Consider Ra^224 -> Rn^220 -> Po^216, where t_1/2(Ra^224) = 3.64 days, t_1/2(Rn^220) = 55 s. Determin — Nuclear Chemistry and Radioactivity Chemistry Question
Question
Consider Ra^224 -> Rn^220 -> Po^216, where t_1/2(Ra^224) = 3.64 days, t_1/2(Rn^220) = 55 s. Determine the N(Ra)/N(Rn) ratio at secular equilibrium in which t_1/2(parent) >> t_1/2(daughter) has been established.
💡 Solution & Explanation
**Step 1: Condition for secular equilibrium.** Secular equilibrium is reached when the parent half-life $t_{1/2}(A)$ is much greater than the daughter half-life $t_{1/2}(B)$. At secular equilibrium, the activities of parent and daughter are equal: $$\lambda_A N_A = \lambda_B N_B$$ **Step 2: Express as a ratio of atom numbers.** $$\frac{N_A}{N_B} = \frac{\lambda_B}{\lambda_A} = \frac{t_{1/2}(A)}{t_{1/2}(B)}$$ **Step 3: Substitute given half-lives.** $$t_{1/2}(A) = 3.64 \text{ days} = 3.64 \times 86400 \text{ s} = 314{,}496 \text{ s}$$ $$t_{1/2}(B) = 55 \text{ s}$$ $$\frac{N_A}{N_B} = \frac{314{,}496}{55} \approx 5727$$ **Physical interpretation:** At secular equilibrium, there are about 5727 atoms of the long-lived parent for every atom of the short-lived daughter — the daughter decays away almost as fast as it is produced. **Answer: A — $N_A/N_B \approx 5727$ at secular equilibrium.**