Consider the following process of decay, _92U^234 -> _90Th^230 + _2He^4; t_1/2 = 2,50,000 years, _90 — Nuclear Chemistry and Radioactivity Chemistry Question
Question
Consider the following process of decay, _92U^234 -> _90Th^230 + _2He^4; t_1/2 = 2,50,000 years, _90Th^230 -> _88Ra^226 + _2He^4; t_1/2 = 80,000 years, _88Ra^226 -> _86Rn^222 + _2He^4; t_1/2 = 1600 years. After the above process has occurred for a long time, a state is reached where for every two thorium atoms formed from _92U^234, one decomposes to form _88Ra^226 and for every two _88Ra^226 formed, one decomposes. The ratio of _90Th^230 to _88Ra^226 formed, will be
💡 Solution & Explanation
Step 1 - Secular Equilibrium in the Chain For the chain: $$\ce{^{234}_{92}U ->[\alpha][250000\ \text{yr}] ^{230}_{90}Th ->[\alpha][80000\ \text{yr}] ^{226}_{88}Ra ->[\alpha][1600\ \text{yr}] ^{222}_{86}Rn}$$ In **secular equilibrium** (all members have constant activity equal to parent's), the rate of decay is the same for all: $$\lambda_1 N_1 = \lambda_2 N_2 = \lambda_3 N_3 = \ldots$$ Step 2 - Find N(Th)/N(Ra) $$\lambda_{Th} N_{Th} = \lambda_{Ra} N_{Ra}$$ $$\frac{N_{Th}}{N_{Ra}} = \frac{\lambda_{Ra}}{\lambda_{Th}} = \frac{t_{1/2}(Th)}{t_{1/2}(Ra)} = \frac{80000}{1600}$$ Step 3 - Evaluate Options - **(A) 250000/80000**: This would be $N(U)/N(Th)$, not $N(Th)/N(Ra)$. Incorrect. - **(B) 80000/1600**: Equal to $N(Th)/N(Ra) = t_{1/2}(Th)/t_{1/2}(Ra)$. **Correct.** - **(C) 250000/1600**: This would be $N(U)/N(Ra)$. Incorrect. - **(D) 251600/8**: Incorrect combination. $$\boxed{\text{Answer: B — }N(\text{Th})/N(\text{Ra}) = 80000/1600 = 50}$$