Question

The activity of a radioactive element decreases to one-third of the original activity $I_0$ in a period of nine years. After a further lapse of nine years, its activity will be

• A. $I_0$
• B. $\left(2/3\right)I_0$
• C. $\left(I_0/9\right)$
• D. $\left(I_0/6\right)$

Answer 1

Answer: (C)

$\left(I_0/9\right)$

Activity $=\frac{dN}{dt}=\lambda N_o{e}^{-lambdtat}$
$I\left(t\right)=I_o{e}^{-\lambda t}$

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

After further $9$ years
$t=18$ years
$I\left(18\right)=I_o{e}^{-18\lambda }$
$\frac{I\left(18\right)}{I_o}=(e^{-9\lambda }{)}^2=(1/3{)}^2$
$I\left(18\right)=\frac{I_o}{9}$