Question

All radioactive processes follow first order kinetics.The $t_{\frac{1}{2}}$ of a radioactive element is 20 minutes. The time interval between the stages of its $34\%$ and $67\%$ decay is:

Answer 1

Since radioactive decay follows first order kinetics,

$k=\frac{ln2}{20}mi{n}^{-1}$
$\therefore \frac{ln2}{20}\Delta t=ln\frac{66}{33}$
$\Rightarrow \Delta t=20$ min