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A hospital uses the radioactive isotope technetium-99m as a tracer - AQA - A-Level Physics - Question 1 - 2021 - Paper 5
Question 1
A hospital uses the radioactive isotope technetium-99m as a tracer. Technetium-99m is produced using a Molybdenum-Technetium generator on site at the hospital. 1.1 ... show full transcript
Worked Solution & Example Answer:A hospital uses the radioactive isotope technetium-99m as a tracer - AQA - A-Level Physics - Question 1 - 2021 - Paper 5
Step 1
Explain why the value of the half-life of technetium-99m makes it suitable for use as a tracer and means that it must be produced in a generator on site.
Answer
The half-life of technetium-99m is approximately 6 hours, which is long enough to allow the necessary procedures to take place but short enough to minimize radiation exposure to the patient. This ensures that the patient is not exposed to excessive radiation for long periods. Additionally, the relatively short half-life means that the isotope must be produced on-site, as its activity decays rapidly.
Step 2
Explain why this makes technetium-99m suitable for use as a tracer.
Answer
Technetium-99m emits only gamma rays, which have high penetrative abilities and cause minimal damage to body tissues compared to other forms of radiation. This allows it to pass through the body without causing significant harm, making it ideal for medical imaging purposes.
Step 3
Describe how the current produced by the photocathode is amplified in the photomultiplier tube.
Answer
In the photomultiplier tube, each visible light photon emitted from the crystal scintillator leads to the release of one electron from the photocathode. These electrons are then accelerated towards the dynode, where each electron collides with the dynode to release additional electrons. This avalanche effect multiplies the current, significantly amplifying the initial signal generated by the incoming gamma radiation.
Step 4
Determine whether the patient can be safely released from hospital after 10 days.
Answer
To find out if the patient can be released safely after 10 days, we first calculate the effective half-life considering biological clearance:
The decay constant,
[ A = A_0 \times e^{- rac{t}{T_{1/2}}} ]
Calculating the number of half-lives in 10 days:
- Initial activity of iodine-131 (A0) = 3.2 GBq and half-life = 8 days.
- 10 days is 1.25 half-lives. After 1 half-life: 1.6 GBq, after 2 half-lives: 0.8 GBq.
Thus, after 10 days, since the activity drops below 1.1 GBq (to 0.8 GBq), the patient can be safely released.