- Home
- Standard 12
- Physics
In a radioactive decay process, the activity is defined as $A=-\frac{\mathrm{d} N}{\mathrm{~d} t}$, where $N(t)$ is the number of radioactive nuclei at time $t$. 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 $3^{\text {rd }}$ and $7^{\text {th }}$ half-lives, respectively, the ratio $A_1 / A_2$ is. . . . . . .
$10$
$12$
$15$
$16$
Solution
$S_1 S_2$
$\mathrm{t}=0 \quad \mathrm{~A}_0 \quad \mathrm{~A}_0$
$t=\tau \quad A_1 \quad A_2$
$\frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{\mathrm{A}_0(0.5)^{t /\left(L_{r i}\right)_2}}{\mathrm{~A}_0(0.5)^{t /\left(t_{\mu z}\right)_2}}=\frac{(0.5)^3}{(0.5)^7}=2^4=16$
