JEE Main 2019 — Nuclear Physics Question with Solution

Given,
Half life of $A=\ln 2$
Initial number of nucleus is, $R_{0A}=R_{0B}=R_0$

Half life is related with decay constant as, ${\left(t_{1/2}\right)}_A=\frac{\ln 2}{\lambda }$
${\lambda }_A=1$

At $t=0 R_A=R_B$
From activity relation, we have, 
$R_0{e}^{-{\lambda }_{A}T}=R_0{E}^{-{\lambda }_{B}T}$

$$\begin{gathered} \text{at} T=t, \\ \Rightarrow \frac{R_B}{R_A}=\frac{R_0{e}^{-{\lambda }_{B}t}}{R_0{e}^{-{\lambda }_{A}t}} \end{gathered}$$

As per given question, $\frac{R_B}{R_A}=e^{-3t}$, On comparing with above relation, we have
$\Rightarrow e^{-\left({\lambda }_B-{\lambda }_A\right)t}=e^{-t}$
$\Rightarrow {\lambda }_{\mathrm{B}}-{\lambda }_{\mathrm{A}}=3$
$\Rightarrow {\lambda }_{\mathrm{B}}=3+1=4$
$\Rightarrow {\left(t_{1/2}\right)}_B=\frac{\lambda \mathrm{n}2}{{\lambda }_{\mathrm{B}}}=\frac{\lambda \mathrm{n}2}{4}$