Explanation:
C Given,
$\frac{\left(\mathrm{N}_{\mathrm{o}}\right)_{\mathrm{x}}}{\left(\mathrm{N}_{\mathrm{o}}\right)_{\mathrm{y}}}=\frac{1}{4}$
$t_{\mathrm{x}}=18 \mathrm{hr}$
$\mathrm{t}_{\mathrm{y}}=12 \mathrm{hr}$
Decay period $(\mathrm{t})=3$ day
$(\mathrm{t})=3 \times 24=72 \mathrm{hr}$
From formula of decay -
$\mathrm{N}=\left(\mathrm{N}_{0}\right)\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{t}_{1 / 2}}$
For substance X,
Then, $\quad \mathrm{N}_{\mathrm{x}}=\left(\mathrm{N}_{0}\right)_{\mathrm{x}}\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{t}_{\mathrm{x}}}$
$\mathrm{N}_{\mathrm{x}}=1\left(\frac{1}{2}\right)^{\frac{72}{18}}$
$\mathrm{~N}_{\mathrm{x}}=\left(\frac{1}{2}\right)^{4}=\frac{1}{2^{4}}$
For substance $\mathrm{Y}$,
And, $\quad \mathrm{N}_{\mathrm{y}}=\left(\mathrm{N}_{0}\right)_{\mathrm{y}}\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{t}_{\mathrm{y}}}$
$\mathrm{N}_{\mathrm{y}}=4\left(\frac{1}{2}\right)^{\frac{72}{12}}$
$\mathrm{~N}_{\mathrm{y}}=4\left(\frac{1}{2}\right)^{6}$
The ratio of the amount of substance $\mathrm{X}$ and $\mathrm{Y}$
So, $\quad \frac{\mathrm{N}_{\mathrm{x}}}{\mathrm{N}_{\mathrm{y}}}=\frac{1}{2^{4}} \times \frac{2^{6}}{4}=1: 1$