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

Answer 1

The correct option is A $5lo{g}_{e}2$

According to activity law

$R=R_0{e}^{-\lambda t}$ $...\left(i\right)$

According to given problem,

$R_0=N$ counts per minute

$R=\frac{N_0}{e}$ counts per minute

t = 5 minutes

Substituting these values in $e{q}^n$ (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 = $T_{\frac{1}{2}}$ the activity R reduces to $\frac{R_0}{2}$

where $T_{\frac{1}{2}}$ = half life of a radioactive sample

From equation (i), we get

$\frac{R_0}{2}=R_0{e}^{-\lambda T_{\frac{1}{2}}}$

$e^{-\lambda T_{\frac{1}{2}}}=2$

Taking natural logarithms of both sides of above equation, we get

$\lambda T_{\frac{1}{2}}=lo{g}_{e}2$

$T_{\frac{1}{2}}=\frac{lo{g}_{e}2}{\lambda }$

$T_{\frac{1}{2}}=\frac{lo{g}_{e}2}{\frac{1}{5}}$

$T_{\frac{1}{2}}=5lo{g}_{e}2$