A hospital uses the radioactive isotope technetium-99m as a tracer - AQA - A-Level Physics - Question 1 - 2021 - Paper 5

Given the short half-life, the technetium-99m must be produced onsite to ensure that it remains active enough for use during medical procedures. Transporting it from a distant facility would result in significant decay, making it ineffective.

Gamma rays can penetrate body tissues with minimal damage, allowing them to escape the body undetected. This reduces radiation exposure to the tissue, making technetium-99m a safer option for imaging. Additionally, the energy and frequency of gamma rays are compatible with the sensitivity of gamma cameras.

When gamma radiation strikes the crystal scintillator, it produces visible light photons. Each photon emitted can cause the photocathode to release an electron. The electrons are accelerated towards the dynode by an applied voltage. As each electron strikes the dynode, it releases multiple secondary electrons. This process creates a cascade effect, significantly amplifying the initial current generated by the action of the photocathode.

To determine if the patient can be released after 10 days, calculate the effective half-life considering biological decay:

The physical half-life is 8.0 days. The biological half-life is 66 days. The effective half-life (T_eff) can be calculated using the formula:

$T_{eff} = \frac{T_{physical} \times T_{biological}}{T_{physical} + T_{biological}} = \frac{8 \times 66}{8 + 66} \approx 7.1 \text{ days}$

After 10 days, the remaining activity can be calculated using the decay formula:

$A = A_0 \left(\frac{1}{2}\right)^{t/T_{eff}}$

where

• $A_0$ = Initial activity = 3.2 GBq,
• $t$ = time = 10 days,
• $T_{eff}$ = 7.1 days.

Substituting the values:

$A = 3.2 \left(\frac{1}{2}\right)^{10/7.1} \approx 1.6 \text{ GBq}$

Since 1.6 GBq is higher than 1.1 GBq, the patient cannot be safely released after 10 days.