Q. In a radioactive disintegration , the ratio of initial number of atoms to the number of atoms present at an instant of time equal to its mean life is
A) $\frac{1}{e^{2}}$
B) $\frac{1}{e}$
C) e
D) $e^{2}$
Explanation
(c) Let the initial number of atoms at time $t = 0\ be\ N_{0}.$
Let $N$ be the number of atoms at any instant t.
Mean life$\tau = \frac{1}{\lambda}$ , where $\lambda$ is disintegration constant.
Given , $t = \ \tau$
According to radioactive disintegration law,
$\ \ \ \ \ \ \ \ \ \ \ \ N = N_{0}e^{- \lambda t}$
$or\ \ \ \ \ \ \ N = N_{0}e^{- \lambda\ \times \ \frac{1}{\lambda}}\ = \ \frac{N_{0}}{e}$
$\ or\ \ \ \ \ \frac{N_{0}}{N}\ = \ e$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.