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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 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 tracer to accumulate in the body but short enough to limit the patient's exposure to radiation. This ensures that the patient is not under excessive radiation for an extended period. Additionally, this relatively short half-life necessitates that the isotope be produced at the hospital in a generator, ensuring a fresh supply is available when needed.
Step 2
Explain why Technetium-99m emits only gamma rays makes it suitable for use as a tracer.
Answer
Technetium-99m emits only gamma rays, which have high penetration power and can pass through the body without causing significant damage to tissues. This characteristic allows for clearer imaging without the adverse effects associated with other types of radiation. In addition, gamma rays are less ionizing compared to alpha and beta radiation, making them safer for diagnostic imaging purposes.
Step 3
Describe how the current produced by the photocatode is amplified in the photomultiplier tube.
Answer
In the photomultiplier tube, the photocatode emits an electron when struck by a visible light photon released from the crystal scintillator. These emitted electrons are then accelerated towards the dynodes by applying a positive voltage. Each electron that collides with a dynode causes the release of more electrons, effectively amplifying the current. This process is repeated across multiple dynodes, leading to a significant increase in the total number of electrons collected, thus amplifying the current generated by the initial light photon.
Step 4
Determine whether the patient can be safely released from hospital after 10 days.
Answer
To calculate the effective half-life considering decay, we use the decay formula:
t_{eff} = rac{t_{physical}}{1 + rac{t_{biological}}{t_{physical}}}
Given that the physical half-life of iodine-131 is 8 days and the biological half-life is 66 days, we find:
t_{eff} = rac{8}{1 + rac{66}{8}} = 1.1 ext{ days}
After 10 days, the number of half-lives passed is:
t = rac{10}{1.1} ext{ which is roughly } 9.1 ext{ half-lives}
The remaining activity would be:
decay = A_0 imes (0.5)^{n}
Where A_0 = 3.2 GBq, and n = 9.1:
decay = 3.2 imes (0.5)^{9.1} ext{ which is approximately } 1.11 ext{ GBq}.
Since 1.11 GBq > 1.1 GBq, the patient cannot be safely released after 10 days.