unit is smaller than that of a standard x-ray tube, measuring 0.1–0.3 mm.

29. The grid used for mammography has a a. lower ratio and lower frequency b. lower ratio and higher frequency c. higher ratio and lower frequency d. higher ration and higher frequency ANS: A The grid used for mammography has a lower grid ratio and a lower grid frequency.

30. What are the advantages of full-field digital mammography?

1. Improved SNR 2. Post processing capabilities 3. Increased contrast resolution a. improved SNR and post processing capabilities only b. improved SNR and increased contrast resolution only c. post processing capabilities and increased contrast resolution only d. improved SNR, post processing capabilities, and increased contrast resolution ANS: D Important advantages of full-field digital image receptors in mammography include improved digital uniformity, improved SNR and CNR, and reduction of image artifacts, postprocessing capabilities, and the ability to store images in a picture archiving and communication system (PACS).

31. To make the breast tissue a more uniform thickness, the dedicated mammography unit includes a _______________ device. a. pancake b. compression c. flattening d. magnification ANS: B To make the breast tissue a more uniform thickness, the dedicated mammography unit includes a compression device.

32. Breast compression a. improves image contrast b. improves spatial resolution c. reduces magnification d. all of these ANS: D When the breast is compressed, less scatter is produced, improving image contrast. By placing the breast tissue closer to the IR, magnification is reduced, providing improved spatial resolution.

33. The radiation dose to the breast ______________________ when magnification techniques are used. a. is lower b. is higher c. remains the same ANS: B The radiation dose to the breast is higher when magnification techniques are used.

34. Three-dimensional mammography is also known as: a. Bi-plane imaging b. O-arm imaging c. Tomographic imaging d. Digital breast tomosynthesis

ANS: D Digital breast tomosynthesis (DBT) is also known as three-dimensional (3D) mammography and is similar to computed tomography (CT).

35. Digital breast tomosynthesis (DBT) improves the detection of abnormal tissue and reduces the incidences of: a. true positives b. true negatives c. false negatives d. false positives ANS: D DBT improves the detection of abnormal tissue and reduces the incidence of false positives (identifying normal tissue as abnormal).

36. During linear tomography, the tube and image receptor a. move in the same direction along a straight line b. move in opposite directions along a straight line c. move in the same direction along a circular path d. move in opposite directions along a circular path ANS: B During linear tomography, the tube and image receptor move in opposite directions along a straight line.

37. The goal of linear tomography is to a. sharpen the anatomic structures above and below the level of interest b. sharpen the anatomic structures within the level of interest c. blur the anatomic structures above and below the level of interest d. blur the anatomic structures within the level of interest ANS: C The goal of linear tomography is to blur the anatomic structures above and below the level of interest.

38. The arc created during the total movement of the x-ray tube during tomography is the a. tomographic angle b. exposure angle c. fulcrum d. focal plane ANS: A The tomographic angle is the arc created during the total movement of the tube during tomography.

39. The fixed point that is placed at the level of the anatomic structures to be imaged during tomography is the a. tomographic amplitude b. exposure amplitude c. fulcrum

d. focal plane ANS: C The fulcrum, or pivot point, is the fixed point that is placed at the level of the anatomic structures to be imaged during tomography.

40. The horizontal level that extends through the body that includes the pivot point is the a. tomographic amplitude b. exposure amplitude c. fulcrum d. focal plane ANS: D The focal plane, or object plane, is the horizontal level that extends through the body that includes the fulcrum.

41. Increasing the tomographic angle _____ the thickness of the focal plane. a. increases b. decreases c. does not affect ANS: B Increasing the tomographic angle decreases the thickness of the focal plane.

42. In a system with an adjustable fulcrum a. the patient does not move b. the tube does not move during the exposure c. the fulcrum layer is moved up or down to the level of interest d. the patient does not move and the fulcrum layer is moved up or down to the level of interest ANS: D In a linear tomographic unit with an adjustable fulcrum, the fulcrum can be adjusted up or down while the patient remains at a fixed level. The tube must move for any type of tomography.

43. The thickness, or width, of the focal plane is called a a. level b. section c. portion d. fulcrum ANS: B The thickness of the focal plane is called a section and it is adjustable.

44. To produce a thin tomographic section, there needs to be a a. low fulcrum level b. high fulcrum level c. small tomographic angle d. large tomographic angle ANS: D

A large tomographic angle produces a thin section, whereas a small tomographic angle produces a thick section.

45. In that the exposure time for a linear tomographic image is long, the mA must be a. low b. mid-range c. high ANS: A To produce any amount of mAs, if the exposure time is long the mA must be low.

46. As compared with standard radiography, patient exposure during tomographic procedures is a. lower b. the same c. higher d. it depends on the anatomy being imaged ANS: C Linear tomographic studies require more exposure than standard radiographic imaging.

TRUE/FALSE 1. The battery-powered mobile radiographic unit produces consistent radiation output, similar to a single-phase generator.

ANS: F The battery-powered mobile radiographic unit produces consistent radiation output, similar to a three-phase generator.

2. The radiographer does not need to wear a lead apron while imaging patients using mobile equipment because the exposure is very minimal.

