Discussion
A radioactive source with half-life \((\tau_{1/2})\) of 69.3 sec has an activity of \(1\mu Ci\). How many radioactive nuclei will remain after time t = 69.3 sec ?
Answer: C
The number remaining is \(1.85 \times 10^{6}\).
The activity fixes the initial number of nuclei through \(A = \lambda N\), with \(\lambda = \ln 2/\tau_{1/2}\):
\[ \begin{aligned} \lambda &= \frac{0.693}{69.3\ \text{s}} = 0.01\ \text{s}^{-1} \\ N_{0} &= \frac{A}{\lambda} = \frac{3.7 \times 10^{4}\ \text{s}^{-1}}{0.01\ \text{s}^{-1}} \\ &= 3.7 \times 10^{6} \end{aligned} \]
using \(1\ \mu Ci = 3.7 \times 10^{4}\) decays per second. After one half-life, 69.3 s, half remain: \(1.85 \times 10^{6}\).
\(3.7 \times 10^{6}\) is the initial number, before any decay. The options with \(10^{5}\) slip a power of ten in the curie conversion.
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