A Level Physics 24.3 PET Scanning MCQs – 30 Practice Questions with Answers (Cambridge 9702)
PET (positron emission tomography) scanning is covered in Cambridge AS & A Level Physics 9702 Section 24.3, an A Level (A2) topic in medical imaging. Work through these 30 original multiple-choice questions on positron emission, annihilation, coincidence detection and the medical uses of PET, and check each worked answer as you go.
Formula recap: rest energy E = m c2; photon energy E = h f; λ = c / f; photon momentum p = E / c. For an electron (or positron), the rest energy is 0.51 MeV, so each annihilation photon carries 0.51 MeV = 8.2 × 10−14 J. Constants: c = 3.00 × 108 m/s, h = 6.63 × 10−34 J s, 1 eV = 1.6 × 10−19 J. Full formula list: A Level Physics Formula Sheet.
Positron Emission & Annihilation
Q1. What is a positron?
- (A) A proton with no charge
- (B) The antiparticle of the electron, with the same mass but a charge of +e
- (C) An electron that has lost energy
- (D) A neutron with a positive charge
Show Answer
Answer: B – A positron is the antiparticle of the electron. It has exactly the same mass (9.11 × 10−31 kg) but the opposite charge, +1.6 × 10−19 C.
Q2. Which isotope is classically used to label the glucose-like tracer FDG in PET scanning?
- (A) Carbon-14
- (B) Technetium-99m
- (C) Fluorine-18
- (D) Iodine-131
Show Answer
Answer: C – FDG (fluorodeoxyglucose) is labelled with fluorine-18, a positron emitter with a half-life of about 110 minutes, so the patient’s cells take it up like ordinary glucose.
Q3. In a PET scan, the tracer is best described as:
- (A) A chemical compound labelled with a positron-emitting isotope, which the body handles like the natural substance
- (B) A strong magnet used to line up nuclei in the body
- (C) A dye that blocks gamma rays so organs show up dark
- (D) A source of positrons placed outside the body
Show Answer
Answer: A – The tracer is a biologically useful molecule (such as glucose) in which one atom has been replaced by a positron-emitting isotope. The body cannot tell the difference, so the tracer follows the normal metabolic pathway while its decays reveal where it has gone.
Q4. After a positron is emitted inside tissue, it typically:
- (A) Travels right across the body before stopping
- (B) Annihilates with a nucleus inside the emitting atom
- (C) Travels several metres through the tissue
- (D) Loses its energy over about a millimetre, then annihilates with an electron
Show Answer
Answer: D – The positron slows down through collisions within about a millimetre of its emission point. Once nearly at rest it meets an electron and the pair annihilate. This small range is one limit on the spatial resolution of PET.
Q5. When a positron annihilates with an electron, the products are:
- (A) One gamma photon carrying all the energy
- (B) Two gamma photons
- (C) Two electrons
- (D) One X-ray photon and one electron
Show Answer
Answer: B – The rest mass of both particles is converted into energy, which appears as two gamma photons. A single photon could not conserve both energy and momentum.
Q6. The energy of each gamma photon produced in positron-electron annihilation is:
- (A) 0.51 MeV
- (B) 1.02 MeV
- (C) 0.255 MeV
- (D) 511 MeV
Show Answer
Answer: A – Each photon carries the rest energy of one particle: E = m c2 = 9.11 × 10−31 × (3.00 × 108)2 = 8.2 × 10−14 J = 0.51 MeV.
Q7. What is the frequency of each annihilation photon? (Photon energy = 0.51 MeV)
- (A) 1.2 × 1014 Hz
- (B) 1.2 × 1017 Hz
- (C) 1.2 × 1020 Hz
- (D) 1.2 × 1023 Hz
Show Answer
Answer: C – E = 0.51 × 106 × 1.6 × 10−19 = 8.16 × 10−14 J. Then f = E / h = 8.16 × 10−14 / 6.63 × 10−34 ≈ 1.2 × 1020 Hz.
Q8. The wavelength of each annihilation photon is approximately:
- (A) 2.4 × 10−9 m
- (B) 2.4 × 10−10 m
- (C) 2.4 × 10−11 m
- (D) 2.4 × 10−12 m
Show Answer
Answer: D – λ = c / f = 3.00 × 108 / 1.23 × 1020 ≈ 2.4 × 10−12 m, about 2.4 pm. These are very short-wavelength, highly penetrating gamma rays.
Q9. The total energy released in one positron-electron annihilation is:
- (A) 1.02 MeV
- (B) 0.51 MeV
- (C) 2.04 MeV
- (D) 0 MeV, because mass is destroyed
Show Answer
Answer: A – Both rest masses are converted: 2 × 0.51 MeV = 1.02 MeV, which is 2 × 8.16 × 10−14 = 1.63 × 10−13 J, shared equally between the two photons. Mass is not destroyed; it is converted into energy.
