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

The half-life of 6 hours necessitates that technetium-99m must be produced on site to ensure a fresh supply for medical use. Any longer transportation times or delays would result in the decay of the isotope, reducing its effectiveness as a tracer.

Technetium-99m emits only gamma rays, which are highly penetrating and can pass through body tissues without causing significant damage. This property allows for the imaging of internal structures while minimizing any harm to the patient, making it a highly suitable tracer for diagnostic procedures.

The photomultiplier tube operates by emitting an electron for each incident photon at the photocathode. These electrons are accelerated towards the dynodes, where each electron collides with a dynode to release more electrons. This cascading effect leads to a significant amplification of the current, allowing for the detection of weak signals.

To determine the effective half-life considering biological half-life:

$t_{effective} = \frac{1}{\frac{1}{t_{physical}} + \frac{1}{t_{biological}}}$

The biological half-life of iodine-131 is 66 days. Thus, we can calculate:

$t_{effective} = \frac{1}{\frac{1}{8.0} + \frac{1}{66}} \approx 7.1 \text{ days}$

Using the decay formula to find the activity after 10 days:

$A = A_0 e^{-\lambda t}$

Where the decay constant ( \lambda ) is given by:

$\lambda = \frac{\ln(2)}{t_{effective}} \approx \frac{\ln(2)}{7.1} \\ A = 3.2 e^{-\lambda imes 10}$

After 10 days, the activity will not have fallen below 1.1 GBq. Hence, the patient should NOT be released from the hospital after 10 days.