Consider the following process of decay, <br> <br> <br> <br> After the above process has occurred fo — Nuclear Chemistry and Radioactivity Chemistry Question
Question
Consider the following process of decay, <br> ${}_{92}\text{U}^{234} \to {}_{90}\text{Th}^{230} + {}_2\text{He}^4; t_{1/2} = 2,50,000 \text{ years}$ <br> ${}_{90}\text{Th}^{230} \to {}_{88}\text{Ra}^{226} + {}_2\text{He}^4; t_{1/2} = 80,000 \text{ years}$ <br> ${}_{88}\text{Ra}^{226} \to {}_{86}\text{Rn}^{222} + {}_2\text{He}^4; t_{1/2} = 1600 \text{ years}$ <br> After the above process has occurred for a long time, a state is reached where for every two thorium atoms formed from ${}_{92}\text{U}^{234}$, one decomposes to form ${}_{88}\text{Ra}^{226}$ and for every two ${}_{88}\text{Ra}^{226}$ formed, one decomposes. The ratio of ${}_{90}\text{Th}^{230}$ to ${}_{88}\text{Ra}^{226}$ will be
💡 Solution & Explanation
Although the text frames the equilibrium rates oddly ("two formed... one decomposes"), for standard radioactive secular equilibrium the ratio of the number of atoms of two successive members is proportional to their half-lives: $N_{\text{Th}} / N_{\text{Ra}} = t_{1/2}(\text{Th}) / t_{1/2}(\text{Ra}) = 80000 / 1600$. Therefore, correct answer is B.