The radioactive elements A and B have half-lives of and respectively. An experiment begins with time — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Let initial A = $N_A$, initial B = $4N_A$. Amounts left after time $t$: $N_A(1/2)^{t/15} = 4N_A(1/2)^{t/5}$. Simplify: $(1/2)^{t/15} = (1/2)^{-2} \times (1/2)^{t/5}$. Equate exponents: $t/15 = t/5 - 2$. Multiply by 15: $t = 3t - 30 \Rightarrow 2t = 30 \Rightarrow t = 15\text{ minutes}$.