Q. A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After 5 min, the rate is 1250 disintegrations per min. Then, the decay constant (per minute) is
A) 0.4 In 2
B) 0.2 In 2
C) 0.1 In 2
D) 0.8 In 2
Explanation
(a) Given $N_{0}\lambda = 5000,\ N\lambda = 1250$
$N = N_{0}e^{- \lambda t}\ = \ N_{0}e^{- 5\lambda}$
Where $\lambda\ $is decay constant per min.
$N\lambda\ \ = \ N_{0}\lambda e^{- 5\lambda}$
$\ \ 1250\ = \ \ N_{0}\lambda e^{- 5\lambda}$
$\therefore\ \ \ \ \ \ e^{- 5\lambda} = \frac{5000}{1250} = 4$
$\ \ \ \ \ \ \ \ \ \ \ {\ e}^{5\lambda} = 4$
$5\lambda\ = \ 2\log_{e}2\ $
$\ \ \ \ \ \ \ \ \ \ \ \lambda = 0.4\ln 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.