Question

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. $log_{e}\frac{2}{5}$
• B. $\frac{5}{log_{e}\,2}$
• C. $5log_{10}\,{2}$
• D. $5log_{e}\,2$

Answer 1

Answer: (D)

$5log_{e}\,2$

Answer 2

According to activity law $R=R_{0}e^{-\lambda\,t}$ $\dots (i)$ where, $R_{0}$ = initial activity at $t = 0$ $R$= activity at time $t$ $\lambda$= decay constant According to given problem, $R_{0}=N_{0}$ counts per minute $R=\frac{N_{0}}{e}$ counts per minute $t = 5$ minutes Substituting these values in equation $(i)$, we get $\frac{N_{0}}{e}=N_{0}e^{-5\lambda}$ $e^{-1}=e^{-5\,\lambda}$ $5\lambda=1$ or $\lambda=\frac{1}{5}$ per minute At $t =T_{1/2}$, the activity $R$ reduces to $\frac{R_{0}}{2}$ where $T_{1/2}$= half life of a radioactive sample From equation $(i)$, we get $\frac{R_{0}}{2}=R_{0}e^{-\lambda T_{1/ 2}}$ $e^{\lambda T_{1 /2}}=2$ Taking natural logarithms on both sides of above equation, we get $\lambda\,T_{1/2}=log_{e}\,2$ or $T_{1 /2}=\frac{log_e \,2}{\lambda}=\frac{log_e \,2}{\left(\frac{1}{5}\right)}$ $=5log_{e}2$ minutes