Test bank for radiographic imaging and exposure 5th edition by fauber

Test Bank for Radiographic Imaging and Exposure 5th Edition by Fauber

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Chapter 06: Exposure Technique Factors Fauber: Radiographic Imaging and Exposure, 5th Edition MULTIPLE CHOICE 1. The product of milliamperage and exposure time is ______________________ to the

quantity of x-rays produced a. directly proportional b. inversely proportional c. not related ANS: A

The product of milliamperage and exposure time is directly proportional to the quantity of x-rays produced.

REF: p.148 2. As mAs increases, the quantity of radiation reaching the IR ____________________. a. increases b. decreases c. is not affected ANS: A

As mAs increases, the quantity of radiation reaching the IR increases.

REF: p.149 3. If you were to select 250 mA and 0.5 s on the control panel, how much mAs would be

created? a. 62.5 mAs b. 250 mAs c. 125 mAs d. 500 mAs ANS: C

mAs is calculated by multiplying mA by time in seconds. In this case, 250  0.5 s = 125 mAs. REF: p.149 4. What effect does the level of mAs have on image brightness when using digital image

receptors? a. Increases image brightness. b. Decreases image brightness. c. Does not directly affect image brightness.

ANS: C

The level of mAs does not directly affect image brightness when using digital IRs. During computer processing, image brightness is maintained when the mAs is too low or too high.

REF: p.150

5. Increased quantum noise is seen in a digital image with: a. severely lower-than-needed mAs. b. a little lower-than-needed mAs. c. a little higher-than-needed mAs. d. severely higher-than-needed mAs. ANS: A

Severely lower-than-needed mAs produces an image with increased quantum noise, and higher-than-needed mAs exposes a patient to unnecessary radiation.

REF: p.152 6. The numerical value that is displayed on the processed image to indicate the level of x-ray

exposure received on the digital image receptor is the_______________. a. image brightness b. exposure indicator c. accession number d. optical density ANS: B

The numerical value that is displayed on the processed image to indicate the level of x-ray exposure received on the digital image receptor is the exposure indicator.

REF: p.150 7. With film-screen imaging, which of the following technique factors directly controls the

density produced on the image? a. mAs b. kVp c. SID d. Central ray angle ANS: A

With film-screen imaging, mAs directly controls the density produced on the image.

REF: p.151 8. With film-screen imaging, when mAs is increased, density is _____________; when mAs

is decreased, density is _____________. a. increased; increased b. increased; decreased c. decreased; decreased d. decreased; increased ANS: B

With film-screen imaging, when mAs is increased, density is increased; when mAs is decreased, density is decreased.

REF: p.151

9. When a film image is too light, it has _________ density, and would require a(n) _______

in mAs by 2 on repeat to obtain a diagnostic image. a. insufficient; increase b. insufficient; decrease c. excessive; decrease d. excessive; increase ANS: A

When a film image is too light, it has insufficient density, and would require an increase in mAs by 2 on repeat to obtain a diagnostic image.

REF: p.151 10. With film-screen imaging, what is the minimum change in mAs that would result in a

visible change in radiographic density? a. 10% b. 15% c. 30% d. 100% ANS: C

A 25% to 30% adjustment in mAs is the minimum needed to demonstrate a visible change in density.

REF: p.151 11. If 20 mAs was used for all exposures, which kVp would produce the film-screen image

with the greatest density? a. 70 kVp b. 80 kVp c. 90 kVp d. 100 kVp ANS: D

The higher the kVp, the greater amount of radiation reaching the image receptor and therefore, with film-screen imaging, the greater density.

REF: p.152 12. Which of the following technical factors affects the exposure to the IR by altering the

amount and penetrating ability of the x-ray beam? a. kVp b. mA c. Seconds d. SID ANS: A

kVp affects the exposure to the IR by altering the amount and penetrating ability of the x-ray beam.

REF: p.152

13. With film-screen imaging, what is the primary controller of image contrast? a. kVp b. mA c. Seconds d. SID ANS: A

kVp, by controlling the penetrating power of the x-ray beam, is considered the primary controller of film-screen radiographic contrast.

