JEE Main 2022 — Nuclear Physics Question with Solution

According to radioactive decay formula, $N=N_0{e}^{-\lambda t}$, here, $N_0$ is the initial number of nuclei in a radioactive material.

Given here,  ${\lambda }_1=25\lambda$ and ${\lambda }_2=16\lambda$.

And $\frac{N_B}{N_A}=e$

For radioactive material $A$, $N_A=N_0{e}^{-{\lambda }_{1}t}$

For radioactive material $B$, $N_B=N_0{e}^{-{\lambda }_{2}t}$

Thus, the ratio $\frac{N_B}{N_A}=\frac{e^{-{\lambda }_{2}t}}{e^{-{\lambda }_{1}t}}=e^{-{\lambda }_{2}t}·e^{{\lambda }_{1}t}$

$\Rightarrow {\mathrm{e}}^1={\mathrm{e}}^{\left({\lambda }_1-{\lambda }_2\right)t}$

$\Rightarrow \left({\lambda }_1-{\lambda }_2\right)t=1$

Time, $t=\frac{1}{{\lambda }_1-{\lambda }_2}=\frac{1}{25\lambda -16\lambda }=\frac{1}{9\lambda }$.

Hence, the value of $a=9$.