A radioactive sample decays with a decay constant $\lambda$. The time required for the activity to decrease from $A_0$ to $A_0/e$ is $t_1$, and the time required for the number of nuclei to decrease from $N_0$ to $N_0/e$ is $t_2$. What is the ratio $t_1/t_2$?

Solution

1. The number of nuclei follows the decay law $N(t)=N_0{e}^{-\lambda t}$, and the activity follows $A(t)=A_0{e}^{-\lambda t}$.
2. For the number of nuclei, setting $N(t_2)=N_0/e$ gives $N_0{e}^{-\lambda t_2}=N_0{e}^{-1}$, which implies $t_2=1/\lambda$.
3. For the activity, setting $A(t_1)=A_0/e$ gives $A_0{e}^{-\lambda t_1}=A_0{e}^{-1}$, which implies $t_1=1/\lambda$.
4. The ratio $t_1/t_2$ is therefore $(1/\lambda )/(1/\lambda )=1$.
5. Hence the answer is (B).