Q 12-13-038JEE MainJEE Main 2025 (23 Jan, Shift 1)Easy
A radioactive nucleus $n_2$ has 3 times the decay constant as compared to the decay constant of another radioactive nucleus $n_1$. If the initial number of both nuclei are the same, what is the ratio of the number of nuclei of $n_2$ to the number of nuclei of $n_1$ after one half-life of $n_1$?
Answer: (D) $1/4$
After one half-life of $n_1$: $N_1 = \dfrac{N_0}{2}$.
$n_2$ decays three times as fast, so in the same time it goes through three of its own half-lives:
$$N_2 = N_0e^{-3\lambda_1 t} = N_0\left(e^{-\lambda_1 t}\right)^3 = N_0\left(\frac{1}{2}\right)^3 = \frac{N_0}{8}$$
$$\frac{N_2}{N_1} = \frac{1/8}{1/2} = \frac{1}{4}$$
Solution by Sreeraj P, M.Sc Physics