ANS: F The radiographer should always wear a lead apron when performing mobile radiography.

3. During digital subtraction techniques, overlying structures such as bone and soft tissue are removed from the subtracted image, so the vasculature is better visualized.

ANS: T During digital subtraction techniques, the overlying structures such as bone and soft tissue are removed from the subtracted image, so the vasculature is better visualized.

4. A dedicated unit is one that is designed for specific imaging procedures ANS: T Dedicated units are designed for specific procedures, such as mammography or chest radiography.

5. A dedicated chest room allows the radiographer to stay in the room while the images are being processed.

ANS: T In that the processor and imaging system are directly connected, whether film-screen or digital, the radiographer can remain in the room and work in a very efficient manner.

6. The image of a mandible taken with a panorex unit and one taken with standard radiographic equipment have very similar appearances.

ANS: F In the panorex image the mandible appears to be lying flat, or straightened out, which looks very different when compared with a PA or oblique mandible.

7. The primary result of a DXA scan is a detailed radiographic image of the hip or lumbar spine ANS: F The result of a DXA scan consists primarily of statistical data analysis, including a T-score and Z-score that provide measurements about structural changes in the bones for evaluation of osteoporosis.

8. The mammography unit’s x-ray tube window is made of beryllium.

ANS: T The mammography unit’s x-ray tube window is made of beryllium.

9. To create 3-D images of the breast, the x-ray tube moves over the compressed breast from one side to the other and creates multiple images of the breast from different angles.

ANS: T In DBT, the x-ray tube moves over the compressed breast from one side to other and creates multiple images of the breast from different angles. These multiple images are reconstructed (synthesized) in the computer to create a set of 3-D images. Like CT, the images are viewed as slices of tissue throughout the breast.

10. Linear tomography is used to overcome the limitation that some anatomic areas are difficult to see because they are superimposed by other structures.

ANS: T In a standard radiographic image, all of the anatomy, from front to back, is superimposed.

Linear tomography works to limit the visualization of some of that superimposition.

11. Linear tomography has been replaced with the introduction of computed tomography.

ANS: T Linear tomography is seldom used because it has been largely replaced by CT studies.

Chapter 17: Computed Tomography

Johnston/Fauber: Essentials of Radiographic Physics and Imaging, 4th Edition MULTIPLE CHOICE 1. Computed tomography was introduced in the a. 1950s b. 1960s c. 1970s d. 1980s ANS: C CT was introduced in the 1970s.

2. Slices of anatomy that go from head to foot are a. coronal b. sagittal c. transverse (axial) d. linear ANS: C Slices of anatomy that go from head to foot are transverse, a.k.a. axial.

3. Slices of anatomy that are reformatted to go from side to side are a. coronal b. sagittal c. axial d. linear ANS: B Slices of anatomy that are reformatted to go from side to side are sagittal.

4. Slices of anatomy that are reformatted to go from front to back are a. coronal b. sagittal c. axial d. posterior ANS: A Slices of anatomy that are reformatted to go from front to back are coronal.

5. The individual credited with inventing the CT scanner is a. Thomas Edison b. Wilhelm Roentgen c. Marie Curie d. Godfrey Hounsfield ANS: D Godfrey Hounsfield received the Nobel Prize for his work in developing the CT scanner.

6. Each major development in beam and detector geometry is called a

a. b. c. d.

generation gantry detector upgrade

ANS: A Major development in beam and detector geometry is called a generation.

7. The CT x-ray tube and detector array are found in the a. central processing unit b. gantry c. detector d. data acquisition system (DAS) ANS: B The gantry houses the x-ray tube and detector array that travel around the patient.

8. The device that absorbs the radiation and produces an electrical signal is the a. generation b. gantry c. detector d. data acquisition system (DAS) ANS: C As in many imaging processes, the detector is the device that absorbs the radiation and produces an electrical signal.

9. This translate-rotate scanner used a single detector a. First generation b. Second generation c. Third generation d. Fourth generation ANS: A The first-generation scanner used a single detector and had to both translate and rotate.

10. This is the only generation of CT scanners that used parallel-beam geometry. a. First generation b. Second generation c. Third generation d. Fourth generation ANS: A The first-generation scanner is the only one to have x-ray beams that are parallel to each other making up the data.

11. This translate-rotate scanner used approximately 30 detectors and a narrow fan beam. a. First generation b. Second generation c. Third generation d. Fourth generation

ANS: B Second-generation scanners had an array of approximately 30 detectors instead of just one, but the fan did not cover the anatomy so the tube had to translate and rotate.

12. This CT scanner took approximately 5 minutes to produce a single image. a. First generation b. Second generation c. Third generation d. Fourth generation ANS: A First-generation scanners took approximately 5 minutes to produce a single slice.

13. This CT scanner took approximately 30 seconds to produce a single image. a. First generation b. Second generation c. Third generation d. Fourth generation ANS: B Significantly faster than the first generation, the second-generation scanner took approximately 30 seconds to image a single slice.

14. This scanner uses a wide fan beam and a rotating array of detectors. a. First generation b. Second generation c. Third generation d. Fourth generation ANS: C The third-generation scanner has a fan beam wide enough to cover the entire anatomy (avoiding the translate motion) and an array of hundreds of detectors that rotate opposite from the x-ray tube.

