A Level Physics: Medical Physics — Revision Notes

Condensed recall notes on ultrasound, X-rays, CT and PET scanning – Topic 24, Medical Physics, of Cambridge International AS & A Level Physics 9702, 2025-2027 series. Condensed for the final weeks; for the full explanation, use the Medical Physics study guide.

Ultrasound

Sound above 20 kHz, generated and detected by the piezoelectric effect in a quartz crystal: an alternating p.d. makes the crystal vibrate, and returning vibrations induce an alternating p.d. The same crystal both transmits and receives.

specific acoustic impedance   Z = rho c        kg m^-2 s^-1

intensity reflection coefficient:
  I_r / I_i = (Z_2 - Z_1)^2 / (Z_2 + Z_1)^2

The coupling gel is the standard question. The impedance of air is enormously different from that of skin, so at an air–skin boundary the reflection coefficient is close to 1 and almost all the ultrasound is reflected before it enters the body. Gel has an impedance close to skin’s, so it displaces the air and allows transmission — this is impedance matching.

Worked example. Soft tissue has Z = 1.63 × 10⁶ kg m⁻² s⁻¹, bone has Z = 6.40 × 10⁶ kg m⁻² s⁻¹.

I_r/I_i = (Z2-Z1)^2 / (Z2+Z1)^2 = (6.40-1.63)^2 / (6.40+1.63)^2 = 22.75/64.48 ≈ 0.353

About 35% of the intensity is reflected at a soft tissue–bone boundary, so relatively little penetrates further — structures lying behind bone are poorly imaged by ultrasound.

It is this pulse-echo reflection at tissue boundaries that provides diagnostic information (the current syllabus no longer names A-scan/B-scan terminology directly, but for context: an A-scan displays a single line of amplitude against time; a B-scan combines many such lines into a two-dimensional image, brightness representing echo amplitude).

Attenuation in tissue: I = I₀ e^(−μx) — the same exponential form as for X-rays, below.

X-rays

Produced by accelerating electrons through a large p.d. onto a metal target. Two components in the spectrum:

• Braking radiation (bremsstrahlung) — the continuous background, from electrons decelerating in the target. Its maximum photon energy equals eV, giving the minimum wavelength λ_min = hc/(eV).
• Characteristic lines — sharp peaks from electron transitions in the target atoms; their positions depend on the target material only.

Well under 1% of the input energy becomes X-rays; the rest is heat, which is why the anode rotates and is cooled.

Attenuation

I = I_0 e^(-mu x)

half-value thickness:   x_1/2 = ln2 / mu

μ depends on photon energy and on the material — strongly on atomic number, which is why bone (calcium, Z = 20) absorbs far more than soft tissue (mostly carbon, oxygen and hydrogen) and appears white on the image.

Contrast media — barium (Z = 56) and iodine (Z = 53) are swallowed or injected to make soft-tissue structures such as the gut or blood vessels visible, because their high atomic number gives them a much larger attenuation coefficient than the surrounding tissue.

Image quality: the current syllabus focuses on contrast, improved by choosing photon energy appropriately and by contrast media (edge “sharpness” is background context only: it improves with a narrower beam and a smaller focal spot).

CT scanning

An X-ray tube rotates around the patient, taking many images of the same section from different angles; a computer combines them into a 2D image of that section. Repeating this for successive sections along an axis and combining the 2D images gives a three-dimensional image.

Background, beyond the 9702 outcome:

	Advantage	Disadvantage
CT vs plain X-ray	3-D image, far better soft-tissue contrast, any slice can be viewed	Much higher radiation dose, slower, more expensive

PET scanning

1. A positron-emitting tracer (commonly fluorine-18 in fluorodeoxyglucose) is injected.
2. Each emitted positron travels a short distance and annihilates with an electron.
3. Annihilation produces two gamma photons of 0.511 MeV travelling in exactly opposite directions — required by conservation of momentum.
4. A ring of detectors registers both; the difference in arrival times locates the annihilation along the line between them.

The 0.511 MeV figure comes from E = mc² using the rest mass of an electron. PET shows function and metabolism, not just structure — active tissue such as a tumour takes up more tracer.

Exam traps

• Omitting the reason gel works: impedance matching, not lubrication.
• Confusing the continuous spectrum’s origin with the characteristic lines’.
• Forgetting that λ_min depends only on the accelerating p.d., not the target.
• Using ln2/μ where μx is required, or vice versa.
• Saying the two PET gamma photons travel in the same direction.
• Claiming CT gives a lower dose than a plain X-ray.

Self-test

1. Why is coupling gel essential in ultrasound scanning?
2. What determines the minimum X-ray wavelength, and what determines the characteristic lines?
3. Why does bone appear white on an X-ray image?
4. Give the attenuation equation and the expression for half-value thickness.
5. Why are two gamma photons produced in PET, and what is the energy of each?

Answers: 1. Air and skin have very different acoustic impedances, so almost all the ultrasound would be reflected at the boundary; the gel has an impedance close to skin’s, displacing the air and allowing transmission. 2. λ_min = hc/(eV), set by the accelerating p.d. alone; the characteristic lines depend only on the target material. 3. Calcium has a much higher atomic number than the elements in soft tissue, giving a larger attenuation coefficient, so more X-rays are absorbed. 4. I = I₀e^(−μx); x_½ = ln2/μ. 5. Momentum must be conserved in electron–positron annihilation, so two photons are emitted in opposite directions, each of 0.511 MeV from E = mc² for the electron rest mass.

Related resources

• Medical Physics study guide — the full explanation behind these condensed notes
• Cambridge AS & A Level Physics hub — syllabus 9702 topics and all Marlbridge Physics resources