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:

$\frac{1}{T_{eff}} = \frac{1}{T_{physical}} + \frac{1}{T_{biological}}$

Given:

• Physical half-life, $T_{physical} = 110 ext{ minutes}$
• Biological half-life, $T_{biological} = 185 ext{ minutes}$

Substituting the values, we have:

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

Calculating, [ \frac{1}{T_{eff}} \approx \frac{0.00909 + 0.00541} \approx 0.01450 ] [ T_{eff} \approx \frac{1}{0.01450} \approx 68.97 \text{ minutes} ] Thus, the effective half-life is approximately 70 minutes.

A suitable length of time for relaxation would typically range from 10 to 70 minutes. This allows for adequate distribution of the radioisotope throughout the body and ensures accurate imaging results by minimizing movement and ensuring proper uptake of the substance.

The decay of the radionuclide results in the emission of a positron, which then collides with an electron in the body. This annihilation results in the production of two gamma photons that are emitted in opposite directions. This is due to conservation of momentum, ensuring that the two photons travel in straight paths away from the annihilation point, making them detectable by the PET scanners.

Given the distance between the detectors is 0.2 m, the time interval Δt can be estimated using the speed of gamma photons:

$v = \frac{distance}{time}$

Rearranging gives: $time = \frac{distance}{v} = \frac{0.2 ext{ m}}{3 \times 10^8 ext{ m s}^{-1}}$

Calculating,

$Δt = 6.67 \times 10^{-10} ext{ seconds}$

Therefore, this is the time it takes for a photon to travel the distance between the detectors, and the scanner should measure time intervals close to this value.