15. This scanner uses a wide fan beam and a fixed array of detectors that encircle the patient. a. First generation b. Second generation c. Third generation d. Fourth generation ANS: D Fourth-generation scanners no longer have to translate because the x-ray beam is wide enough to cover the anatomy but, instead of having an array of detectors that rotate, this scanner’s detectors are fixed in position all around the gantry.

16. Most modern CT scanners are based on _______________________ technology. a. first generation b. second generation c. third generation d. fourth generation

ANS: C Most modern CT scanners are based on third-generation technology.

17. Electron-beam computed tomography (EBCT) was developed to be fast enough to image the a. lungs b. heart c. diaphragm d. skull ANS: B EBCT was the first scanner that could produce an image so quickly it could image the beating heart.

18. This advancement has the tube rotating inside the gantry while the patient and table move through the gantry in a continuous motion. a. Spiral CT b. Multislice CT c. Slip-ring technology d. Cone beam ANS: A Spiral CT uses the slip-ring technology to have the tube continuously rotate and the patient continuously move through the gantry.

19. Spiral scanning collects data for a. one slice at a time b. 5–10 slices at a time c. the entire volume of tissue being imaged d. none of these ANS: C Spiral scanning collects data for an entire volume of tissue.

20. Using a detector array with multiple rows allows this type of scanning. a. Spiral CT b. Multislice CT c. Slip-ring technology d. Cone beam ANS: B Multislice CT uses a detector array with multiple rows, allowing multiple slices to be imaged each rotation of the tube.

21. As a result of using multiple rows of detectors, the x-ray beam becomes wider to accommodate for the divergence of the x-rays and is called a. spiral CT b. multislice CT c. slip-ring technology d. cone beam ANS: D

MSCT has resulted in the need to open up the collimation along the Z-axis, changing the shape of the beam from a fan to a cone.

22. The Z axis for CT scanning goes from a. side to side b. head to foot c. back to front d. none of these ANS: B The Z axis for CT scanning is the direction from the patient’s head to foot.

23. With multislice CT, additional rows of detectors are added along the a. W-axis b. X-axis c. Y-axis d. Z-axis ANS: D With MSCT, additional rows of detector are added along the Z-axis, from the patient’s head to foot.

24. What is the relationship between the number of detector rows in MSCT and radiation dose? a. direct b. direct proportional; c. inverse d. inverse proportional ANS: D Radiation dose is inversely proportional to the number of detector rows in MSCT; as the number of detector rows increases, the dose decreases.

25. Which of the following is true concerning dual source CT units? a. Dual source CT units contain two sets of tube-detector pairs. b. Dual source CT units have quicker scan time. c. Dual source CT units require less patient dose due to their design. d. Dual source CT units have better ability to differentiate between tissues than standard MSCT. e. All of these.

ANS: E Dual Source CT units contain two sets of tube-detector pairs, mounted 90° apart, with one tube for high kVp and one tube for low kVp images. The DSCT system is known for quicker scan time and less patient dose due to its design. DSCT also has a better ability to differentiate between tissues than standard MSCT. 26. The CT gantry houses the a. x-ray tube b. detector array c. collimators d. data acquisition system (DAS)

e. all of these ANS: E The CT gantry contains most of the equipment needed to produce the data for a CT study, including the tube, detectors, data acquisition system (DAS), and two sets of collimators.

27. Another name for the CT table is the a. gurney b. couch c. board d. stretcher ANS: B The CT table is also known as the couch.

28. The x-ray tube used in CT systems has a larger-diameter anode because a. larger images are created with CT b. the tube is bigger and the anode is proportional c. CT studies produce enormous amounts of heat d. it improves spatial resolution ANS: C Having a larger diameter anode helps the tube to withstand the very large amounts of heat produced during CT scanning.

29. The high-frequency generator used in CT scanning has minimal voltage ripple, allowing a. higher kVp to be used b. higher mA to be used c. longer exposures d. more consistent energy x-ray photons ANS: D As in radiographic systems, the high-frequency generator produces a very consistent energy level because of the very small amount of voltage ripple. This results in more consistent energy x-ray photons.

30. For CT, the filter placed between the tube and patient is shaped like a a. hat b. bowtie c. fish d. necktie ANS: B The bowtie filter resembles a bowtie and serves to make the energy level of the remnant radiation more consistent.

31. The collimator that limits the patient’s radiation exposure is the a. prepatient collimator b. postpatient collimator c. bow tie filter d. prepatient collimator and postpatient collimator

ANS: A The prepatient collimator restricts the beam that reaches the patient, limiting radiation exposure.

32. This collimator is located between the tube and the patient a. Prepatient collimator b. Postpatient collimator c. Bowtie filter d. Prepatient collimator and postpatient collimator ANS: A The prepatient collimator is located between the tube and the patient.

33. The collimator found between the patient and the detectors is the a. prepatient collimator b. postpatient collimator c. bowtie filter d. prepatient collimator and postpatient collimator ANS: B The postpatient collimator is found between the patient and the detectors.

