Thank you for reporting, we will resolve it shortly
Q.
Half-life of a radioactive substance is $20$ minutes. Difference between points of time when it is $33 \%$ disintegrated and $67 \%$ disintegrated is approximately
Nuclei
Solution:
Half-life, $T_{1 / 2}=20$ minutes
First disintegration $=33 \%$
Second disintegration $=67 \%$
$\lambda=\frac{\ln 2}{T_{1 / 2}}=\frac{0.693}{T_{1 / 2}}$
$=\frac{0.693}{20}=0.03465$ per minute
As, $N=N_{0} e^{-\lambda t}$
Therefore, time of decay, $t=\frac{1}{\lambda} \ln \frac{N_{0}}{N}$
Therefore, time of decay for $33 \%$ disintegration
$t_{1}=\frac{1}{0.03465} \ln \frac{100}{67}=11.6$ minutes
Similarly, the time of decay for $67 \%$ disintegration
$t_{2}=\frac{1}{0.03465} \ln \frac{100}{33}=32$ minutes
The difference between points of time $= t_2 - t_1$
$ = 32 - 11.6 =20.4 \approx 20$ minutes