Half-life of a radioactive substance is `20` minutes. Difference between points of time when it is `33 %` disintegrated and `67 %` disintegrated is approximate.

Options

A. 10 minutes
B. 20 minutes
C. 30 minutes
D. 40 minutes

Answer

B

Explanation

Half - life, `T_(1//2)=20` minutes
First disintegration `=33%`
Second desintegration `=67%`
`lambda=(ln2)/(T_(1//2))=(0.693)/(T_(1//2))=(0.693)/(20)="0.03465 per minute"`
As, `N=N_(0)e^(0lambdat)`
Therefore, time of decay, `t=(1)/(lambda)ln(N_(0))/(N)`
Therefore, time of decay for `33%` disintegration
`t_(1)=(1)/(0.03465)ln""(100)/(67)="11.6 minutes"`
Similarly, the time of decay for `67%` disintegration
`t_(2)=(1)/(0.03465)ln""(100)/(33)="32 minutes"`
The difference between points of time `t_(2)-t_(1)`
`=32-11.6=20.4~~"20 minutes"`