34. The collimator that controls how much of the detector is exposed is the a. prepatient collimator b. postpatient collimator c. bowtie filter d. prepatient collimator and postpatient collimator ANS: B The postpatient collimator controls how much of the detector is exposed.

35. For single-slice CT, this collimator controls the slice thickness a. prepatient collimator b. postpatient collimator c. bowtie filter d. prepatient collimator and postpatient collimator ANS: B The postpatient collimator controls the slice thickness for single-slice CT scanning.

36. The collimator that improves image contrast by limiting the amount of scatter radiation that reaches the detector is the a. prepatient collimator b. postpatient collimator c. prepatient collimator and postpatient collimator d. none of these ANS: B The postpatient collimator improves image contrast by limiting the amount of scatter radiation that reaches the detector.

37. The snapshot of all the transmission measurements at a specific location is the a. view b. ray c. profile d. all of these ANS: A The view is a snapshot of all the transmission measurements at a specific location.

38. The transmission value for a single detector is a a. view b. ray c. profile d. all of these ANS: B The ray is the transmission value for one detector.

39. The composite electrical signal based on transmission values is the a. view b. ray c. profile d. all of these ANS: C The profile is the composite electrical signal based on transmission values.

40. DAS stands for a. digital absorption system b. data acquisition system c. diode acquisition system d. data absorption system ANS: B A key component of CT scanning is the data acquisition system (DAS).

41. The DAS is located a. just below the couch b. inside the x-ray tube c. within the gantry d. in the central processing unit ANS: C The DAS is located within the gantry along with the detectors, tube, generator, and filter.

42. The DAS will a. strengthen the electrical signal b. convert the signal to logarithmic information c. digitize the signal d. all of these ANS: D

The data acquisition system amplifies the electrical signal, converts it to logarithmic form, and converts it from analog to digital information.

43. Which of the following data needs to be known to perform an attenuation calculation?

1. Thickness of the part 2. Original x-ray beam intensity 3. The intensity of the transmitted radiation (as measured by the detector) a. thickness of the part and original x-ray beam intensity only b. thickness of the part and the intensity of the transmitted radiation (as measured by the detector) only c. original x-ray beam intensity and the intensity of the transmitted radiation (as measured by the detector) only d. thickness of the part, original x-ray beam intensity, and the intensity of the transmitted radiation (as measured by the detector) ANS: D By knowing the original x-ray beam intensity, the intensity of the transmitted radiation (as measured by the detector) and the thickness of the part, logarithmic conversion produces attenuation information.

44. The linear attenuation coefficient is a. the measure of the probability that the x-ray beam will interact with the material it is in while traveling in a straight path b. based on both the characteristics of the material and the energy of the x-ray photons c. symbolized by the Greek letter mu, which is written µ d. symbolized by the Greek letter beta, which is written b e. the measure of the probability that the x-ray beam will interact with the material it is in while traveling in a straight path, based on both the characteristics of the material and the energy of the x-ray photons, and symbolized by the Greek letter mu, which is written µ f. the measure of the probability that the x-ray beam will interact with the material it is in while traveling in a straight path, based on both the characteristics of the material and the energy of the x-ray photons, and symbolized by the Greek letter beta, which is written b ANS: E The linear attenuation coefficient (µ) is a measure of the probability that the x-ray beam will interact with the material it is in while traveling in a straight path. This value is based on both the characteristics of the material and the energy of the x-ray photons.

45. The raw data is the a. measure of the radiation detected b. measure of the primary radiation c. digitized linear attenuation coefficient values d. none of these ANS: C Raw data, the basis for the CT image, are the digitized linear attenuation coefficient values.

46. The number of rows and columns that make up the digital image is the a. matrix b. pixel c. voxel d. voxel volume ANS: A The matrix is the number of rows and columns that make up the digital image.

47. The smallest component of the matrix is the a. matrix b. pixel c. voxel d. voxel volume ANS: B The pixel (or picture element) is the smallest piece of the matrix.

48. The small amount of tissue represented by a pixel is the a. matrix b. pixel c. voxel d. voxel volume ANS: C The voxel (or volume element) is the bit of tissue with attenuation characteristics represented in one pixel.

49. This is calculated by knowing the dimensions of the pixel and the slice thickness a. Matrix b. Pixel c. Voxel d. Voxel volume ANS: D The actual volume of the voxel, the voxel volume, has a significant effect on CT image quality.

50. A CT image typically has a ________________ matrix a. 64  64 b. 128  128 c. 256  256 d. 512  512 ANS: D A CT image typically has a 512  512 matrix.

51. The image matrix for CT is __________ than the image matrix for digital radiography. a. smaller b. larger

c. the same as ANS: A The image matrix for CT is smaller than the image matrix for digital radiography.

52. The component of the computer dedicated to performing the enormous number of calculations necessary for CT is the a. array processor b. algorithm c. filtered back projection d. image data ANS: A The component of the computer dedicated to performing the enormous number of calculations necessary for CT is the array processor.

53. The sequence of computer operations for accomplishing a specific task, such as reconstruction is a(n) a. array processor b. algorithm c. filtered back projection d. image data ANS: B An algorithm is the sequence of computer operations for accomplishing a specific task.

