A patient is going to have a PET scan - AQA - A-Level Physics - Question 3 - 2017 - Paper 5

The effective half-life ($T_{eff}$) can be calculated using the formula: $\frac{1}{T_{eff}} = \frac{1}{T_{physical}} + \frac{1}{T_{biological}}$

Where:

• $T_{physical} = 110$ minutes
• $T_{biological} = 185$ minutes

Calculating: $\frac{1}{T_{eff}} = \frac{1}{110} + \frac{1}{185}$

Finding a common denominator and solving leads to: $T_{eff} \approx 68.68 \text{ minutes}$

This value rounds to approximately 70 minutes.

A suitable length of time for the patient to relax would be between 10 to 70 minutes. Specifically, allowing around 30 minutes would enable sufficient distribution of the radioisotope throughout the body, ensuring optimal imaging, while considering patient comfort.

When a radionuclide decays, it emits a positron, which is an antiparticle of an electron. When this positron encounters an electron in the body, they annihilate each other. This annihilation results in the release of energy in the form of two gamma photons. These photons are emitted in opposite directions to conserve momentum, allowing detectors to measure their arrival time.

The scanner measures the time interval of Δt between the triggering of the first detector and the triggering of the second detector. Given the distance of 0.2 m between the two detectors, and assuming the speed of gamma photons in the head is approximately $3 \times 10^8$ m/s, Δt can be calculated as follows: $Δt = \frac{distance}{speed} = \frac{0.2 \, m}{3 \times 10^8 \, m/s} \approx 6.67 \times 10^{-10} \, s$

Therefore, the scanner must be able to measure time intervals less than or approximately equal to this value to ensure accurate positioning of the gamma photons.