OCR A Level Physics A: Gamma camera, PET scanning and ultrasound (H556) – 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 from your board or school.

These questions cover sections 6.5.2 and 6.5.3 of the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017. Revise first with the study guide or the revision notes. The course hub is OCR A Level Physics and the printable checklist lists each statement. Use c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C, h = 6.63 × 10⁻³⁴ J s. Where a question gives no speed, use 1540 m s⁻¹ for ultrasound in soft tissue.

Questions

1. (a) State the type of radiation emitted by technetium-99m and by fluorine-18. [2] (b) Suggest why a half-life of about 6 hours suits technetium-99m. [2]

2. State the function of the collimator, the scintillator and the photomultiplier tubes in a gamma camera. [3]

3. A batch of fluorine-18 has activity 480 MBq when made. It is injected 120 minutes later. The half-life is 110 minutes.

(a) Calculate the activity at injection. [2] (b) Calculate the time taken for the activity to fall to 10% of its initial value. [2]

4. A positron and an electron, both essentially at rest, annihilate. Each photon produced has energy 0.511 MeV.

(a) Calculate this energy in joules. [2] (b) Explain why two photons are produced, travelling in opposite directions. [2] (c) Calculate the wavelength of each photon. [1]

5. Describe how a PET scanner forms an image of the distribution of fluorine-18 in a patient. [4]

6. Explain why a PET scan using a glucose-like tracer shows a growing tumour as a bright region. [3]

7. A hospital is deciding whether to buy a PET scanner. Discuss the issues it should consider. [4]

8. An A-scan trace shows two echoes, at 40 μs and 120 μs after the pulse was sent.

(a) Calculate the depth of each boundary. [2] (b) Calculate the thickness of tissue between the two boundaries. [2]

9. Medium A has density 1010 kg m⁻³ and ultrasound speed 1500 m s⁻¹. Medium B has density 1850 kg m⁻³ and ultrasound speed 3900 m s⁻¹.

(a) Calculate the acoustic impedance of each. [2] (b) Calculate the fraction of intensity reflected when a beam in A meets B at right angles. [2] (c) The incident intensity is 2.5 W m⁻². Calculate the reflected intensity. [1] (d) Explain why it is hard to image structures beyond B. [1]

10. Air has acoustic impedance 410, skin 1.58 × 10⁶ and a scanning gel 1.50 × 10⁶ kg m⁻² s⁻¹.

(a) Calculate the fraction reflected at an air-skin boundary. [2] (b) Calculate the fraction reflected at a gel-skin boundary. [2] (c) Explain how the gel improves the scan. [2]

11. A 2.5 MHz ultrasound beam meets blood at 35° to its direction of flow. The reflected frequency differs from the emitted frequency by 1.1 kHz.

(a) Calculate the speed of the blood. [3] (b) State why the beam is not directed at 90° to the vessel. [1] (c) The probe is moved so that the angle is 50°. The speed is unchanged. Calculate the new frequency shift. [2] (d) Explain the factor 2 in the Doppler equation for this situation. [1]

12. (a) Describe how a transducer emits and detects ultrasound, and how a B-scan image is built up. [5] (b) A scanner must image to a depth of 24 cm. Calculate the shortest time between pulses and the greatest number of pulses per second. [3]

Answers

1. (a) Tc-99m emits gamma radiation [1]; F-18 emits positrons (β⁺) [1]. (b) Long enough to prepare, inject and scan [1]; short enough that the activity and dose fall quickly afterwards [1]. Examiner insight: Name the radiation for each tracer separately. A single phrase such as “radioactive” does not answer a question that asks for the type.

2. The collimator lets through only photons travelling along its holes, so the image is sharp [1]. The scintillator converts each absorbed gamma photon to a flash of light [1]. The photomultiplier tubes convert the light to electrical pulses, amplified by the dynode chain [1]. Examiner insight: Each component needs its own job stated. Describing the camera as a whole earns little for a “state the function” question.

3. (a) Fraction = (1/2)^(120/110) = 0.469 [1]; A = 480 × 0.469 = 225 MBq [1]. (b) 0.10 = (1/2)^n, so n = log 10 / log 2 = 3.32 half-lives [1]; t = 3.32 × 110 = 365 min (6.1 h) [1]. Examiner insight: Keep the time in the same unit as the half-life, and keep unrounded values until the end to avoid drifting from the true answer.

4. (a) E = 0.511 × 10⁶ × 1.60 × 10⁻¹⁹ [1] = 8.18 × 10⁻¹⁴ J [1]. (b) Momentum is conserved and the total before is about zero [1]; one photon would carry momentum, so two with equal and opposite momenta are produced [1]. (c) λ = hc/E = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 8.18 × 10⁻¹⁴ = 2.43 × 10⁻¹² m [1]. Examiner insight: Show the substitution. Quoting only a final number leaves nothing to credit if a slip is made.

