A2 · Practice questions · PET scanning

Two photons, one line.

Six original Cambridge-style questions on positron-emitting tracers, electron-positron annihilation into two 0.511 MeV gamma photons, why they travel in opposite directions, coincidence detection and lines of response.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Annihilation conserves energy and momentum.

Tracer: a positron-emitting (β⁺) isotope with a short half-life, attached to a molecule such as glucose, gathers in the tissue that uses it.
Annihilation: each positron meets an electron and both vanish; their rest energy 2m_ec² becomes two gamma photons of 0.511 MeV each, travelling in opposite directions so that momentum is conserved.
Detection: detectors in a ring record the two photons in coincidence; the line joining them (the line of response) passes through the annihilation point, and many lines reconstruct the tracer distribution.

Analysis

[3 marks]

Describe what happens to a positron emitted by the tracer in a PET scan, state the energy of each photon produced, and explain why the photons travel in opposite directions.

The positron travels a very short distance in the tissue, meets an electron, and the two annihilate: both particles disappear and their mass is converted into energy. ✓
Two gamma photons are produced, each of energy 0.511 MeV, the rest energy m_ec² of an electron. ✓
The electron and positron are almost at rest, so the total momentum before is about zero; two photons of equal momentum must therefore leave in opposite directions for momentum to be conserved. ✓

Calculation

[3 marks]

Show that each photon produced in electron-positron annihilation has an energy of about 0.51 MeV, and calculate the wavelength and frequency of this photon. (Electron mass = 9.11 × 10⁻³¹ kg.)

E = m_ec² = 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 8.20 × 10⁻¹⁴ J = 8.20 × 10⁻¹⁴ / 1.60 × 10⁻¹³ = 0.51 MeV. ✓
λ = hc / E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 8.20 × 10⁻¹⁴ = 2.4 × 10⁻¹² m. ✓
f = E / h = 8.20 × 10⁻¹⁴ / 6.63 × 10⁻³⁴ = 1.2 × 10²⁰ Hz. ✓

Analysis

[2 marks]

Explain what is meant by coincidence detection in a PET scanner and how it leads to a line of response, and explain why a single detected photon on its own gives no useful information about where the tracer is.

Two detectors on opposite sides of the ring register photons at the same instant (within a few nanoseconds); these are taken to be the pair from one annihilation, and the straight line joining the two detectors, the line of response, must pass through the point where it occurred. ✓
One photon alone fixes only the detector it struck, not its direction of travel, so the annihilation could have been anywhere along any line through that detector; only the pair defines a line. ✓

Calculation

[3 marks]

Two detectors A and B face each other across the centre of a PET ring, 0.80 m apart. A pair of annihilation photons is recorded in coincidence, the photon reaching A arriving 0.80 ns before the photon reaching B. Calculate how far from the centre of the ring, and on which side, the annihilation occurred.

If the annihilation is a distance x from the centre toward A, the photon to A travels 0.40 − x and the photon to B travels 0.40 + x; the path difference is 2x. ✓
2x = cΔt = 3.00 × 10⁸ × 0.80 × 10⁻⁹ = 0.24 m. ✓
x = 0.12 m: the annihilation was 12 cm from the centre, on the side nearer detector A (the photon that arrived first had the shorter path). ✓

Calculation

[3 marks]

Nitrogen-13, a positron emitter with a half-life of 10.0 minutes, is produced in a cyclotron with an activity of 1.2 GBq. It takes 25 minutes to prepare the tracer and inject it into the patient. Calculate the decay constant of nitrogen-13 in s⁻¹ and the activity at the moment of injection, and explain what this implies about where the tracer must be made.

λ = ln 2 / t½ = 0.693 / (10.0 × 60) = 1.2 × 10⁻³ s⁻¹. ✓
A = A₀e^(−λt) = 1.2 × 10⁹ × e^(−0.693 × 25 / 10.0) = 1.2 × 10⁹ × 0.177 = 2.1 × 10⁸ Bq. ✓
More than 80% of the activity is lost in 25 minutes, so the tracer must be produced in a cyclotron close to the scanner and used immediately. ✓

Calculation

[4 marks]

In a PET scanner each annihilation produces two gamma photons. (Electron mass = 9.11 × 10⁻³¹ kg.)

(a) Explain why the total energy of the two photons is 1.02 MeV. (b) Calculate the momentum of each photon. (c) Explain why the scanner only accepts photons whose energy is close to 0.511 MeV.

(a) Both the electron and the positron are destroyed, so the energy released is the rest energy of both particles, 2m_ec² = 2 × 0.511 = 1.02 MeV, shared equally between the two photons. ✓
(b) Each photon has E = 0.511 MeV = 8.2 × 10⁻¹⁴ J, so p = E / c = 8.2 × 10⁻¹⁴ / 3.00 × 10⁸. ✓
p = 2.7 × 10⁻²² N s, equal and opposite for the two photons. ✓
(c) A photon that has been scattered in the body has lost energy and changed direction, so its line of response would be wrong; rejecting photons below 0.511 MeV removes these and keeps the image sharp. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.