\(R=R_{0} e^{-\lambda t}\)  ..... \((i)\)

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