Q. Activity of radioactive element decreased to one third of original activity $R_{0}$ in 9 years. After further 9 years, its activity will be
A) $R_{0}$
B) $\frac{2}{3}R_{0}$
C) $R_{0}/9$
D) $R_{0}/6$
Explanation
(c) Activity $R = R_{0}e^{- \lambda t}$
$\frac{R_{0}}{3} = R_{0}e^{- \lambda \times 9} \Rightarrow e^{- 9\lambda} = \frac{1}{3}$ …(i)
After further 9 years $R^{'} = Re^{- \lambda t} = \frac{R_{0}}{3} \times e^{- \lambda \times 9}$ …(ii)
From equation (i) and (ii), $R^{'} = \frac{R_{0}}{9}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.