At a state of radioactive equilibrium, the atomic ratio of a highly stable parent element A to its r — Nuclear Chemistry and Radioactivity Chemistry Question
Question
At a state of radioactive equilibrium, the atomic ratio of a highly stable parent element A to its radioactive daughter element B is found to be $3.1 \times 10^9 : 1$. If the half-life of parent A is $2 \times 10^{10}\text{ years}$, what is the approximate half-life of daughter element B?
Answer: A
💡 Solution & Explanation
At equilibrium, $T_{1/2(B)} = T_{1/2(A)} \times (N_B/N_A) = (2 \times 10^{10}) \times (1 / 3.1 \times 10^9) = 20 / 3.1 \approx 6.45\text{ years}$.
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