Q 12-13-121JEE MainJEE Main 2019 (9 Jan, Shift 2)Medium
At a given instant, say $t = 0$, two radioactive substances $A$ and $B$ have equal activities. The ratio $\dfrac{R_B}{R_A}$ of their activities after time $t$ itself decays with time $t$ as $e^{-3t}$. If the half-life of $A$ is $\ln2$, the half-life of $B$ is:
Answer: (C) $\dfrac{\ln2}{4}$
With equal initial activities, $\dfrac{R_B}{R_A} = e^{-(\lambda_B-\lambda_A)t} = e^{-3t}$, so $\lambda_B - \lambda_A = 3$.
$\lambda_A = \dfrac{\ln2}{\ln2} = 1$, so $\lambda_B = 4$ and
$$T_B = \frac{\ln2}{4}$$
Solution by Sreeraj P, M.Sc Physics