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Chapter 32: Medical imaging Learning outcomes You should be able to: ■■ ■■ ■■

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explain how X-ray beams are produced and controlled explain how ultrasound is produced and detected explain how ultrasound images are produced, revealing internal structures describe how conventional and CT scan X-ray images are produced explain the principles of magnetic resonance imaging

Chapter 32: Medical imaging

Applying physics In this book, you have learned many important ideas from physics. You may have noticed that the same big ideas keep reappearing – for example, the idea of a field of force (magnetic, electric, gravitational), or the idea of energy transmitted as waves, or the idea that matter is made of particles with forces acting between them.

This is an important characteristic of physics; ideas that are used in one area prove to be useful in another.

Hopefully, you will see many of these connections now that you are approaching the end of your course.

Physics is also useful. It is applied in many areas of life. In this chapter, we look at one of these areas: medical imaging. This topic covers a range of techniques which doctors use to see inside our bodies. The best known is X-rays, good for showing up bones (Figure 32.1), and the subject of the first part of this chapter. The sections that follow will look at the physics behind two other medical diagnostic techniques: ultrasound scanning and magnetic

Figure 32.1 A radiographer and a doctor examine X-ray images of a patient’s leg at a hospital in Uganda.

resonance imaging. In this chapter, you will make use of several important aspects of physics that you have studied earlier in the course, including sound as a wave, electromagnetic radiation, the behaviour of charged particles, and magnetic fields.

The nature and production of X-rays X-rays are a form of electromagnetic radiation. They belong to the short-wavelength, high-frequency end of the electromagnetic spectrum, beyond ultraviolet radiation (Figure 32.2). They have wavelengths in the range 10−8 m to 10−13 m and are effectively the same as gamma-rays (γ-rays), the difference being in the way they are produced: ■■

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X-rays are produced when fast-moving electrons are rapidly decelerated. As the electrons slow down, their kinetic energy is transformed to photons of electromagnetic radiation. γ-rays are produced by radioactive decay. Following alpha (α) or beta (β) emission, a gamma photon is often emitted by the decaying nucleus (see Chapter 16).

ultraviolet

γ-rays 10–12

The X-rays used in medical applications are usually described as soft X-rays, because their energy is not very great, usually less than the energies of γ-rays produced by radioactive substances.

As with all electromagnetic radiation, we can think of X-rays either as waves or as photons (see Chapter 30).

X-rays travel in straight lines through a uniform medium.

X-ray tube

Figure 32.3a shows a patient undergoing a pelvic X-ray to check for bone degeneration. The X-ray machine is above the patient; it contains the X-ray tube that produces the X-rays which pass downwards through the patient’s body.

Below the patient is the detection system. In this case an electronic detector is being used, but often photographic

visible

X-rays 10–9

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infrared 10–6

microwaves 10–3

radio waves 1

103

Figure 32.2 The electromagnetic spectrum; X-rays and γ-rays lie at the high-frequency, short-wavelength end of the spectrum.

Wavelength / m

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Typically 50 to 200 kV – +

a

vacuum

electrons

filament (cathode)

rotating anode X-rays

Figure 32.4 A simplified diagram of an X-ray tube.

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Figure 32.3 a A general-purpose X-ray system. b A typical X-ray image produced by such a machine, showing the region around the pelvis.

film is used in the detection system. Figure 32.3b shows the resulting image.

Figure 32.4 shows the principles of the modern X-ray tube. The tube itself is evacuated, and contains two electrodes: ■■

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Cathode – the heated filament acts as the cathode (negative) from which electrons are emitted.

Anode – the rotating anode (positive) is made of a hard metal such as tungsten. (The anode metal is often referred to as the ‘target metal’.)

An external power supply produces a voltage of up to 200 kV between the two electrodes. This accelerates a beam of electrons across the gap between the cathode and the anode. The kinetic energy of an electron arriving at the anode is 200 keV. When the electrons strike the anode at high speed, they lose some of their kinetic energy in the

X-ray spectrum

The X-rays that emerge from an X-ray tube have a range of energies, as represented in the X-ray spectra shown in Figure 32.5. Each spectrum has two components, the

characteristic X-rays Relative intensity

b

form of X-ray photons, which emerge in all directions.