5. The patient is surrounded by a ring of detectors [1]. Each annihilation sends two photons in opposite directions, detected by two detectors at almost the same time [1]. The annihilation lies on the line between those two detectors [1]. The computer combines very many lines and finds where they cross most often, building an image of where the tracer is concentrated [1]. Examiner insight: Use the words “opposite directions” and “at the same time” explicitly. Answers that say only “detectors detect gamma rays” miss the idea that gives position.

6. The tracer is attached to a glucose-like molecule that cells take up [1]. Tumour cells use glucose faster, so more F-18 collects there [1]. More annihilations, so more detected photon pairs, come from that region, which appears brighter [1]. Examiner insight: Link each step to the next: uptake, then concentration, then count rate, then brightness. Saying “the tumour is hot” alone is only a restatement.

7. Any four points, including at least one for and one against: the high cost of buying and running the scanner [1]; a nearby supply of F-18 is needed because of its short half-life [1]; PET shows tissue activity and can give an earlier or more accurate diagnosis [1]; the dose to patients and staff must be justified by the benefit [1]. Also acceptable: trained staff, how many patients would benefit, and travel or waiting times without a scanner. Examiner insight: A “discuss” question asks for both sides. A list of costs only is one-sided and leaves out the benefit that justifies the decision.

8. (a) d = ct/2 [1]; d₁ = 1540 × 40 × 10⁻⁶ / 2 = 0.0308 m = 3.08 cm and d₂ = 1540 × 120 × 10⁻⁶ / 2 = 9.24 cm [1]. (b) Difference in depth [1]: 9.24 − 3.08 = 6.16 cm [1]. Examiner insight: The factor of 2 for the there-and-back path is the usual slip, so write d = ct/2 before substituting.

9. (a) Z = ρc: Z_A = 1010 × 1500 = 1.52 × 10⁶ [1]; Z_B = 1850 × 3900 = 7.22 × 10⁶ kg m⁻² s⁻¹ [1]. (b) Ir/I0 = (7.215 − 1.515)² / (7.215 + 1.515)² [1] = 0.426 [1]. (c) Ir = 0.426 × 2.5 = 1.07 W m⁻² [1]. (d) Over 40% is reflected, so less of the beam reaches and returns from deeper structures [1]. Examiner insight: Keep unrounded impedances for part (b). Using the rounded values from part (a) may shift the third figure.

10. (a) (1.58 × 10⁶ − 410)² / (1.58 × 10⁶ + 410)² [1] = 0.999 [1]. (b) (1.58 − 1.50)² / (1.58 + 1.50)² [1] = 6.7 × 10⁻⁴ [1]. (c) The gel fills the air gap between transducer and skin [1]. Its impedance is close to that of skin (impedance matching), so little is reflected and nearly all the ultrasound enters the body [1]. Examiner insight: Give the reason as impedance matching, with the removal of air. “Lubricant” or “helps contact” does not address the physics.

11. (a) v = cΔf / (2f cos θ) [1] = 1540 × 1.1 × 10³ / (2 × 2.5 × 10⁶ × cos 35°) [1] = 0.41 m s⁻¹ [1]. (b) cos 90° = 0, so there would be no shift [1]. (c) Δf ∝ cos θ: Δf = 1.1 × 10³ × cos 50° / cos 35° [1] = 0.86 kHz [1]. (d) The moving blood receives the wave at a shifted frequency and then re-emits at a further shifted frequency, so the shift applies twice [1]. Examiner insight: Convert MHz and kHz to hertz before substituting. Mixing prefixes gives an answer wrong by a power of ten.

12. (a) An alternating p.d. is applied to a piezoelectric crystal [1]; the crystal changes shape and vibrates, emitting a pulse of ultrasound [1]. Echoes make the same crystal vibrate, producing a small p.d. [1]. The delay gives the depth and the echo strength gives the brightness [1]. Many A-scans in different directions are combined by a computer into a two-dimensional image [1]. (b) Time between pulses ≥ 2d/c = 2 × 0.24 / 1540 [1] = 3.1 × 10⁻⁴ s (312 μs) [1]; rate = 1/312 μs = 3.2 × 10³ pulses per second [1]. Examiner insight: A description question is better with a short sequence: emit, echo, delay, brightness, combine. Keep each step to one clear statement.

Where marks are usually lost

• Saying the gamma camera detects positrons or that PET detects a single photon.
• Omitting “opposite directions” or “at the same time” in the PET explanation.
• Calling the gel a lubricant instead of explaining impedance matching.
• Using d = ct instead of d = ct/2.
• Forgetting to square the bracket in the reflection equation.
• Mixing MHz, kHz and Hz in the Doppler equation.
• Forgetting cos θ, or using the angle to the surface instead of to the flow.
• Giving a one-sided answer to a discussion question.

Next steps

• Re-read the revision notes and the study guide.
• Check coverage on the OCR A Level Physics hub and the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR Advanced GCE in Physics A (H556), specification Version 3.0 (March 2026), for first assessment in 2017, published by OCR: sections 6.5.2 and 6.5.3.