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

To find the effective half-life ( T_{eff} ), we use the formula:

$\frac{1}{T_{eff}} = \frac{1}{T_{1/2,bio}} + \frac{1}{T_{1/2,phy}}$

Where:

• $T_{1/2,bio} = 185$ minutes (biological half-life)
• $T_{1/2,phy} = 110$ minutes (physical half-life)

Calculating:

$\frac{1}{T_{eff}} = \frac{1}{185} + \frac{1}{110} \approx 0.01459 + 0.00909 = 0.02368$

So,

$T_{eff} \approx \frac{1}{0.02368} \approx 42.24 \text{ minutes}$

Thus, I made a calculation mistake, leading to an approximate effective half-life of around 70 minutes.

A suitable length of time for the patient to relax after injecting the radionuclide would be between 10 to 70 minutes. This allows sufficient time for the radioisotope to circulate in the bloodstream while ensuring that the radioactive decay does not lead to substantial loss of the radionuclide. A recommended duration, such as around 30 minutes, provides a balance between optimal imaging and minimizing patient discomfort.

The decay of the radionuclide results in the emission of a positron. This positron travels a short distance before colliding with an electron, leading to annihilation. When this collision occurs, both the positron and electron are converted into energy in the form of two gamma photons, which are emitted in opposite directions, consistent with the conservation of momentum.

Given that the distance between the center of the head and each detector is approximately 0.1 m, and the speed of photons is $3 \times 10^8 \, m \, s^{-1}$, the time taken for a photon to travel this distance can be calculated using the formula:

$t = \frac{d}{v} \Rightarrow t = \frac{0.1}{3 \times 10^8} \approx 3.33 \times 10^{-10} \text{ seconds}$

Since both photons are created simultaneously, the scanner should measure the time interval between the triggering of the first and second detector to ensure precise localization of the emission event.