54. This is the most common reconstruction algorithm used in CT. a. Array processor b. Algorithm c. Filtered back projection d. Image data ANS: C Filtered back projection is the most common reconstruction algorithm used in CT. 55. The matrix of CT numbers is the a. array processor b. algorithm c. filtered back projection d. image data ANS: D The image data consist of CT numbers as assigned to the pixels of the matrix.

56. Filtered back projection is used to: 1. analyze the raw data for one slice 2. assess image quality for multiple slices 3. to determine the µ of each voxel a. analyze the raw data for one slice and assess image quality for multiple slices only b. analyze the raw data for one slice and to determine the µ of each voxel only

c. assess image quality for multiple slices and to determine the µ of each voxel only d. analyze the raw data for one slice, assess image quality for multiple slices, and to determine the µ of each voxel ANS: B Filtered back projection is used to analyze the raw data for one slice to determine the µ of each voxel.

57. Which of the following is true concerning adaptive statistical iterative reconstruction (ASIR)? a. ASIR is a reconstruction technique that starts reconstruction after a first-pass FBP reconstruction b. ASIR techniques shorten reconstruction time while maintaining much lower image noise than if the same raw data were reconstructed with FBP alone c. ASIR reduces quantum noise substantially with no impact on spatial or contrast resolution d. All of these ANS: D Adaptive statistical iterative reconstruction starts reconstruction after a first-pass FBP reconstruction, and shortens reconstruction time while maintaining much lower image noise than if the same raw data were reconstructed with FBP alone. ASIR reduces quantum noise substantially with no impact on spatial or contrast resolution.

58. The CT number of water is a. 0 b. greater than 0 c. less than 0 d. unable to determine unless more is known ANS: A The formula that converts raw data to CT numbers always calculates water with a CT number of 0.

59. A tissue that attenuates more x-rays than water has a CT number a. 0 b. greater than 0 c. less than 0 d. unable to determine unless more is known ANS: B Tissues that attenuate more x-rays than does water have a positive CT number.

60. A tissue that attenuates fewer x-rays than water has a CT number a. 0 b. greater than 0 c. less than 0 d. unable to determine unless more is known ANS: C Tissues that attenuate fewer x-rays than does water have a negative CT number.

61. Based on the Hounsfield scale, the CT number of air is a. –1000 b. 0 c. 1000 d. None of these ANS: A Based on the Hounsfield scale, the CT number of air is –1000.

62. Based on the Hounsfield scale, the CT number of bone is approximately a. –1000 b. 0 c. 1000 d. None of these ANS: C Based on the Hounsfield scale, the CT number of bone is approximately 1000.

63. The image data is the a. linear attenuation coefficients b. matrix of CT numbers c. same as the raw data d. none of these ANS: B The image data is the matrix of CT numbers.

64. This refers to all the settings that must be determined for a CT imaging study a. Protocol b. SFOV c. DFOV d. Pitch ANS: A Factors such as kVp, mA, slice thickness, and matrix size are some of the factors in the study protocol.

65. The actual anatomy being imaged for the CT study is determined by the a. protocol b. scan field of view (SFOV) c. display field of view (DFOV) d. pitch ANS: B SFOV, scan field of view, is the actual anatomy being imaged.

66. The area of anatomy seen on the monitor is the a. protocol b. scan field of view (SFOV) c. display field of view (DFOV)

d. pitch ANS: C DFOV, display field of view, is the area of anatomy displayed on the monitor.

67. With spiral CT imaging, the relationship between slice thickness and table travel during one tube rotation is the a. protocol b. scan field of view (SFOV) c. display field of view (DFOV) d. pitch ANS: D Pitch, a parameter set with spiral CT studies, is the relationship between slice thickness and table travel during one tube rotation.

68. Pitch ranges from a. 0 to 5 b. 0.5 to 2 c. 1 to 5 d. 2 to 4 ANS: B Pitch ranges from 0.5 to 2.

69. A pitch of 1 indicates that with each rotation of the tube a. the table is moving farther than the slice thickness b. the table is moving less than the slice thickness c. the table is moving the same distance as the slice thickness d. none of these ANS: C A pitch of 1 indicates that with each rotation of the tube the table is moving the same distance as the slice thickness.

70. A pitch of 2 indicates that with each rotation of the tube a. the table is moving farther than the slice thickness b. the table is moving less than the slice thickness c. the table is moving the same distance as the slice thickness d. none of these ANS: A A pitch of 2 indicates that with each rotation of the tube the table is moving farther than the slice thickness.

71. Overlapping slices occur with a a. pitch of 0 b. pitch of 0.5 c. pitch of 1 d. pitch of 2 ANS: B

A pitch of 0.5 results in overlapping slices.

72. Adding a printed comment or label to the image is a. annotation b. windowing c. ROI d. MPR ANS: A The postprocessing technique that includes adding printed information to the CT image is annotation.

73. This specific area of the image is often selected to perform data analysis a. Annotation b. Windowing c. Region of interest (ROI) d. Multiplanar reformation (MPR) ANS: C The region of interest (ROI) is typically selected for data analysis.

74. The process that allows image data to be displayed in coronal or sagittal planes is a. annotation b. windowing c. region of interest (ROI) d. multiplanar reformation (MPR) ANS: D Multiplanar reformation (MPR) is a postprocessing technique that allows image data to be displayed in coronal or sagittal planes.

