The activity of a radioactive sample is measured as N _ 0 counts...

Fraction remains after $n$ half-lives

$\dfrac{N}{{N}_{0}} = {\left(\dfrac{1}{2}\right)}^{n} = \left(\dfrac{1}{2}\right)\dfrac{t}{T}$

Given, $N = \dfrac{{N}_{0}}{e} ⇒ \dfrac{{N}_{0}}{e{N}_{0}} = \left(\dfrac{1}{2}\right)\dfrac{5}{T}$

$⇒ \dfrac{1}{e} = \left(\dfrac{1}{2}\right)\dfrac{5}{T}$

Taking log on both sides, we get

$\log 1 - loge = \dfrac{5}{T}log\dfrac{1}{2}$

$⇒ - 1 = \dfrac{5}{T}\left(- \log 2\right)$

$⇒ T = 5\log_{e}2$

Now, let $t$ be the time after which activity reduces to half

\left(\dfrac{1}{2}\right) = {\left(\dfrac{1}{2}\right)}^{t / 5\log_{e}2}$</p><p>$⇒ {t}^{'} = 5\log_{e}2