REF: p.152 14. Increasing the kVp will _________ the IR exposure and density produced on the film

image. a. increase b. decrease c. not affect ANS: A

Increasing the kVp increases the IR exposure and density produced on a film image, and decreasing the kVp decreases the IR exposure and density produced on a film image.

REF: p.153 15. With film-screen imaging, if the kVp were decreased by 15% and no changes were made

to the mAs: a. density would be unchanged, and contrast would decrease. b. density would increase, and contrast would decrease. c. density would decrease, and contrast would increase. d. there would be no change in either contrast or density.

ANS: C

Decreasing kVp results in photons with lower penetrating power. Fewer photons are transmitted, resulting in decreased density, and even fewer penetrate the more dense parts, resulting in a film-screen image with increased contrast.

REF: p.154 16. In order to reduce patient exposure, _____ kVp and _____ mAs should be used when

possible. a. lower, higher b. higher, higher c. higher, lower d. It makes no difference.

ANS: C

Using a higher kVp and lower mAs is best, because the higher kVp provides more penetration, requiring less patient exposure.

REF: p.155

17. With film-screen imaging, what would be the appropriate change in mAs if the kVp were

decreased by 15% and the density needed to be maintained? a. Double the mAs. b. Halve the mAs. c. Use one fourth of the mAs. d. No change would be necessary.

ANS: A

Decreasing the kVp by 15% would require two times the mAs in order to maintain film-screen image density.

REF: p.154 18. With film-screen imaging, which of the following technique factors will produce an image

with the greatest density? a. 300 mA, 0.2 s b. 400 mA, 0.2 s c. 100 mA, 1 s d. 800 mA, 0.001 s ANS: C

mA and exposure time, also expressed as their product (mAs), control radiographic density. 100 mA  1 s (100 mAs) is the highest amount listed and will produce the greatest density.

REF: p.149 19. With film-screen imaging, which of the following technique factors will produce an image

with the greatest density? a. 300 mA, 0.5 s b. 400 mA, 0.5 s c. 100 mA, 0.5 s d. 800 mA, 0.01 s ANS: B

mA and exposure time, also expressed as their product (mAs), control radiographic density. 400 mA  0.5 s (200 mAs) is the highest amount listed and will produce the greatest density.

REF: p.149 20. When a film-screen image needs to be repeated because it is too dark, the minimum

change in mAs needed is: a. reduce the mAs by 30%. b. reduce the mAs by 50%. c. increase the mAs by 30%. d. increase the mAs by a factor of two.

ANS: B

When the image density is so far off that the image must be repeated, the minimum amount of change needed is to either double or halve the mAs.

REF: p.151 21. With film-screen imaging, an abdomen image is done using 60 kVp and 30 mAs. Image

density is appropriate, but the image has too high (short scale) contrast. Which of these exposure factors would be the best change to make? a. 51 kVp and 60 mAs b. 51 kVp and 30 mAs c. 69 kVp and 30 mAs d. 69 kVp and 15 mAs ANS: D

In order to lower the contrast, a higher kVp is needed. However, because the density was appropriate, the mAs has to be reduced to maintain image density.

REF: p.157 22. A high kVp results in: a. less absorption and more transmission in anatomic tissues. b. less variation in the x-ray intensities exiting the patient. c. increase interactions from Compton scattering. d. low-contrast image. e. all of the above.

ANS: E

A high kVp results in less absorption and more transmission in anatomic tissues, which results in less variation in the x-ray intensities exiting the patient (lower subject contrast), producing a low-contrast image. High kVp results in a greater proportion of interactions from Compton scattering than x-ray absorption (photoelectric effect).

REF: p.157 23. During selection of the focal spot size, the radiographer is really determining the: a. angle of the anode used. b. actual size of the filament used. c. energy of electrons available for tube current. d. distance the electrons travel from cathode to anode.

ANS: B

When selecting large or small focal spot at the console, what’s really being selected is the large or small cathode filament.

REF: p.158 24. The distance between the radiation source and the image receptor is the: a. magnification factor. b. SID. c. OID. d. SOD. ANS: B

The distance between the radiation source and the image receptor is the SID, or source to image receptor distance.

REF: p.158 25. The relationship between distance and x-ray beam intensity, specifically that the intensity

of the x-ray beam is inversely proportional to the square of the distance from the source, is the: a. direct square law. b. Gurney-Mott theory. c. inverse square law. d. magnification law.

