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. $lo{g}_e\frac{2}{5}$
• B. $\frac{5}{lo{g}_{e}2}$
• C. $5lo{g}_{10}2$
• D. $5lo{g}_{e}2$

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
$\lambda T_{1/2}=lo{g}_{e}2$
or $T_{1/2}=\frac{lo{g}_{e}2}{\lambda }=\frac{lo{g}_{e}2}{\left(\frac{1}{5}\right)}$
$=5lo{g}_{e}2$ minutes