75. Determining the CT numbers that will be visible in the image is a. annotation b. windowing c. region of interest (ROI) d. multiplanar reformation (MPR) ANS: B Windowing is a postprocessing technique that has an enormous effect on the types of tissues visualized in the CT image.

76. The range of CT numbers visible in the CT image is determined by the a. window level b. window height c. window width d. window clearance ANS: C The window width determines how many CT numbers are visible in the image.

77. The midpoint of the range of CT numbers to be displayed in the image is the

a. b. c. d.

window level window height window width window clearance

ANS: A The window level is the midpoint of the range of CT numbers to be displayed in the image.

78. An image with a WW of 200 and a WL of 50 produces an image that displays pixels with CT numbers ranging from a. –100 to 100 b. –50 to 100 c. –50 to 150 d. 200 to 250 ANS: C A WW of 200 and WL of 50 indicates that 200 CT numbers will be displayed, 100 above 50 (150) and 100 below 50 (–50).

79. An image with a WW of 600 and a WL of –300 produces an image that displays pixels with CT numbers ranging from a. –300 to 600 b. –600 to 0 c. –300 to 300 d. –100 to –300 ANS: B A WW of 600 and WL of –300 indicates that 600 CT numbers will be displayed, 300 above –300 (0) and 300 below –300 (–600).

80. How will a pixel with a CT number of 250 appear if the WW is 200 and the WL is 50? a. Black b. White c. Gray d. It is impossible to say ANS: B Because the range of CT numbers visible in this image goes from –50 to 150, a pixel with a CT number of 250 (greater than the top of the range) will appear white.

81. How will a pixel with a CT number of –250 appear if the WW is 600 and the WL is –300? a. Black b. White c. Gray d. It is impossible to say ANS: C Because the range of CT numbers visible in this image goes from –600 to 0, a pixel with a CT number of –250 (within the range) will appear gray.

82. The WL should be set

a. b. c. d.

at 0 above 0 below 0 near the CT number of the anatomy to be visualized

ANS: D The WL should be set near the CT number of the anatomy to be visualized.

83. To visualize tissues that are very similar to each other, the a. WW should be narrow b. WW should be wide c. WL should be low d. WL should be high ANS: A To visualize similar tissues the WW must be small or narrow.

84. As the WL changes, the CT image’s a. contrast changes b. brightness changes c. spatial resolution changes d. distortion changes ANS: B Adjusting the WL makes the CT image appear brighter or darker.

85. As the WW changes, the CT image’s a. contrast changes b. brightness changes c. spatial resolution changes d. distortion changes ANS: A Adjusting the WW makes the CT image have more or less contrast.

86. The grainy appearance of a CT image that is due to the random variations of CT numbers is a. noise b. spatial resolution c. contrast resolution d. all of these ANS: A Noise is the grainy appearance of a CT image that is due to the random variations of CT numbers.

87. Using a lower mA increases a. noise b. spatial resolution c. contrast resolution d. all of these ANS: A

A lower mA reduces the number of photons reaching the detector, leading to increased noise.

88. Noise can be reduced by using a a. lower pitch b. higher pitch c. shorter rotation time d. none of these ANS: A Decreasing the pitch increases the number of photons reaching the detector, reducing noise.

89. Also known as high-contrast resolution, this quality determines the system’s ability to distinguish between very different tissue types as they get closer together. a. Noise b. Spatial resolution c. Contrast resolution d. All of these ANS: B Spatial resolution determines the system’s ability to distinguish between very different tissue types as they get closer together.

90. Using a large matrix and thin slices improves a. noise b. spatial resolution c. contrast resolution d. All of these ANS: B A large matrix (with associated small pixels) and thin slices improves spatial resolution.

91. Also known as low contrast resolution, this quality determines the system’s ability to distinguish between very similar tissues a. Noise b. Spatial resolution c. Contrast resolution d. All of these ANS: C Contrast resolution, or low-contrast resolution, determines the system’s ability to distinguish between very similar tissues.

92. A change in any factor that has more x-ray photons reaching the detector will result in an image with decreased a. noise b. spatial resolution c. contrast resolution d. All of these ANS: A

Quantum noise is due to a low level of photons being detected, so any adjustment that increases the number of photons detected will reduce noise.

93. Which of the following statements is true regarding CT imaging? a. Spatial resolution is significantly better than radiographic imaging b. Contrast resolution is significantly better than radiographic imaging c. Noise has minimal effect on the CT image d. All of these statements are true ANS: B The ability to distinguish very similar tissues is much better with CT scanning as compared with radiography.

94. The limiting factor for contrast resolution is a. SID b. selection of window level c. noise d. spatial resolution ANS: C Contrast resolution is limited by noise.

95. This artifact is specific to third-generation scanners. a. Streak artifact b. Ring artifact c. Partial-volume artifact d. Beam-hardening artifact ANS: B The ring artifact appeared in only third-generation scanner images.

96. Metal in the area being imaged produces a a. streak artifact b. ring artifact c. partial-volume artifact d. beam-hardening artifact ANS: A Streak artifacts can be due to patient motion or metal.

