Question

The half life period of ${}_{53}{\text{I}}^{125}$ is 60 days. what % of radioactivity would be present after 240 days. The activity of a radioactive sample decreases to 1/3 of the original activity 9$A_0$ in a period of 9 years. After 9 years more, its activity is $\frac{A_0}{x}$. Find the value of $x$.

Answer 1

$N=N_0{e}^{-\lambda t}$

$\lambda =\frac{ln2}{60}$, $t=240$ days

$N=N_0{e}^{-4In2}$

$=\frac{N_0}{16}$

% radioactivity=$\frac{{\lambda }_{1}N}{\lambda N_0}\times 100=\frac{1}{16}\times 100=$6.25%

Given $\lambda N=\frac{1}{3}\lambda N_0$

$\frac{1}{3}=e^{-\lambda 9}$

$e^{\lambda 9}=3$

$\lambda =\frac{In3}{9}$

$\lambda N=\lambda N_0{e}^{-\frac{ln3}{9}\times 18}$=$\frac{\lambda N_0}{3^2}=\frac{\lambda N_0}{9}$