Q. The activity of a radioactive sample is measured as $N_{0}$ counts per minute at $t = 0$ and $N_{0}/e$ counts per minute at $t = 5$ minutes. The time (in minutes) at which the activity reduces to half its value is
A) $5\log_{e}2$
B) $\log_{e}{2/5}$
C) $\frac{5}{\log_{e}2}$
D) $5\log_{10}2$
Explanation
(a) $N = N_{0}e^{- \lambda t} \Rightarrow \frac{N_{0}}{e} = N_{0}e^{- \lambda(5)} \Rightarrow \lambda = \frac{1}{5}$
Now $\frac{N_{0}}{2} = N_{0}e^{- \lambda(t)} \Rightarrow t = \frac{1}{\lambda}\ln 2 = 5\ln 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.