Q. A radioactive sample $S_{1}$ having the activity $A_{1}$ has twice the number of nuclei as another sample $S_{2}$ of activity$A_{2}$. If $A_{2} = 2A_{1},$ then the ratio of half-life of $S_{1}$ to the half-life of $S_{2}$ is
A) 4
B) 2
C) 0.25
D) 0.75
Explanation
(a) Activity, $A = \lambda N = \frac{0.693}{T_{1/2}}\ N$
Where $T_{1/2}\ $is the half-life of a radioactive sample,
$\therefore\ \ \ \ \ \ \ \ \ \ \ \frac{A_{1}}{A_{2}} = \frac{N_{1}}{T_{1}}\ \times \frac{T_{2}}{N_{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{T_{1}}{T_{2}} = \frac{A_{2}}{A_{1}}\ \times \frac{N_{1}}{N_{2}}$
$= \frac{{2A}_{1}}{A_{1}}\ \times \frac{{2N}_{2}}{N_{2}} = \frac{4}{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.