Question

The activity of a radioactive sample decreases to 1/3rd of the original activity $A_0$ in a period of 9 year. After 9 more year, its activity will be :

• A. $\frac{A_0}{2}$
• B. $\frac{A_0}{9}$
• C. $\frac{2A_0}{9}$
• D. $\frac{3A_0}{9}$

Answer 1

The correct option is B $\frac{A_0}{9}$

As we know,
$t=\frac{2.303}{\lambda }\log \frac{r_0}{r}$
$9=\frac{2.303}{\lambda }\log 3$
$\therefore \lambda =0.122$
Now, $18=\frac{2.303}{0.122}\log \frac{r_0}{r_2}$
$\frac{r_2}{r_0}=\frac{1}{9}$