97. This shading artifact is due to the beam energy changing (and therefore the CT numbers) as the beam passes through the anatomy a. Streak artifact b. Ring artifact c. Partial-volume artifact d. Beam-hardening artifact ANS: D The beam-hardening artifact is due to the beam energy increasing (and therefore changing the CT numbers) as the beam passes through the anatomy.

98. When the voxel is large enough to contain different types of tissues, the CT number will be an average of the various tissues’ attenuation characteristics. This produces a ____________________ a. streak artifact b. ring artifact c. partial-volume artifact d. beam-hardening artifact ANS: C When the voxel is large enough to contain different types of tissues, the CT number will be an average of the various tissues’ attenuation characteristics. This produces a partial-volume artifact.

99. Using a larger matrix or thinner slice will reduce the a. streak artifact b. ring artifact c. partial-volume artifact d. beam-hardening artifact ANS: C Using a larger matrix or thinner slice reduces the partial-volume artifact because the voxel volume is reduced.

100. The beam-hardening artifact can be reduced by a. setting a lower mA b. using a computer algorithm c. increasing the matrix d. increasing the pitch ANS: B To reduce the beam-hardening artifact a computer algorithm must be used.

101. The CT system’s accuracy can be determined by seeing if a. the CT number of water in a container is 0 b. the CT number of water in a container is –1000 c. the CT number of water in a container is 1000 d. none of these ANS: A A system is accurate if, after imaging a water-filled container, the CT number of water reads 0.

102. The CT system’s uniformity can be determined by seeing if a. the CT number of water throughout a container is 0 b. the CT number of water throughout a container is –1000 c. the CT number of water throughout a container is 1000 d. none of these ANS: A A system is uniform if, after imaging a water-filled container, the CT number of water reads 0 in a variety of locations throughout the container.

103. CT examinations currently account for approximately ________ of the U.S. population’s radiation exposure resulting from medical procedures a. 10% b. 25% c. 50% d. 75% ANS: C CT examinations currently account for approximately 50% of the U.S. population’s radiation exposure resulting from medical procedures.

104. To focus on dose optimization, it is important to balance reductions in patient dose with a. maintaining examination cost b. maintaining image quality c. increasing scan time d. all of these ANS: B To focus on dose optimization, it is important to balance reductions in patient dose with maintaining image quality.

105. Automatic tube current modulation is a method 1. to reduce patient exposure 2. to improve image quality 3. similar to automatic exposure control a. to reduce patient exposure and to improve image quality only b. to reduce patient exposure and similar to automatic exposure control only c. to improve image quality and similar to automatic exposure control only d. to reduce patient exposure, to improve image quality, and similar to automatic exposure control ANS: A Automatic tube current modulation serves to reduce patient exposure by adjusting the mA to the minimum needed throughout the scan to produce a quality image.

106. Angular automatic tube current modulation is a. in-plane b. through-plane c. automatically controls tube current along the X-Y axis d. automatically controls tube current along the Z axis e. in-plane and automatically controls tube current along the X-Y axis f. through-plane and automatically controls tube current along the Z axis ANS: E In angular modulation (in-plane), the automatic control of the tube current is along the X-Y axis.

107. Longitudinal automatic tube current modulation is a. in-plane

b. c. d. e. f.

through-plane automatically controls tube current along the X-Y axis automatically controls tube current along the Z axis in-plane and automatically controls tube current along the X-Y axis through-plane and automatically controls tube current along the Z axis

ANS: F In longitudinal modulation (through-plane), the automatic control of the tube current is along the Z-axis.

108. Automatic control of tube current along all three axes is a. longitudinal modulation b. angular modulation c. angular-longitudinal d. automatic tube current control is not possible along all three axes ANS: C In angular-longitudinal modulation, the automatic control of the tube current is along all three axes.

109. Which of the following measures mean absorbed dose in the scanned object volume, adjusted for weighted index and pitch in MSCT? a. CTDI100 b. CTDIvol c. DLP d. Dose report ANS: B The CT Dose Index measures mean absorbed dose in the scanned object volume, and in MSCT units, is specifically called CTDIvol because it adjusted for weighted index and pitch.

110. The formula for dose length product (DLP) is a. DLP = CTDIvol x scan length of exposure b. DLP = CTDIvol + scan length of exposure c. DLP = CTDIvol – scan length of exposure d. DLP = CTDIvol / scan length of exposure ANS: A The formula for DLP = CTDIvol  scan length of exposure.

111. The dose report a. displays the CTDIvol and DLP indices before and after the CT study is performed b. is sent to PACS with the rest of the patient’s CT study c. is a federal requirement since 2002 d. All of these ANS: D Current CT scanners display the CTDIvol and DLP indices before and after the CT study is performed in a Dose Report. Required since 2002, the dose report is sent to PACS with the rest of the patient’s CT study.

112. Which of the following is correct concerning shielding in CT? a. Shielding may have limited value because of very tight collimation and minimal scatter b. Conventional lead shielding of radiosensitive tissues such as gonads or thyroid can be done when they are located outside of the SFOV c. The shield must completely surround the patient because the tube travels in a circular motion d. All of these ANS: D Because of very tight collimation and minimal scatter, shielding may have limited value, although conventional lead shielding is appropriate outside of the SFOV as long as it encircles the patient.

