Question

Half-life of a radioactive substance is $20$ min. Difference between points of time when it is $33$ % disintegrated and $67$ % disintegrated is approximately:

Answer 1

Given,

$t_{1/2}=20min$

Half-life of a radioactive substance is given by

$t_{1/2}=\frac{0.693}{\lambda }$

$\lambda =\frac{0.693}{20min}=0.03464mi{n}^{-1}$

We, know that ,

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

When, $N=33N_0$

$\frac{33N_0}{100}=N_0{e}^{-\lambda t_1}$

$0.33=e^{-\lambda t_1}$

$-\lambda t_1=ln\left(0.33\right)$

$t_1=\frac{ln\left(0.33\right)}{-\lambda }=\frac{ln\left(0.33\right)}{-0.03465}$

$t_1=32min$

When, $N=67N_0$%

$\frac{67}{100}{N}_0=N_0{e}^{-\lambda t_2}$

$0.67=e^{-\lambda t_2}$

$\lambda t_2=ln\left(0.67\right)$

$t_2=\frac{0.67}{-\lambda }=\frac{0.67}{-0.03465}$

$t_2=11.6min$

Difference point of time is

$\Delta t=t_1-t_2=20.4min$