Q. A radioactive sample is $\alpha$-emitter with half life 138.6 days is observed by a student to have 2000 disintegration/s. The number of radioactive nuclei for given activity are
A) $3.45 \times 10^{10}$
B) $1 \times 10^{10}$
C) $3.45 \times 10^{15}$
D) $2.75 \times 10^{11}$
Explanation
(a) Activity of substance that has 2000 disintegrations/sec
$= \frac{2000}{3.7 \times 10^{10}} = 0.054 \times 10^{- 6}ci = 0.054\ \mu ci$
The number of radioactive nuclei having activity $A$
$N = \frac{A}{\lambda} = \frac{2000 \times T_{1/2}}{\log_{e}2}$
$= \frac{2000 \times 138.6 \times 24 \times 3600}{0.693} = 3.45 \times 10^{10}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.