In a radioactive decay process, the activity is defined as $A=-\frac{dN}{dt}$, where $N\left(t\right)$ is the number of radioactive nuclei at time $\left(t\right)$. Two radioactive sources, $S_1$ and $S_2$ have same activity at time $t=0$. At a later time, the activities of $S_1$ and $S_2$ are $A_1$ and $A_2$, respectively. When $S_1$ and $S_2$ have just completed their $3^{\mathrm{rd}}$ and $7^{\mathrm{th}}$ half-lives, respectively, the ratio $\frac{A_1}{A_2}$ is _____.

Activity for both the cases can be written as, $A_1=A_0{e}^{-{\lambda }_1{t}_1}$

also $A_2=A_0{e}^{-{\lambda }_2{t}_2}$.

Half-life for the reaction can be written as, ${\left(t_{\frac{1}{2}}\right)}_1=\frac{\ln 2}{{\lambda }_1}$ and hence

at $t_1=\frac{3\ln 2}{{\lambda }_1}$.

$A_1=A_0{e}^{-{\lambda }_1\left(\frac{3\ln 2}{{\lambda }_1}\right)}$

$\Rightarrow A_1=A_0{e}^{-3\ln 2} ... \left(\mathrm{i}\right)$

Similarly, for the second case, we can write

$t_2=7\left(\frac{\ln 2}{{\lambda }_2}\right)$ and therefore,

$A_2=A_0{e}^{-{\lambda }_2\left(\frac{7\ln 2}{{\lambda }_2}\right)}$

$A_2=A_0{e}^{-7\ln 2} ... \left(\mathrm{ii}\right)$

From $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, we get

$$\begin{gathered} \frac{A_1}{A_2}=\frac{A_0{e}^{-3\ln 2}}{A_0{e}^{-7\ln 2}} \\ =\frac{A_0{e}^{\ln 2^{-3}}}{A_0{e}^{\ln 2^{-7}}}=\frac{2^{-3}}{2^{-7}}=16 \end{gathered}$$

Therefore, required ration would be$\therefore \frac{A_1}{A_2}=16$