ANS: C

The relationship between distance and x-ray beam intensity, specifically that the intensity of the x-ray beam is inversely proportional to the square of the distance from the source, is the inverse square law.

REF: p.158 26. If a person stands 3 feet from the source of exposure, receives an exposure of 160 mR, and

then moves to 6 feet from the source of exposure, what would be the new exposure according to the inverse square law? a. 10 mR b. 40 mR c. 640 mR d. 2560 mR ANS: B

According to the inverse square law, increasing the distance from the x-ray source from 3 feet to 6 feet will result in a reduction of exposure from 160 to 40 mR. REF: p.158 27. If a person stands 12 feet from the source of exposure, receives an exposure of 60 mR, and

then moves to 3 feet from the source of exposure, what would be the new exposure according to the inverse square law? a. 15 mR b. 45 mR c. 240 mR d. 960 mR ANS: D

According to the inverse square law, decreasing the distance from the x-ray source from 12 to 3 feet (factor of four) will result in an increase of exposure from 60 to 960 mR. REF: p.158 28. The distance between the object being radiographed and the IR is the: a. magnification factor. b. SID. c. OID. d. SOD.

ANS: C

The distance between the object being radiographed and the IR is the OID or object to image distance.

REF: p.162 29. When the SID is divided by the SOD, what is the result called? a. Magnification factor b. Degree of unsharpness c. OID d. Image width ANS: A

The magnification factor formula is SID divided by SOD. REF: p.163 30. If the first radiograph of a chest is done using 72 in and 12 mAs, and a second radiograph

is done using 40 in, how much mAs should be used to maintain exposure to the IR? a. 2 mAs b. 4 mAs c. 7 mAs d. 9 mAs ANS: B

According to the mAs/distance compensation formula, decreasing the distance from 72 to 40 in will require a decrease of mAs from 12 to 4 in order to maintain exposure to the IR. REF: p.160 31. A quality radiograph is done using 10 mAs, 70 kVp, and a 12:1 ratio grid. How much mAs

is needed to produce an image with the same exposure to the IR when the grid is removed? a. 2 mAs b. 5 mAs c. 15 mAs d. 50 mAs ANS: A

According to the grid conversion chart, when a 12:1 grid is removed, only one fifth of the mAs is needed.

REF: p.169 32. The misrepresentation of the size of an object is: a. shape distortion. b. magnification. c. foreshortening. d. elongation.

ANS: B

One aspect of distortion is magnification, or size distortion.

REF: p.161 33. In order to image a structure that is located anteriorly in the body, it is best radiographed to

minimize magnification by doing a(n) _____ projection. a. posterior-anterior b. anterior-posterior c. lateral d. oblique ANS: A

The posterior-anterior (PA) projection will put the anterior surface of the body closest to the image receptor, reducing OID and magnification.

REF: p.164 34. Magnification is affected by: a. focal spot size. b. OID. c. SID. d. B and C. ANS: D

Both the OID and SID affect magnification, although OID has the greatest effect.

REF: p.165 35. If the image size measures 4.6 in, and the magnification factor is 1.42, what is the size of

the object? a. 2.43 in b. 3.24 in c. 5.63 in d. 6.53 in ANS: B

In order to determine object size, the image size is divided by the magnification factor; in this case 4.6 in is divided by 1.42.

REF: p.165 36. Shape distortion can be created by: a. off-centering the central ray. b. angling the central ray. c. increasing the SID. d. A and B only.

ANS: D

Shape distortion results from misalignment or angulation of the tube, part, or image receptor.

REF: p.165 37. The SOD can be determined by:

a. b. c. d.

adding SID and OID. adding the MF to the OID. subtracting the OID from SID. subtracting the SID from the MF.

ANS: C

The source-to-object distance (SOD) can be calculated by subtracting the OID from the SID. REF: p.163 38. What effect will increasing SID have on recorded detail? a. Increased recorded detail b. Decreased recorded detail c. No effect on recorded detail ANS: A

Increasing SID increases recorded detail by reducing the amount of geometric unsharpness and decreasing size distortion.

