A Level Physics: Medical Physics — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Medical Physics revision notes

---

Questions

1. Explain how the piezoelectric effect is used to both generate and detect ultrasound. [3]

2. Explain why a coupling gel is essential in ultrasound scanning. [4]

3. The acoustic impedance of soft tissue is 1.63 × 10⁶ and of bone 6.40 × 10⁶ kg m⁻² s⁻¹.

(a) Calculate the intensity reflection coefficient at a soft tissue–bone boundary. [3] (b) Comment on what this means for imaging structures behind bone. [2]

4. X-rays are produced by accelerating electrons through 80 kV onto a metal target.

(a) Calculate the maximum photon energy in joules. [2] (b) Calculate the minimum wavelength produced. [3] (c) Explain how the hardness (penetrating power) and the intensity of the X-ray beam produced can each be controlled independently in the tube. [3]

5. A beam of X-rays passes through 4.5 cm of tissue with attenuation coefficient 0.32 cm⁻¹.

(a) Calculate the fraction of the intensity transmitted. [3] (b) Calculate the half-value thickness. [2]

6. Explain why two gamma photons of 0.511 MeV are produced in a PET scan, and how their detection locates the annihilation. [4]

7. The intensity of an ultrasound beam falls to 60% of its initial value after passing through 4.0 cm of a particular tissue.

(a) Write down the equation for the exponential attenuation of ultrasound intensity with distance travelled. [1] (b) Calculate the attenuation coefficient of the tissue. [2]

8. Outline how a CT scanner builds up a three-dimensional image of an internal structure. [3]

9. X-rays of initial intensity I₀ pass through 5.0 cm of tissue with attenuation coefficient 0.20 cm⁻¹. Calculate the transmitted intensity as a fraction of I₀. [2]

10. State what determines a tissue’s specific acoustic impedance, and explain why a larger impedance mismatch at a boundary produces a stronger ultrasound reflection. [2]

---

Answers

1. An alternating p.d. applied to a piezoelectric crystal makes it vibrate, producing ultrasound [1]. Returning ultrasound makes the same crystal vibrate, which generates an alternating p.d. [1]. The same crystal therefore acts as both transmitter and receiver [1].

2. The acoustic impedance of air differs enormously from that of skin [1], so at an air–skin boundary the intensity reflection coefficient is close to 1 and almost all the ultrasound is reflected [1]. The gel has an impedance close to that of skin [1], displacing the air and allowing the ultrasound to be transmitted into the body — this is impedance matching [1].

3. (a) (Z₂ − Z₁)² ÷ (Z₂ + Z₁)² = (6.40 − 1.63)² ÷ (6.40 + 1.63)² [1] = (4.77)² ÷ (8.03)² = 22.75 ÷ 64.48 [1] = 0.353 [1]. (b) About 35% of the intensity is reflected at the boundary [1], so relatively little penetrates further — structures behind bone are poorly imaged by ultrasound [1].

4. (a) E = eV = 1.60 × 10⁻¹⁹ × 80 000 [1] = 1.28 × 10⁻¹⁴ J [1]. (b) λ_min = hc ÷ E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ 1.28 × 10⁻¹⁴ [1] [1] = 1.55 × 10⁻¹¹ m [1]. (c) Increasing the accelerating p.d. increases the energy of the incoming electrons, and hence the energy (and penetrating power, or hardness) of the X-ray photons produced [1]. Increasing the filament current increases the rate of thermionic emission of electrons, increasing the number of X-ray photons produced per second, i.e. the intensity, without changing their energy [1]. The two controls are therefore independent: p.d. controls hardness, filament current controls intensity [1].

5. (a) I ÷ I₀ = e^(−μx) = e^(−0.32 × 4.5) [1] = e^(−1.44) [1] = 0.237 or 23.7% [1]. (b) x½ = ln2 ÷ μ = 0.693 ÷ 0.32 [1] = 2.17 cm [1].

6. A positron from the tracer annihilates with an electron [1]. Momentum must be conserved, and the initial total momentum is approximately zero, so two photons are emitted in opposite directions [1]. Their energy comes from the rest mass of the electron and positron via E = mc², giving 0.511 MeV each [1]. A ring of detectors registers both, and the difference in their arrival times locates the annihilation point along the line joining the detectors [1].

7. (a) I = I₀e^(−μx) [1]. (b) 0.60 = e^(−μ × 4.0) → ln(0.60) = −4.0μ [1] → μ = −ln(0.60)/4.0 ≈ 0.128 cm⁻¹ [1].

8. The X-ray tube and detectors rotate around the patient so that many X-ray images of the same thin section are taken from different angles [1]; a computer combines these images to obtain a 2-D image of that section [1]; the process is repeated for successive sections along an axis, and the 2-D images of the sections are combined to give a 3-D image [1]. (Background, not required by 9702: compared with a plain X-ray image, CT gives 3-D information and better soft-tissue contrast, but at a much higher radiation dose.)

9. I/I₀ = e^(−μx) = e^(−0.20 × 5.0) [1] = e^(−1.0) ≈ 0.37 [1].

10. A tissue’s specific acoustic impedance depends on its density and the speed of ultrasound through it [1]. The intensity reflection coefficient depends on the difference in impedance between the two tissues at a boundary, so a larger mismatch reflects a greater proportion of the incident intensity [1] — which is exactly why coupling gel, matched closely to skin’s own impedance, is needed before an ultrasound scan can even begin.

---

Where marks are usually lost

• Explaining the gel as a lubricant rather than impedance matching.
• Saying λ_min depends on the target material — it depends only on the p.d.
• Confusing which control affects hardness versus intensity — accelerating p.d. controls hardness (penetrating power), filament current controls intensity.
• Saying the two PET photons travel in the same direction.
• Forgetting that ultrasound attenuation, like X-ray attenuation, is exponential rather than linear.
• Claiming CT scanning gives a lower radiation dose than a plain X-ray image — it is substantially higher.