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

The effective half-life ($T_e$) can be calculated using the formula:

$\frac{1}{T_e} = \frac{1}{T_{h}} + \frac{1}{T_{b}}$ where:

• $T_{h}$ is the physical half-life (110 minutes)
• $T_{b}$ is the biological half-life (185 minutes)

Calculating:

$\frac{1}{T_e} = \frac{1}{110} + \frac{1}{185}$

This results in:

$T_e \approx 68.98 \text{ minutes} \approx 70 \text{ minutes}$

A suitable length of time for the patient to relax could range from 10 to 70 minutes. This allows the radioisotope to circulate and localize in the body while ensuring the patient is comfortable and adequately prepared for the scan.

When a positron emitted from the decay of the radionuclide collides with an electron in the body, they annihilate each other, resulting in the release of energy in the form of two gamma photons. These photons are emitted in opposite directions, in accordance with the conservation of momentum. This process is integral to the detection mechanism in PET scanning.

The scanner must account for the distance the gamma photons travel and the speed of light. Given that the head is 0.2 m across, the time interval ($riangle t$) can be calculated:

Using the speed of light ($c = 3 imes 10^8 ext{ m/s}$), the time taken to travel across the head would be:

$riangle t = \frac{0.2 \text{ m}}{3 \times 10^8 ext{ m/s}} = 6.67 \times 10^{-10} \text{ s}$

The range for the scanner could be adjusted from this value to account for measurement precision and timing variations, thus ensuring accurate positioning information.