Two radioactive substances $A$ and $B$ have decay constants ' $5 \lambda$ ' and ' $\lambda$ ' respectively.…
Two radioactive substances $A$ and $B$ have decay constants ' $5 \lambda$ ' and ' $\lambda$ ' respectively. At $\mathrm{t}=0$, they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be $\left(\frac{1}{\mathrm{e}}\right)^2$ after a time interval
$\frac{1}{4 \lambda}$
$4 \lambda$
$2 \lambda$
$\frac{1}{2 \lambda}$
Solution
Number of nuclei remained after time $t$ can be written as $\mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}}$
$\mathrm{N}_1=\mathrm{N}_0 \mathrm{e}^{-5 \lambda \mathrm{t}}...(i)$
and $\mathrm{N}_2=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}}$...(ii)
Dividing equation (i) by equation (ii), we get,
$\begin{aligned}
& \frac{N_1}{N_2}=e^{(-5 \lambda+\lambda) t}=e^{-4 \lambda t}=\frac{1}{e^{4 \lambda t}} \\
& \frac{N_1}{N_2}=\left(\frac{1}{e}\right)^2=\frac{1}{e^2} ...[Given]\\
\therefore \quad & \frac{1}{e^2}=\frac{1}{e^{4 \lambda t}} \\
\therefore \quad & 2=4 \lambda t \Rightarrow t=\frac{2}{4 \lambda}=\frac{1}{2 \lambda}
\end{aligned}$