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
A) $10\ min$
B) $20\ min$
C) $30\ min$
D) $40\ min$
Explanation
(b) $\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{20} = 0.03465$
Now time of decay $t = \frac{2.303}{\lambda}\log\frac{N_{0}}{N}$
$\Rightarrow t_{1} = \frac{2.303}{0.03465}\log{\frac{100}{67} = 11.6\min\min}$
and $t_{2} = \frac{2.303}{0.03465}\log{\frac{100}{33} = 32min}$
Thus time difference between points of time
$= t_{1} - t_{2} = 32 - 11.6 = 20.4\ min = 20\ min$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.