Explanation:
A Given that,
Decay constant to first sample $\left(\lambda_{1}\right)=15 \mathrm{x}$
Decay constant to second sample $\left(\lambda_{2}\right)=3 \mathrm{x}$
We know that, number of nuclei at time $t$ is given by,
$\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$
The number of nuclei at time $\frac{1}{6 \mathrm{x}}$ for first sample,
So, $\quad \mathrm{N}_{1}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{1} t}$
$\mathrm{N}_{1}=\mathrm{N}_{0} \mathrm{e}^{-\frac{15 \mathrm{x}}{6 \mathrm{x}}}$
$\mathrm{N}_{1}=\mathrm{N}_{0} \mathrm{e}^{-2.5}$
Similarly, $\mathrm{N}_{2}=\mathrm{N}_{0} \mathrm{e}^{-\frac{3 \mathrm{x}}{6 \mathrm{x}}}$
Dividing equation (i) by (ii), we get-
$\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{N}_{0} \mathrm{e}^{-2.5}}{\mathrm{~N}_{0} \mathrm{e}^{-0.5}}$
$\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-2.5} \times \mathrm{e}^{+0.5}$
$\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-2}$