REF: p.173 39. For every _____ of part thickness, the radiographer should adjust the mAs by a factor of

two. a. 2 to 3 cm b. 4 to 5 cm c. 6 to 7 cm d. 8 to 9 cm ANS: B

For every increase of 4 to 5 cm of part thickness, the mAs should be doubled. It should be halved when the part is 4 to 5 cm thinner.

REF: p.172 40. If the first radiograph of a foot is done using 4 mAs at 40 in, and a second radiograph is

done using 50 in, how much mAs should be used to maintain exposure to the IR? a. 1 mAs b. 2 mAs c. 6 mAs d. 12 mAs ANS: C

According to the mAs/distance compensation formula, increasing the distance from 40 to 50 in will require an increase of mAs from 4 to 6 in order to maintain exposure to the IR. REF: p.160 41. In film-screen imaging, reducing the area of exposure (increasing collimation) _____

image contrast. a. increases b. decreases

c. has no effect on ANS: A

Increased collimation (a smaller field size) reduces the amount of scatter produced by the tissue and therefore increases the image contrast, because less scatter radiation reaches the image receptor.

REF: p.169 42. Assuming all produce appropriate density images, which of the following would be the

best exposure technique choice when performing a chest radiograph on an infant? a. Use 40 ms exposure time. b. Use 60 ms exposure time. c. Use 200 ms exposure time. d. Use at least 80 kVp. ANS: A

When imaging an infant for a chest radiograph, a short exposure time is critical in order to stop patient motion.

REF: p.169 43. With all other factors remaining the same, a patient with which body habitus would

require the highest exposure factors? a. Asthenic b. Sthenic c. Hypersthenic d. Hyposthenic ANS: C

The hypersthenic patient has a large, stocky build and will require the highest exposure factors.

REF: p.170 44. The property of the x-ray beam that impacts the unsharpness on the radiograph is beam: a. quality. b. quantity. c. divergence. d. restriction.

ANS: C

Due to the divergence of the x-ray beam, a geometric relationship exists among the source of x-rays, the object, and the image receptor.

REF: p.158 45. Once the part has been penetrated, a doubling of the mAs results in ____________ the

exposure to the IR. a. one fourth b. one half c. two times

d. four times ANS: C

Only after the part has been adequately penetrated, doubling mAs results in twice the exposure to the IR. REF: p.172 46. Which of the following mA and exposure time combinations does not produce 12 mAs? a. 200 mA at 0.06 s b. 400 mA at 0.03 s c. 600 mA at 20 ms d. 800 mA at 0.15 s ANS: D

800 mA at 0.15 s results in 120 mAs. REF: p.149 47. For film-screen imaging, density is controlled by: a. mA. b. exposure time. c. mAs. d. all of the above.

ANS: D

mAs, consisting of mA and exposure time, control film-screen density.

REF: p.151 48. With film-screen imaging, increasing the kVp results in an image with ___________

density and ____________contrast. a. decreased; lower b. decreased; higher c. increased; lower d. increased; higher ANS: C

With film-screen imaging, increasing the kVp results in an image with increased density and lower (more gray) contrast.

REF: p.152 49. With digital imaging, if lower than needed kVp is set, the image will have

_______________ brightness and _________________ quantum mottle. a. very low; increased b. appropriate; increased c. excessive; decreased d. unacceptable; decreased ANS: B

With digital imaging, if too low kVp is set, the image will have appropriate brightness (due to the computer making the adjustment) and increased quantum mottle.

REF: p.152 50. With digital imaging, using twice as much mAs as needed will result in an image with

_______________ brightness and _________________ patient exposure. a. appropriate; increased b. appropriate; appropriate c. excessive; increased d. excessive; appropriate ANS: A

With digital imaging, using twice as much mAs as needed will result in an image with appropriate brightness (due to the computer making the adjustment) and increased patient exposure.

REF: p.150 51. Knowing that 6 mAs is appropriate for a knee examination done at 40 in SID, how much

mAs should be used at 56 in SID? a. 1.5 mAs b. 3 mAs c. 12 mAs d. 24 mAs ANS: C

As a rule of thumb, changing from 40 in SID to 56 in SID requires twice as much mAs. REF: p.161 52. Which of the following is not affected by a change in SID? a. Recorded detail b. Image contrast c. Amount of radiation reaching the IR d. Magnification ANS: B

Image contrast is not affected by SID changes.

