Q. Activity of a radioactive element decreased to one-third of original activity $I_{0}$ in 9 yr. After further 9 yr, its activity will be
A) $I_{0}$
B) $\frac{2}{3}I_{0}$
C) $I_{0}/9$
D) $I_{0}/3$
Explanation
(c) In 9 years, activity becomes $I = \frac{I_{0}}{3}$
In further 9 years, activity would becomes
$I^{'} = \frac{I}{3} = \frac{I_{0}}{3 \times 3} = \frac{I_{0}}{9}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.