The activity of a radioactive sample is measured as $${N_0}$$ counts per minute at $$t = 0$$  and $$\frac{{{N_0}}}{e}$$ counts per minute at $$t = 5\,min.$$   The time (in minute) at which the activity reduces to half its value is

Solution :
Fraction remains after $$n$$ half lives
$$\eqalign{ & \frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^n} = {\left( {\frac{1}{2}} \right)^{\frac{t}{T}}} \cr & {\text{Given}}\,\,N = \frac{{{N_0}}}{e} \Rightarrow \frac{{{N_0}}}{{e{N_0}}} = {\left( {\frac{1}{2}} \right)^{\frac{5}{T}}} \cr & {\text{or}}\,\,\frac{1}{e} = {\left( {\frac{1}{2}} \right)^{\frac{5}{T}}} \cr} $$
Taking log on both sides, we get
$$\eqalign{ & \log 1 - \log e = \frac{5}{T}\log \frac{1}{2} \cr & \Rightarrow - 1 = \frac{5}{T}\left( { - \log 2} \right) \cr & \Rightarrow T = 5{\log _e}2 \cr} $$
Now, let $$t’$$ be the time after which activity reduces to half
$$\left( {\frac{1}{2}} \right) = {\left( {\frac{1}{2}} \right)^{\frac{{t'}}{{5{{\log }_e}2}}}} \Rightarrow t' = 5{\log _e}2$$