A radioactive source (Phosphorus-32) has an initial activity 10 µCi, its half life is 14 days. What time elapses before the activity falls to 1.25 µCi?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 42 days
The activity of a radioactive source falls by half in every half-life, so after \( n \) half-lives \( A = A_0\left(\frac{1}{2}\right)^{n} \), with \( n = t/T_{1/2} \).
\( \displaystyle \frac{A}{A_0} = \frac{1.25}{10} = \frac{1}{8} = \left(\frac{1}{2}\right)^{3} \Rightarrow n = 3 \)
\( t = nT_{1/2} = 3\times14\,\mathrm{days} = 42\,\mathrm{days} \)