Consider , where , . Determine the N(Ra)/N(Rn) ratio at secular equilibrium in which has been establ — Nuclear Chemistry and Radioactivity Chemistry Question
Question
Consider $\text{Ra}^{224} \to \text{Rn}^{220} \to \text{Po}^{216}$, where $t_{1/2}(\text{Ra}^{224}) = 3.64 \text{ days}$, $t_{1/2}(\text{Rn}^{220}) = 55 \text{ s}$. Determine the N(Ra)/N(Rn) ratio at secular equilibrium in which $t_{1/2}(\text{parent}) \gg t_{1/2}(\text{daughter})$ has been established.
💡 Solution & Explanation
In secular equilibrium, the activities of parent and daughter are equal: $\lambda_1 N_1 = \lambda_2 N_2$. Therefore, the ratio of atoms is inversely proportional to their decay constants, or directly proportional to their half-lives: $\frac{N_1}{N_2} = \frac{t_{1/2}(1)}{t_{1/2}(2)}$. Converting days to seconds, $t_{1/2}(\text{Ra}) = 3.64 \times 24 \times 3600 = 314496 \text{ s}$. The ratio is $\frac{314496}{55} \approx 5718$. The closest option is 5727. Therefore, correct answer is A.