Q. The activity of a sample is $64 \times 10^{- 5}Ci$. Its half-life is $3\ days$. The activity will become $5 \times 10^{- 6}Ci$ after
A) $12\ days$
B) $7\ days$
C) $18\ days$
D) $21\ days$
Explanation
(d) $A = A_{0}\left( \frac{1}{2} \right)^{t/T_{1/2}} \Rightarrow 5 \times 10^{- 6} = 64 \times 10^{- 5}\left( \frac{1}{2} \right)^{t/3}$
$\Rightarrow \frac{1}{128} = \left( \frac{1}{2} \right)^{t/3} \Rightarrow t = 21\ days$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.