decays to by emission. The resulting is in its excited state and comes to the ground state by emitti — Nuclear Chemistry and Radioactivity Chemistry Question
Question
$\text{Co}^{57}$ decays to $\text{Fe}^{57}$ by $\beta+$ emission. The resulting $\text{Fe}^{57}$ is in its excited state and comes to the ground state by emitting $\gamma$-rays. The half-life of $\beta^+$ decay is 270 days and that of the $\gamma$-emission is $10^{-8} \text{ s}$. A sample of $\text{Co}^{57}$ gives $5.0 \times 10^9$ gamma rays per second. How much time will elapse before the emission rate of gamma rays drops to $2.5 \times 10^9$ per second?
💡 Solution & Explanation
The half-life of the parent isotope ($\text{Co}^{57}$) is vastly greater than the daughter's half-life ($270 \text{ days} \gg 10^{-8} \text{ s}$), establishing secular equilibrium. In secular equilibrium, the rate of gamma emission (daughter activity) tracks the parent's activity. The parent's activity drops to half its original value ($5.0 \times 10^9$ to $2.5 \times 10^9$) in one half-life, which is 270 days. Therefore, correct answer is B.