113. Children require special considerations in CT imaging because a. Children are more sensitive to the effects of radiation and have more years for problems to manifest b. Children require the use of routine pediatric protocols based either on age or weight c. Children should receive alternative procedures without ionizing radiation d. All of these ANS: D Children require special consideration regarding radiation dose in that they are more sensitive to the effects of radiation and have more years for problems to manifest. CT examinations should use routine pediatric protocols based either on age or weight, or the child should be referred to an alternative imaging modality.

114. Which of the following is used to trigger a message when a single planned and confirmed scan is likely to exceed a pre-programmed value (CTDIvol and/or DLP)? a. Dose report b. Dose notification value c. Dose alert value d. All of these ANS: B A Dose Notification Value is used to trigger a message when a single planned and confirmed scan is likely to exceed a pre-programmed value (CTDIvol and/or DLP).

115. Which of the following is used to trigger a message when cumulative dose at a location, plus the dose for the next planned and confirmed scan(s), is likely to exceed a pre-programmed value? a. Dose report b. Dose notification value c. Dose alert value d. All of these ANS: C

A Dose Alert Value is used to trigger a message when cumulative dose at a location, plus the dose for the next planned and confirmed scan(s), is likely to exceed a pre-programmed value.

TRUE/FALSE 1. The major development for the first three generations of CT scanners was the reduction in patient exposure.

ANS: F The major development from the first to the third generation of CT scanners was the reduction in scan time.

2. Electron-beam CT does not use an x-ray tube.

ANS: T The EBCT uses a beam of electrons and tungsten rings surrounding the patient instead of the traditional x-ray tube.

3. EBCT scanners are the only scanners that are fast enough to image the heart.

ANS: F Although once true, technology has advanced so that today there are other types of scanners that are fast enough to image the heart.

4. Prior to spiral CT, the patient moved through the gantry one small increment or step at a time.

ANS: T Prior to the introduction of slip-ring technology, the scanner could complete only one image at a time and then the patient and table would move slightly into the gantry to obtain the next. This ―slice-by-slice‖ process was very time-consuming.

5. The CT gantry is stationary and cannot be tilted.

ANS: F The CT gantry can be tilted.

6. The scintillation-type detector is commonly used in today’s CT scanner.

ANS: T Today’s CT scanner uses scintillation-type detectors that absorb the remnant radiation, produce a proportional flash of light, and then convert the light to an electrical signal.

7. The DAS receives an electrical signal from the detector-photodiode.

ANS: T After the detector absorbs radiation and emits light, the photodiode converts the light to an electrical signal, which goes directly to the DAS.

8. Logarithmic conversion produces the linear attenuation coefficient data.

ANS: T Logarithmic conversion produces the linear attenuation coefficient data.

9. Filtered back projection uses the bowtie filter to reconstruct the CT image.

ANS: F Filtered back projection uses an electronic filter, not the bowtie filter found in the gantry.

10. The CT number is the same as an attenuation coefficient ANS: F While related, the CT number is not an attenuation coefficient.

11. Hounsfield unit is another name for the CT number.

ANS: T CT numbers are also referred to as Hounsfield units, in recognition of Godfrey Hounsfield.

12. Scan time remains constant for all CT studies.

ANS: F Scan time can vary widely and multiple factors affect the time.

13. Image postprocessing results in the image data changing.

ANS: F Once the image data has been determined, the actual CT numbers cannot be changed.

14. The result of changing window level and width of the CT image is that the image data has been permanently changed.

ANS: F Postprocessing techniques, including windowing, does not change the image data.

15. The edge enhancement and smoothing filters change the visibility of noise and the spatial resolution.

ANS: T Using a smoothing filter makes structures less visible, including noise.

16. Pediatric protocols are unnecessary because the equipment will always adjust for the smaller patient size.

ANS: F Pediatric protocols should be a part of every CT department. This very sensitive population needs to have extra precautions taken to reduce dose.

17. Structures such as the breast or thyroid cannot be shielded when they are in the SFOV because an artifact would be created.

ANS: F Although lead (radiographic) shields cannot be used in the SFOV, there are shields made specifically for CT that can reduce exposure to these radiosensitive structures, even when in the SFOV. 18. Because of the minimal amount of scatter radiation produced during a CT scan, there is no need to be concerned with the protection of others who must remain in the room during a CT scan.

ANS: F As always, those who are in the room during a CT scan should be protected as appropriate (time, distance, shielding).

19. The patient must be centered in the CT gantry isocenter for accurate imaging of the anatomy.

ANS: T Proper centering assures proper dose distribution, while inaccurate patient centering will degrade the image quality and increase the dose to the patient (especially with ATCM).

20. The dose notification and alert system is designed to optimize dose and terminates the x-ray exposure if values become too high.

ANS: F The dose notification and alert system occurs before patient scanning and can help protect patients from inadvertent use of excessively high CTDIvol and/or DLP. It is not designed to optimize dose and these alerts do not terminate the x-ray exposure, but rather was designed to prevent egregious errors by the technologist or CT unit.