Part of the outer casing, the window, is thinner than the rest and allows X-rays to emerge into the space outside the tube. The width of the X-ray beam can be controlled using metal tubes beyond the window to absorb X-rays. This produces a parallel-sided beam called a collimated beam.

Only a small fraction, about 1%, of the kinetic energy of the electrons is converted to X-rays. Most of the incident energy is transferred to the anode, which becomes hot. This explains why the anode rotates; the region that is heated turns out of the beam so that it can cool down by radiating heat to its surroundings. Some X-ray tubes have water circulating through the anode to remove this excess heat.

0

120kV 60kV 0

90kV

50 100 Energy of X-ray photon / keV

150

Figure 32.5 X-ray spectra for a tungsten target with accelerating voltages of 60 kV, 90 kV and 120 kV. The continuous curve shows the braking radiation while the sharp spikes are the characteristic X-rays.

Chapter 32: Medical imaging

broad background ‘hump’ of braking radiation (also known as Bremsstrahlung radiation) and a few sharp ‘lines’ of characteristic radiation. These arise from the different ways in which an individual electron loses its energy when it crashes into the anode.

When an electron strikes the anode, it will be attracted towards the nucleus of an atom in the anode. This will cause it to change both speed and direction – in other words, it decelerates. A decelerating electron (or any other charged particle) loses energy by emitting electromagnetic radiation. The result is a single X-ray photon or, more usually, several photons. The electron interacts with more nuclei until it has lost all its energy and it comes to a halt. The X-rays emitted in this way all contribute to the background braking radiation.

The energy E gained by the electron when it is accelerated through a potential difference of V between the cathode and the anode is given by E = eV. This is the maximum energy that an X-ray photon can have, and so the maximum X-ray frequency fmax that can be produced can be calculated using the formula E = hf. Hence: eV h An electron may cause a rearrangement of the electrons in an anode atom, with an electron dropping from a high energy level to a lower energy level. As it does so, it emits a single photon whose energy is equal to the difference in energy levels. You should recall from Chapter 30 that this is how a line spectrum arises and the photon energies are characteristic of the atom involved. So the characteristic spectral lines of X-rays from a tungsten anode have different energies from those of a molybdenum or copper target. In practice, these characteristic X-rays are relatively unimportant in medical applications. fmax =

QUESTIONS 1 a Summarise the energy changes that take place

in an X-ray tube. b An X-ray tube is operated with a potential difference of 80 kV between the cathode and the tungsten anode. Calculate the kinetic energy (in electronvolts and joules) of an electron arriving at the anode. Estimate the impact speed of such an electron (assume that the electron is non-relativistic).

2 Determine the minimum wavelength of X-rays

emitted from an X-ray tube operated at a voltage of 120 kV.

We can also see from Figure 32.5 that X-rays of a whole range of energies are produced. The lowest energy X-rays will not have sufficient energy to penetrate through the body, so will have no effect on the resulting image.

However, they will contribute to the overall X-ray dose that the patient receives. These X-rays must be filtered out; this is done using aluminium absorbers across the window of the tube.

Controlling intensity and hardness

The intensity of an X-ray beam is a measure of the energy passing through unit area (see the next section). To increase the intensity of a beam, the current in the X-ray tube must be increased. Since each electron that collides with the anode produces X-rays, a greater current (more electrons per second) will produce a beam of greater intensity (more X-ray photons per second). A more intense beam means that the X-ray image will be formed in a shorter time.

Another consideration is the hardness of the X-rays. An X-ray may be thought of as ‘hard’ or ‘soft’. Soft X-rays have lower energies and hence longer wavelengths than hard X-rays. Soft X-rays are less penetrating (they are more easily absorbed) and so they contribute more to the patient’s exposure to hazardous radiation. It is often better to use hard X-rays, which pass through the body more easily.

The hardness of an X-ray beam can be increased by increasing the voltage across the X-ray tube, thereby producing X-rays of higher energies (see Figure 32.5).

Another method is to use a filter which absorbs the lower energy soft X-rays so that the average energy of the X-rays is higher.

