Medical physics

All three subtopics here are printed under "A Level subject content" in the 9702 syllabus, and every objective carries the tier "A Level". None of it is AS. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper, and the whole 9702 bank on this site comes from Paper 1, so no practice is tagged to this topic.

Syllabus points

24.1 Production and use of ultrasound

• Understand that a piezo-electric crystal changes shape when a p.d. is applied across it and that the crystal generates an e.m.f. when its shape changes.
• Understand how ultrasound waves are generated and detected by a piezo-electric transducer.
• Understand how the reflection of pulses of ultrasound at boundaries between tissues can be used to obtain diagnostic information about internal structures.
• Define the specific acoustic impedance of a medium as Z = ρc, where c is the speed of sound in the medium.
• Use I_R/I₀ = (Z₁ - Z₂)²/(Z₁ + Z₂)² for the intensity reflection coefficient of a boundary between two media.
• Recall and use I = I₀e^(-μx) for the attenuation of ultrasound in matter.

24.2 Production and use of X-rays

• Explain that X-rays are produced by electron bombardment of a metal target and calculate the minimum wavelength of X-rays produced from the accelerating p.d.
• Understand the use of X-rays in imaging internal body structures, including an understanding of the term contrast in X-ray imaging.
• Recall and use I = I₀e^(-μx) for the attenuation of X-rays in matter.
• Understand that computed tomography (CT) scanning produces a 3D image of an internal structure by first combining multiple X-ray images taken in the same section from different angles to obtain a 2D image of the section, then repeating this process along an axis and combining 2D images of multiple sections.

24.3 PET scanning

• Understand that a tracer is a substance containing radioactive nuclei that can be introduced into the body and is then absorbed by the tissue being studied.
• Recall that a tracer that decays by beta-plus decay is used in positron emission tomography (PET scanning).
• Understand that annihilation occurs when a particle interacts with its antiparticle and that mass-energy and momentum are conserved in the process.
• Explain that, in PET scanning, positrons emitted by the decay of the tracer annihilate when they interact with electrons in the tissue, producing a pair of gamma-ray photons travelling in opposite directions.
• Calculate the energy of the gamma-ray photons emitted during the annihilation of an electron-positron pair.
• Understand that the gamma-ray photons from an annihilation event travel outside the body and can be detected, and an image of the tracer concentration in the tissue can be created by processing the arrival times of the gamma-ray photons.

The piezo-electric transducer

A piezo-electric crystal, usually quartz or lead zirconate titanate, has two complementary properties:

• Apply a potential difference across it and it changes shape.
• Change its shape and it generates an e.m.f. across it.

That symmetry is what makes one device serve as both source and detector.

Generating ultrasound. Apply an alternating p.d. across the crystal and it expands and contracts at the frequency of the supply, so it acts as an oscillating surface producing a longitudinal sound wave. The crystal is cut to a thickness such that its natural frequency matches the applied frequency, so it resonates and the amplitude of the ultrasound is large. Frequencies of 1 to 15 MHz are typical, which is well above the 20 kHz limit of human hearing.

Detecting ultrasound. A returning pulse compresses and stretches the same crystal, which generates an alternating e.m.f. of the same frequency. The size of that e.m.f. indicates the intensity of the reflected pulse, and the time delay indicates how far the pulse travelled.

The transducer works in pulses, not continuously, and the reason is worth stating: it must be silent to listen. A short pulse is sent, then the crystal switches to detection until the echoes have returned, then the next pulse is sent. That is also why the pulse must be short compared with the time between pulses.

How an ultrasound scan produces an image

1. A short pulse of ultrasound is directed into the body.
2. At every boundary between two tissues part of the pulse is reflected and part is transmitted.
3. The reflected pulses return at different times, according to the depth of the boundary that produced them.
4. The time delay gives the depth, since distance is speed multiplied by time, halved because the pulse travels there and back.
5. The amplitude of each echo gives information about the nature of the boundary.
6. Scanning the transducer across the body and combining the returns builds up a two-dimensional image.

Worked example. An echo returns 60 μs after the pulse was sent. The speed of ultrasound in soft tissue is 1500 m s⁻¹.

Total distance travelled:

1500 × 60 × 10⁻⁶ = 0.09

The boundary is half that away:

0.09 / 2 = 0.045

so 4.5 cm below the surface. Forgetting to halve is the standard error here.

