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)

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