A radioactive isotope phosphorous-32 has an initial activity 10 μCi and has half life of 14 days. What is the activity of the source after 42 days?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.25 μCi
The activity of a radioactive source is its number of decays per second; it is proportional to the number of undecayed nuclei, so it halves in every half-life.
\( \displaystyle n = \frac{t}{T_{1/2}} = \frac{42\,\mathrm{days}}{14\,\mathrm{days}} = 3 \text{ half-lives} \)
\( \displaystyle R = \frac{R_0}{2^{n}} = \frac{10\,\mu\mathrm{Ci}}{2^{3}} = \frac{10}{8}\,\mu\mathrm{Ci} = 1.25\,\mu\mathrm{Ci} \)