Specific acoustic impedance

The specific acoustic impedance of a medium is

Z = ρc

where ρ is the density and c the speed of sound in the medium. Its unit is kg m⁻² s⁻¹.

The fraction of the intensity reflected at a boundary between two media depends only on their impedances:

I_R / I₀ = (Z₁ - Z₂)² / (Z₁ + Z₂)²

Read that expression before using it. The numerator is the difference squared, so:

• If the two impedances are almost equal, the numerator is nearly zero and almost nothing is reflected. The pulse passes on into the deeper tissue, and the boundary is faint in the image.
• If they are very different, the fraction approaches 1 and almost everything is reflected, so nothing gets past the boundary at all.

Both extremes matter clinically. Soft tissues have similar impedances, so boundaries between them reflect weakly, which is why an image needs sensitive detection. The impedance of air is about 400 while that of soft tissue is about 1.6 × 10⁶, so the difference is enormous and effectively all the ultrasound is reflected at a skin-air boundary before it ever enters the body.

That is exactly why a coupling gel is used. Its impedance is close to that of skin, so it displaces the air and allows the ultrasound to enter. Without it a scan shows nothing at all.

Worked example. Fat has Z = 1.34 × 10⁶ and muscle Z = 1.71 × 10⁶ kg m⁻² s⁻¹.

Difference:

1.71 - 1.34 = 0.37

Sum:

1.71 + 1.34 = 3.05

I_R / I₀ = 0.37² / 3.05² = 0.01472

so about 1.5 per cent of the intensity is reflected, and 98.5 per cent carries on deeper. A weak echo, but enough to locate the boundary, and the fact that most of the pulse continues is what allows deeper structures to be seen at all.

Attenuation

As ultrasound passes through matter its intensity falls exponentially, because energy is absorbed and scattered:

I = I₀e^(-μx)

where μ is the linear attenuation coefficient, in m⁻¹, and x the thickness traversed. A large μ means the beam is absorbed in a short distance.

The half-value thickness x½ is the thickness that reduces the intensity to half, and by the same argument as for half-life:

x½ = 0.693 / μ

Producing X-rays

In an X-ray tube, electrons are emitted from a heated cathode by thermionic emission, accelerated through a large potential difference, and strike a metal target, usually tungsten. The electrons decelerate abruptly, and the energy lost appears as X-ray photons.

Most of the energy, over 99 per cent, becomes heat, which is why the target rotates and is cooled.

The maximum photon energy occurs when a single electron loses all its kinetic energy in one interaction. That sets the minimum wavelength:

eV = hc / λ_min, so λ_min = hc / (eV)

Note what this does and does not depend on. The minimum wavelength depends only on the accelerating potential difference, and not on the target material or the tube current. Raising the current makes more X-rays of the same range of energies; raising the voltage makes more energetic ones.

Worked example. An X-ray tube operates at 80 kV.

λ_min = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / (1.60 × 10⁻¹⁹ × 80000) = 1.554 × 10⁻¹¹

so about 1.55 × 10⁻¹¹ m, or 0.016 nm.

X-ray imaging and contrast

An X-ray image is a shadow. X-rays pass through the body and are attenuated by different amounts in different tissues, and what reaches the detector forms the image.

Contrast is the difference in degree of blackening between adjacent regions of the image. Good contrast means neighbouring structures are easy to distinguish.

Bone contains calcium and phosphorus, which have high proton numbers and therefore a large attenuation coefficient, so bone absorbs strongly and appears white. Soft tissues have similar and much smaller attenuation coefficients, so their boundaries show poor contrast.

Two ways of improving it are examined:

• A contrast medium, such as a barium meal for the digestive tract or an iodine-based dye for blood vessels. Both barium and iodine have high proton numbers, so they attenuate strongly and make an organ visible that would otherwise be invisible against its surroundings.
• Adjusting the tube voltage. A lower accelerating voltage gives softer, less penetrating X-rays and therefore greater differences in attenuation between tissues, so better contrast, at the cost of a higher absorbed dose to the patient.

Sharpness is a different property from contrast, and the two are often confused. Sharpness is how well defined an edge is, and it is improved by using a small anode area so the beam comes from close to a point source.

Worked example. Bone has μ = 600 m⁻¹ for a certain beam. What fraction of the intensity passes through 5.0 mm of bone?

μx = 600 × 0.0050 = 3

so I/I₀ = e⁻³, which is 0.0498, meaning about 5 per cent is transmitted and 95 per cent absorbed.