Q10. The two annihilation photons travel in (almost) exactly opposite directions mainly because:
- (A) They repel each other electrically
- (B) The detectors force them apart
- (C) Momentum must be conserved, and the positron-electron pair has almost zero total momentum before annihilation
- (D) Gamma photons always travel in pairs
Show Answer
Answer: C – The positron has slowed to near rest before annihilating, so the pair’s total momentum is almost zero. The two photons have equal energies, hence equal momenta (p = E / c), so their momenta cancel only if they leave in opposite directions, 180° apart.
Detecting the Annihilation Photons
Q11. In a PET scanner, the gamma detectors are arranged:
- (A) In a single flat panel behind the patient
- (B) In a ring surrounding the patient
- (C) Inside the patient’s body
- (D) In two rows above the patient only
Show Answer
Answer: B – Thousands of scintillation detectors form a complete ring around the patient, so photon pairs travelling in any direction across the ring can be caught by detectors on opposite sides.
Q12. Coincidence detection in PET means that an annihilation event is recorded when:
- (A) One detector records two photons in a row
- (B) A photon is absorbed by the patient’s tissue
- (C) The tracer emits a positron and a gamma photon together
- (D) Two photons are detected at opposite detectors at (almost) the same time
Show Answer
Answer: D – The two annihilation photons leave together in opposite directions, so they reach detectors on opposite sides of the ring within a few nanoseconds of each other. Only such near-simultaneous pairs are accepted as genuine events.
Q13. A recorded coincidence event tells us that the annihilation took place:
- (A) Somewhere along the straight line joining the two detectors that fired
- (B) Exactly at the centre of the ring
- (C) At the position of one of the detectors
- (D) Somewhere on the skin surface of the patient
Show Answer
Answer: A – The straight line between the two firing detectors is called the line of response. The annihilation must have occurred somewhere along that line, but a basic coincidence measurement alone does not say exactly where along it.
Q14. How can the position of the annihilation along the line of response be pinned down more precisely?
- (A) By using a stronger tracer
- (B) By measuring the energies of the two photons
- (C) By measuring the tiny difference in arrival times of the two photons
- (D) By counting how many photons each detector receives
Show Answer
Answer: C – If the annihilation is off-centre, the photon heading for the nearer detector arrives slightly earlier. This time-of-flight difference (a fraction of a nanosecond) locates the event along the line: the offset from the centre is c × Δt / 2.
Q15. The detector ring has a diameter of 0.80 m. An annihilation occurs 0.10 m from the centre, towards one detector, so one photon travels 0.30 m and the other 0.50 m. The difference in their arrival times is:
- (A) 6.7 × 10−9 s
- (B) 6.7 × 10−10 s
- (C) 3.3 × 10−10 s
- (D) 1.3 × 10−9 s
Show Answer
Answer: B – Δt = (0.50 − 0.30) / c = 0.20 / 3.00 × 108 = 6.7 × 10−10 s. Measuring sub-nanosecond timing differences like this is what makes time-of-flight PET possible.
Q16. The final PET image is produced by:
- (A) A computer combining millions of recorded lines of response to reconstruct the tracer distribution
- (B) Photographic film exposed by the gamma photons
- (C) The detectors printing out the position of each decay directly
- (D) A single detector scanning across the patient once
Show Answer
Answer: A – Each coincidence event adds one line of response. With millions of such lines at all angles, a computer reconstructs cross-sectional slices, and stacking the slices gives a 3D map of where the tracer has concentrated.
Q17. A gamma photon is detected with no matching partner photon arriving at the opposite side. This detection is:
- (A) Used to mark the edge of the image
- (B) Stored and used on its own
- (C) Counted as half an annihilation
- (D) Rejected, as it cannot define a line of response and is likely background or a scattered photon
Show Answer
Answer: D – Single, unpaired detections carry no positional line information and usually come from background radiation or photons scattered inside the body. The coincidence requirement filters them out, which is a major reason PET images are so clean.
Q18. The momentum of each 0.51 MeV annihilation photon is:
- (A) 2.7 × 10−28 kg m/s
- (B) 8.2 × 10−22 kg m/s
- (C) 2.7 × 10−22 kg m/s
- (D) 2.4 × 10−5 kg m/s
Show Answer
Answer: C – p = E / c = 8.16 × 10−14 / 3.00 × 108 = 2.7 × 10−22 kg m/s. The two photons carry equal momenta in opposite directions, so the total momentum after annihilation is zero, matching the pair before.
Q19. PET can locate annihilations without heavy lead collimators (unlike a gamma camera) because:
- (A) Gamma photons are not affected by lead
- (B) The tracer emits photons in only one direction
- (C) The opposite directions of the two photons mean the pair of detectors that fire already defines the line along which the event occurred
- (D) The ring of detectors is made of lead
Show Answer
Answer: C – This is sometimes called electronic collimation. Because the photons travel back-to-back, the line of response is fixed by the two detectors recording the coincidence, so no physical collimator is needed to select a direction. This makes PET far more sensitive than collimated imaging.
Q20. Two photons from unrelated decays happen to reach opposite detectors at nearly the same time by chance. This causes:
- (A) No problem, because it can never happen
- (B) The scanner to stop working
- (C) A double exposure on the film
- (D) A false (random) coincidence, which adds background noise to the image
Show Answer
Answer: D – Random coincidences create lines of response where no annihilation occurred, adding background to the image. Keeping the coincidence time window very short (a few nanoseconds) keeps this effect small.
