The radioactive elements and have half-lives of and respectively. The experiment begins with times t — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Let the initial amount of $A$ be $N_A$ and $B$ be $N_B = 4N_A$. The amount remaining after time $t$ is $N(t) = N_0 (1/2)^{t/T_{1/2}}$. Equating the amounts: $N_A (1/2)^{t/15} = 4N_A (1/2)^{t/5}$. Simplifying: $(1/2)^{t/15} = (2^2) (1/2)^{t/5} = (1/2)^{-2} (1/2)^{t/5}$. Equating the exponents: $t/15 = t/5 - 2$. Solving for $t$: $2 = \frac{3t}{15} - \frac{t}{15} = \frac{2t}{15}$, which gives $t = 15\text{ minutes}$.