REF: p.161 53. The radiographic grid is located: a. inside the x-ray tube. b. between the tube and patient. c. between the patient and the IR. d. beneath the IR. ANS: C

In order to absorb the scattered radiation exiting the patient, the grid is located between the patient and IR. REF: p.167

54. With film-screen imaging, decreasing collimation (using a larger field size) results in

____________image contrast and ____________ patient exposure. a. lower; higher b. lower; lower c. higher; higher d. higher; lower ANS: A

With film-screen imaging, decreasing collimation (using a larger field size) results in lower image contrast (due to increased scatter) and higher patient exposure.

REF: p.169 55. Increasing tube filtration results in an x-ray beam with __________________ quantity and

___________ energy. a. less; lower b. less; higher c. more; lower d. more; higher ANS: B

Increasing tube filtration results in an x-ray beam with less quantity (due to increased attenuation by the filter) and higher energy (because the lower-energy photons were absorbed).

REF: p.170 56. As compared to a patient with a 24 cm thick abdomen, imaging a patient with a 20 cm

thick abdomen will result in _____________ scatter being produced and ______________ image contrast. a. less; lower b. more; lower c. less; higher d. more; higher ANS: C

As compared to a patient with a 24 cm thick abdomen, imaging a patient with a 20 cm thick abdomen will result in less scatter being produced and therefore higher image contrast.

REF: p.172 TRUE/FALSE 1. With film-screen imaging, for a given exposure technique, increasing part thickness

decreases radiographic density.

ANS: T

A thicker part absorbs more radiation, resulting in less exit radiation and reduced density on a film-screen image.

REF: p.172 2. Kilovoltage does not affect the quantity of exit radiation until adequate penetration is

reached.

ANS: T

The area of interest must be adequately penetrated before kilovoltage can be adjusted to affect the quantity of exit radiation.

REF: p.152 3. The amount of mAs used has a direct effect on the density of a film-screen image.

ANS: T

Increasing the mAs used with film-screen imaging will increase image density, and decreasing mAs will reduce image density.

REF: p.152 4. Changes in kVp do not affect the digital exposure indicator.

ANS: F

Increasing kVp will increase beam transmission, resulting in more radiation reaching the IR and a change in the exposure indicator.

REF: p.152 5. In order to see the effect of changing kVp, a greater change is needed when operating at

low kVp as compared to high kVp. ANS: F

In order to see the effect of changing kVp, a greater change is needed when operating at high kVp as compared to low kVp. REF: p.152 6. Film-screen images should be produced with very high kVp and very low mAs in order to

reduce patient exposure.

ANS: F

While high kVp and low mAs will reduce patient exposure, film-screen image contrast is dependent on the kVp being used. Using a higher kVp will reduce image contrast, which may negatively impact image quality.

REF: p.155 7. Any increase in magnification results in a decrease in recorded detail.

ANS: T

The greater the magnification, the less the recorded detail.

REF: p.165

Test Bank for Radiographic Imaging and Exposure 5th Edition by Fauber

8. When choosing a grid for an examination, the grid with the highest grid ratio should

always be selected in order to most improve image contrast.

ANS: F

The highest grid ratio will result in a requirement for more mAs. The image contrast improvement and patient exposure need to be balanced.

REF: p.167 9. As compared to a three-phase x-ray unit, a single-phase unit requires higher exposure

techniques to produce a comparable image.

ANS: T

A single-phase x-ray unit operates less efficiently than does a three-phase, resulting in higher exposure factors being used to produce comparable images.

REF: p.169 10. With digital imaging, the same mAs and kVp should be used for an AP lumbar spine and a

lateral lumbar spine.

ANS: F

The thickness of the part is significantly greater for the lateral lumbar spine as compared to the AP projection. Therefore, more radiation is needed in order to produce a quality image.

REF: p.172 11. With digital imaging, exposure factors do not need to be adjusted in order to visualize soft

tissue.

ANS: T

Because computer postprocessing techniques can be used to adjust the digital image brightness and contrast, exposure factors do not need to be adjusted in order to visualize soft tissue.

REF: p.172

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