CT scanning

A CT scan builds a three-dimensional image, and the syllabus wants the process in order:

1. The X-ray tube and detectors rotate around the patient in one plane, taking many images of the same section from different angles.
2. A computer combines those images into a single two-dimensional image of that section, in which the position of each feature within the slice is known, which a single X-ray image cannot give.
3. The patient is moved along the axis and the process is repeated for the next section.
4. The stack of two-dimensional sections is combined into a three-dimensional image, which can then be viewed from any direction or sliced in any plane.

The advantages over a single X-ray image are that structures are not superimposed, that soft tissues can be distinguished because the computer can enhance small differences in attenuation, and that the image can be rotated and re-sectioned after the scan. The disadvantage is a much higher radiation dose, since many images are taken, and a longer scan time.

PET scanning

A tracer is a substance containing radioactive nuclei that is introduced into the body, usually by injection, and taken up by the tissue being studied. The standard tracer is fluorine-18 attached to a glucose molecule, which accumulates where metabolic activity is high, such as in a tumour or an active region of the brain.

The tracer decays by beta-plus decay, emitting a positron.

Annihilation

Annihilation occurs when a particle meets its antiparticle: both are destroyed and their mass-energy appears as photons. Both mass-energy and momentum are conserved.

Momentum conservation is why there must be two photons rather than one. The electron and positron are effectively at rest, so the total momentum before is nearly zero. A single photon would carry momentum in some direction, so two photons must be emitted in exactly opposite directions for the momenta to cancel.

The energy of the photons

Each photon carries the rest energy of one particle. The rest mass of an electron and of a positron is 9.11 × 10⁻³¹ kg each.

E = 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 8.199 × 10⁻¹⁴

so about 8.2 × 10⁻¹⁴ J per photon. In electronvolts:

8.199 × 10⁻¹⁴ / (1.60 × 10⁻¹⁹) = 5.124 × 10⁵

which is about 0.51 MeV, the figure quoted for annihilation photons. The total energy released in the event is twice that, about 1.02 MeV.

Building the image

1. The tracer is absorbed by the tissue and its nuclei decay, emitting positrons.
2. Each positron travels a very short distance, a few millimetres at most, before meeting an electron in the surrounding tissue.
3. They annihilate, producing two gamma-ray photons travelling in opposite directions.
4. The photons leave the body and are detected by a ring of detectors around the patient.
5. The difference in arrival times at two opposite detectors locates the annihilation event along the line joining them. A photon arriving earlier at one detector means the event was closer to it.
6. Repeating this for many events builds an image of the concentration of the tracer, and therefore of metabolic activity.

Worked example. Two detectors on opposite sides of a ring of diameter 0.80 m record photons 0.40 ns apart. How far is the annihilation from the centre?

The extra distance travelled by the later photon is

3.00 × 10⁸ × 0.40 × 10⁻⁹ = 0.12

That extra distance is twice the displacement from the centre, because one path is that much longer and the other that much shorter:

0.12 / 2 = 0.06

so the event was 6.0 cm from the centre, on the side of the detector that recorded first.

The point of PET is that it shows function rather than structure. An X-ray or CT scan shows what is there; a PET scan shows what is metabolically active, which is why it is used to find tumours and to study the brain.

Common mistakes

• Forgetting to halve the distance when converting an ultrasound echo time into a depth.
• Saying the coupling gel lubricates the probe. It matches the acoustic impedance and excludes air.
• Saying a large difference in acoustic impedance gives a good image. It reflects nearly everything, so nothing deeper can be seen.
• Forgetting to square both the difference and the sum in the reflection coefficient.
• Saying the minimum X-ray wavelength depends on the target metal or on the tube current. It depends only on the accelerating p.d.
• Confusing contrast with sharpness.
• Saying a contrast medium works because it is radioactive. It works because its atoms have a high proton number and attenuate strongly.
• Describing CT as taking one image from many angles without saying that sections are stacked along an axis to give the third dimension.
• Saying a PET tracer emits gamma rays directly. It emits positrons, and the gamma rays come from annihilation.
• Saying annihilation produces one photon, or two photons in the same direction.
• Using the mass of only one particle when finding the total energy released in annihilation, or using both masses for the energy of a single photon.
• Saying the detectors locate the event by the direction the photons arrive from, rather than by the difference in arrival times along a known line.