148067
The activity of a radioactive element decreases to one-third of the original activity $A_{0}$ in a period of $9 \mathrm{yr}$. After a further lapes of $9 \mathrm{yr}$, its activity will be
1 $\mathrm{A}_{0}$
2 $\frac{2}{3} \mathrm{~A}_{0}$
3 $\frac{\mathrm{A}_{0}}{9}$
4 $\frac{\mathrm{A}_{0}}{6}$
5 $\frac{\mathrm{A}_{0}}{18}$
Explanation:
C Activity of radioactive elements given by - $\mathrm{A}=\mathrm{A}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ According to question - $\mathrm{A}=\frac{\mathrm{A}_{\mathrm{o}}}{3}, \mathrm{t}=9 \text { years }$ $\therefore \quad \frac{\mathrm{A}_{\mathrm{o}}}{3}=\mathrm{A}_{\mathrm{o}} \mathrm{e}^{-9 \lambda}$ $\frac{1}{3}=\mathrm{e}^{-9 \lambda}$ When, $\mathrm{t}=18$ years Then, $\mathrm{A}^{\prime}=\mathrm{A}_{\mathrm{o}} \mathrm{e}^{-18 \lambda}$ $\mathrm{A}^{\prime}=\mathrm{A}_{\mathrm{o}}\left(\mathrm{e}^{-9 \lambda}\right)^{2}$ From equation (i) and (ii) we get - $\mathrm{A}^{\prime}=\mathrm{A}_{\mathrm{o}}\left(\frac{1}{3}\right)^{2}$ $\mathrm{~A}^{\prime}=\frac{\mathrm{A}_{\mathrm{o}}}{9}$