Question

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

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

Answer 1

Answer: (B)

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