PET Imaging in Medicine
Q21. FDG accumulates most strongly in tissues that:
- (A) Use a lot of glucose, i.e. have high metabolic activity
- (B) Contain the most fat
- (C) Have the lowest blood supply
- (D) Are made of bone
Show Answer
Answer: A – FDG is treated like glucose by cells, so it builds up where glucose use is greatest, such as the brain, the heart muscle and many tumours. Those regions then emit more positrons and appear bright on the scan.
Q22. The main difference between a PET scan and a CT scan is that PET shows:
- (A) Sharper outlines of bones
- (B) Only the surface of the body
- (C) Function and metabolic activity, whereas CT mainly shows anatomical structure
- (D) Images in colour rather than grey
Show Answer
Answer: C – PET maps where the tracer is being used, so it reveals how tissue is functioning. CT uses X-ray attenuation to show structure. A diseased area may be metabolically abnormal before its structure changes, which is why PET can detect some conditions earlier.
Q23. In a combined PET-CT scanner, the two images are used together so that:
- (A) The CT image deletes the PET image
- (B) The metabolic (function) information from PET is overlaid on the detailed anatomical image from CT
- (C) The patient only needs half the tracer dose of X-rays
- (D) The scanner can produce ultrasound images as well
Show Answer
Answer: B – PET alone has rather poor spatial detail, so its bright “hot spots” can be hard to place. Overlaying them on the CT scan’s sharp anatomy shows exactly which structure is abnormally active.
Q24. Fluorine-18 has a half-life of only about 110 minutes. A direct consequence is that the tracer:
- (A) Can be stored for months before use
- (B) Gives the patient no radiation dose
- (C) Cannot be used in humans
- (D) Must be produced close to the scanner, usually in a nearby cyclotron, and used soon after manufacture
Show Answer
Answer: D – With a half-life under two hours, the activity falls too fast for the isotope to be shipped over long distances. Hospitals therefore produce fluorine-18 on site or nearby in a cyclotron shortly before each scanning session.
Q25. A sample of fluorine-18 tracer is prepared at 09:00. What fraction of its activity remains at 14:30, which is 330 minutes later? (Half-life = 110 minutes)
- (A) 25%
- (B) 12.5%
- (C) 6.25%
- (D) 37.5%
Show Answer
Answer: B – 330 / 110 = 3 half-lives. The fraction remaining is (1/2)3 = 1/8 = 12.5%. This rapid decay limits working time but also means the patient’s dose falls quickly after the scan.
Q26. The annihilation gamma photons can be detected outside the patient’s body because they:
- (A) Are high-energy photons that are only weakly absorbed by tissue, so most escape the body
- (B) Are reflected by the skin surface
- (C) Travel through the scanner cables
- (D) Have a very long wavelength
Show Answer
Answer: A – At 0.51 MeV the photons are highly penetrating and interact only weakly with soft tissue, so a large fraction pass out of the body and reach the detector ring. It is this escaping radiation that carries the information used to build the image.
Q27. A tracer in a patient has an activity of 2.0 × 108 Bq, and each decay leads to one annihilation. The number of gamma photons produced per second is:
- (A) 2.0 × 108
- (B) 1.0 × 108
- (C) 4.0 × 108
- (D) 8.0 × 108
Show Answer
Answer: C – An activity of 2.0 × 108 Bq means 2.0 × 108 decays (and annihilations) per second. Each annihilation produces two photons, so 2 × 2.0 × 108 = 4.0 × 108 gamma photons per second.
Q28. Which is a genuine limitation of PET scanning?
- (A) It cannot produce cross-sectional images
- (B) It uses no ionising radiation at all
- (C) It can only image bones
- (D) Its spatial resolution is poorer than CT or MRI, and it is expensive
Show Answer
Answer: D – PET resolution is limited (among other things by the short range of the positron before annihilation) to a few millimetres, worse than CT or MRI. The scanner and on-site tracer production also make PET one of the most expensive imaging methods.
Q29. The radiation dose to the patient from a PET tracer falls quickly after the scan mainly because:
- (A) The gamma photons are harmless
- (B) The tracer isotope has a short half-life, so its activity drops rapidly (and the tracer is also excreted)
- (C) The detectors absorb the radiation from the patient
- (D) The positrons are electrically neutral
Show Answer
Answer: B – Fluorine-18 decays with a half-life of about 110 minutes, so within hours only a small fraction of the activity remains, and the tracer is also cleared from the body. A short half-life is deliberately chosen to keep the dose down while still giving enough counts during the scan.
Q30. Many tumours appear as bright “hot spots” on an FDG PET scan because cancer cells often:
- (A) Have a higher metabolic rate and take up more glucose than normal tissue
- (B) Contain more calcium than normal tissue
- (C) Block gamma photons completely
- (D) Emit positrons naturally
Show Answer
Answer: A – Rapidly dividing cancer cells typically consume glucose at a much higher rate than healthy tissue. They therefore accumulate more FDG, produce more annihilations, and show up as regions of high count rate on the scan.
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