Q. The counting rate observed from a radioactive source at $t =$9s was 1600 $counts^{- 1}$ and at $t =$8 sit was 100 $counts\ s^{- 1}$. The counting rate observed as counts per second at $t =$6s,will be
A) 400
B) 300
C) 250
D) 200
Explanation
(d) The half-life of source $= \ \frac{8}{4}\ = \ 2s$
Now, $\ \ \ \ \ \ \ \ \ \ R = R_{0}\left( \frac{1}{2} \right)^{n}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R = 1600\left( \frac{1}{2} \right)^{\frac{6}{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ 1600\left( \frac{1}{2} \right)^{3} = \ 200$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.