What is the alpha activity of a 10-kg sample of { }^{235} U that is used in a nuclear reactor?
Strategy We find the number of radioactive atoms by using Avogadro’s number and the gram-molecular weight. We find in Appendix 8 that { }^{235} U has a half-life for emitting α particles of t_{1 / 2}=7.04 \times 10^{8} y. Then we use Equation (12.22) to find the activity.
R=\lambda N(t) (12.22)
The number of { }^{235} U atoms in a 10-kg sample is
N=M \frac{N_{ A }}{M\left({ }^{235} U \right)}
\begin{aligned}N &=(10 kg )\left(\frac{10^{3} g }{1 kg }\right)\left(\frac{6.02 \times 10^{23} \text { atoms } / mol }{235 g / mol }\right) \\&=2.56 \times 10^{25} \text { atoms }=2.56 \times 10^{25} \text { nuclei }\end{aligned}
The activity is
\begin{aligned}R=\lambda N &=\frac{\ln (2) \cdot N}{t_{1 / 2}} \\&=\frac{\ln (2) \cdot\left(2.56 \times 10^{25} \text { nuclei }\right)}{7.04 \times 10^{8} y } \\&=2.52 \times 10^{16} \text { decays } / y =8.0 \times 10^{8} Bq\end{aligned}