Explanation:
B Given that,
Activity $\left(\mathrm{A}_{1}\right)=600 \mathrm{dps}$, time $\left(\mathrm{t}_{1}\right)=28$ days
Activity $\left(\mathrm{A}_{2}\right)=300 \mathrm{dps}$, time $\left(\mathrm{t}_{2}\right)=14$ days
Let, the initial activity be $\mathrm{A}_{0}$
Now, for the internal $\left(0-t_{1}\right)$
$\mathrm{A}_{1}=\mathrm{A}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}$
$600=\mathrm{A}_{0} \mathrm{e}^{-28 \lambda}$
And for the interval $\left(t_{1} \rightarrow t_{2}\right)$
$\mathrm{A}_{2}=\mathrm{A}_{1} \mathrm{e}^{-\lambda \mathrm{t}_{2}}$
$300=600 \mathrm{e}^{-14 \lambda}$
Or $\quad \mathrm{e}^{-14 \lambda}=\frac{1}{2}$
On dividing equation (i) by (ii), we get-
$\frac{600}{300}=\frac{\mathrm{A}_{0} \mathrm{e}^{-28 \lambda}}{600 \mathrm{e}^{-14 \lambda}}$
$2=\frac{\mathrm{A}_{0} \mathrm{e}^{-14 \lambda}}{600}$
Putting the value of $\mathrm{e}^{-14 \lambda}=1 / 2$, we get-
$2=\frac{\mathrm{A}_{0}}{600} \times \frac{1}{2}$
$\mathrm{~A}_{0}=4 \times 600$
$\mathrm{~A}_{0}=2400 \mathrm{dps}$