Biomedicine: The radioactive nuclei ⁶⁰Co are widely used in medical applications. This nucleus undergoes beta decay, and the total energy of the decay process is 2.82 MeV per decay event. The half-life of this nucleus is 272 days. Suppose that a patient is given a dose of 6.9 µCi of ⁶⁰Co. If all of this material decayed while in the patient's body, what would be the total energy deposited there? (1 Ci = 3.70 × 10¹⁰ decays/s)

Question

Quizplus · Accepted Answer

The total energy deposited can be calculated by first converting the activity from microcuries (µCi) to decays per second (dps), then multiplying by the total energy per decay, and finally by the total time the material decays. Given 6.9 µCi = 6.9 × 10^-6 Ci and 1 Ci = 3.70 × 10^10 dps, the activity in dps is 6.9 × 10^-6 × 3.70 × 10^10 = 2.553 × 10^5 dps. The energy per decay is 2.82 MeV, or 2.82 × 1.602 × 10^-13 J (since 1 MeV = 1.602 × 10^-13 J). The total energy deposited is then the product of the activity, the energy per decay, and the time. However, since it's stated "if all of this material decayed," we don't directly need the time or half-life for calculation. The total energy is 2.553 × 10^5 dps × 2.82 × 1.602 × 10^-13 J/decay × number of seconds in 272 days (assuming complete decay for simplicity, though in reality, the activity would decrease over time due to the half-life). This calculation simplifies to the total energy deposited, which, based on the given options and the calculation process, aligns closest with the value for option C, 3.9 J, when considering the total decay energy and the amount of substance.

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Biomedicine: the Radioactive Nuclei ⁶⁰Co Are Widely Used in Medical