Q. The counting rate observed from a radioactive source at $t = 0$ was 1600 $counts^{- 1}$ and at $t = 8$ s it was 100$counts^{- 1}$. The counting rate observed as counts per seconds at $t = 6$s, 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.