X-ray attenuation As you can see if you look back to Figure 32.1, bones look white in an X-ray photograph. This is because they are good absorbers of X-rays, so that little radiation arrives at the photographic film to cause blackening. Flesh and other soft tissues are less absorbing, so the film is blackened.

Modern X-ray systems use digital detectors instead of photographic films. The digital images are easier to process, store and transmit using computers.

X-rays are a form of ionising radiation; that is, they ionise the atoms and molecules of the materials they pass through. In the process, the X-rays transfer some or all of their energy to the material, and so a beam of X-rays is gradually absorbed as it passes through a material.

The gradual decrease in the intensity of a beam of X-rays as it passes through matter is called attenuation.

We will now look at the pattern of attenuation of X-rays as they travel through matter.

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Decreasing intensity

You should recall from Chapter 13 that the intensity of a beam of radiation indicates the rate at which energy is transferred across unit cross-sectional area. Intensity is defined thus: Intensity is the power per unit cross-sectional area.

WORKED EXAMPLE 1 The attenuation (absorption) coefficient of bone is

600 m−1 for X-rays of energy 20 keV. A beam of such X-rays has an intensity of 20 W m−2. Calculate the intensity of the beam after passing through a 4.0 mm thickness of bone.

I0 = 20 W m−2

We can determine the intensity I using the equation: P A where P is power and A is the cross-sectional area normal to the radiation. The unit of intensity is W m−2.

The intensity of a collimated beam of X-rays (i.e. a beam with parallel sides, so that it does not spread out) decreases as it passes through matter. Picture a beam entering a block of material. Suppose that, after it has passed through 1 cm of material, its intensity has decreased to half its original value. Then, after it has passed through 2 cm, the intensity will have decreased to one quarter of its original value (half of a half), and then after 3 cm it will be reduced to one eighth. You should recognise this pattern (1, 12 , 14 , 18 , …) as a form of exponential decay.

We can write an equation to represent the attenuation of X-rays as they pass through a uniform material as follows:

x = 4.0 mm = 0.004 m

I=

I = I0 e−µx where I0 is the initial intensity (before absorption), x is the thickness of the material, I is the transmitted intensity and µ is the attenuation (or absorption) coefficient of the material. Figure 32.6 shows this pattern of absorption. It also shows that bone is a better absorber of X-rays than flesh; it has a higher attenuation coefficient. (The attenuation coefficient also depends on the energy of the X-ray photons.)

The unit of the attenuation coefficient µ is m−1 (or cm−1 etc.).

Now look at Worked example 1.

Intensity of X-rays, I

510

0

air

Step 1 Write down the quantities that you are given; make sure that the units are consistent.

µ = 600 m−1

Step 2 Substitute in the equation for intensity and solve.

Hint: Calculate the exponent (the value of –µx) first.

I = I0 e−µx = 20 × e−(600 × 0.04) = 20 × e−2.4 = 1.8 W m−2

So the intensity of the X-ray beam will have been reduced to about 10% of its initial value after passing through just 4.0 mm of bone.

Half thickness

If we compare the graphs (or equations) for the attenuation of X-rays as they pass through a material with the decay of a radioactive nuclide we see that they are both exponential decays. From Chapter 31, you should become familiar with the concept of the half-life of a radioactive isotope (the time taken for half the nuclei in any sample of the isotope to decay). In a similar manner we refer to the halfthickness of an absorbing material. This is the thickness of material which will reduce the transmitted intensity of an X-ray beam of a particular frequency to half its original value.

QUESTIONS 3 Use the equation I = I0 e−µx to show that the

body

flesh

half-thickness x½ is related to the attenuation coefficient µ by: ln 2 x ½ = μ

4 An X-ray beam transfers 400 J of energy through an

bone 0 Distance into tissue, x

Figure 32.6 The absorption of X-rays follows an exponential pattern.

area of 5.0 cm2 each second. Calculate its intensity in W m−2.

5 An X-ray beam of initial intensity 50 W m−2 is incident

on soft tissue of attenuation coefficient 1.2 cm−1.

Calculate its intensity after it has passed through a 5.0 cm thickness of tissue.