Question

A solution contains a mixture of isotopes of $X^{A_1}(t_{1/2}=14days)$ and $X^{A_2}(t_{1/2}=25days)$. Total activity is $1$ curie at $t=0$. The activity reduces by $50\%$ in $20$ days. Find:
(a) The initial activities of $X^{A_1}$ and $X^{A_2}$
(b) The ratio of their initial no. of nuclei.

Answer 1

Let activity of ${}_x{A}_1$ and ${}_x{A}_2$ be $m$ and $n$ curie respectively at $t=0$
$\therefore m+n=1$ ... (i)
Also, $\because$ Rate $\propto$ No. of atoms
$\because$ For ${}_x{A}_1$ decay
$t=\frac{2.303}{\lambda }\log \frac{N_0}{N}=\frac{2.303}{\lambda }\log \frac{r_0}{r}$
$20=\frac{2.303\times 14}{0.693}\log \frac{m}{r_1}$
$\therefore r_1=0.3716m$
Similarly, for ${}_x{A}_2$ decay
$t=\frac{2.303}{\lambda }\log \frac{r_0}{r}$;
$20=\frac{2.303\times 25}{0.693}\log \frac{n}{r_2}$
$\therefore r_2=0.5744n$
Given that activity after 20 day remains $\frac{1}{2}$ of original activity
$\therefore 0.3716m+0.5744n=\frac{1}{2}$ ... (ii)
Solving eqs. (i) and (ii) $m=0.3669curie;n=0.6331curie$
For ratio of atoms, i.e., $\left(N_0^{A_1}/N_0^{A_2}\right)$, we can write
$\frac{r_0^{A_1}}{r_0^{A_2}}=\frac{{\lambda }^{A_1}}{{\lambda }^{A_2}}\times \frac{N_0^{A_1}}{N_0^{A_2}}(\because r=\lambda .N_0)$
or $\frac{0.3669}{0.6331}=\frac{0.693\times 25}{14\times 0.693}\frac{N_0^{A_1}}{N_0^{A_2}}$
or $\frac{N_0^{A_1}}{N_0^{A_2}}=0.3255$