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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
$log_e \;\frac{2}{5}$
$\frac{5}{{\log_e 2}}$
$5$$\log_{10}2$
$5$$\log_e2$
Solution
According to activity law
$R=R_{0} e^{-\lambda t}$ ….. $(i)$
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 of both sides of above equation, we get
$\lambda T_{1 / 2}=\log _{e} 2$
or $T_{1 / 2}=\frac{\log _{e} 2}{\lambda}=\frac{\log _{e} 2}{\left(\frac{1}{5}\right)}=5 \log _{e} 2$ minutes
