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

To find the effective half-life, we use the formula: $\frac{1}{T_{eff}} = \frac{1}{T_{physical}} + \frac{1}{T_{biological}}$

Given:
$T_{physical} = 110 \text{ min}$
$T_{biological} = 185 \text{ min}$

Calculating the effective half-life: $\frac{1}{T_{eff}} = \frac{1}{110} + \frac{1}{185}$
Computing this gives: $\frac{1}{T_{eff}} \approx \frac{0.00909 + 0.00541}{1} \approx 0.0145$
Therefore, $T_{eff} \approx 68.97 \text{ min} \approx 70 \text{ min}$

A suitable length of time for the patient to relax would be between 10 to 70 minutes, which allows sufficient time for the radioisotope to distribute throughout the bloodstream and ensure optimal imaging results. This also accounts for both the physical and biological half-lives, ensuring a consistent concentration of the isotope in the area of interest.

During the decay of the radionuclide, a positron is emitted. When this positron encounters an electron, they annihilate each other, resulting in the production of two gamma photons. These photons are emitted in opposite directions, conserving momentum, which can then be detected by the sensors positioned around the patient.

The time interval Δt depends on the distance the gamma photons travel and the speed of light. Assuming the distance across the head is 0.2 m and taking the speed of gamma photons as $3 \times 10^8 \mathrm{m/s}$:

$\Delta t = \frac{distance}{speed} = \frac{0.2 \text{ m}}{3 \times 10^8 \text{ m/s}} \approx 6.67 \times 10^{-10} \text{ s}$

The scanner would therefore measure Δt values ranging from 0 up to approximately $6.67 \times 10^{-10} s$, accounting for the maximum time taken between detection events.