The half-life of radioisotopic bromine - 82 is 36 hours. The...

Half life of bromine $- 82 = 36$ hours

${t}_{1/2} = \dfrac{0 . 693}{K}$

$K = \dfrac{0 . 693}{36} = 0 . 01925{hr}^{- 1}$

$K = \dfrac{2 . 303}{t}log\dfrac{{N}_{0}}{{N}_{t}}$

 \left(t = 1day = 24hr\right)$</p><p>$log\dfrac{{N}_{0}}{{N}_{t}} = \dfrac{24hr \times 0 . 01925{hr}^{- 1}}{2 . 303}$</p><p>$= 0 . 2006$</p><p>$Or\dfrac{{N}_{0}}{{N}_{t}} =$anti \log \left(0 . 2006\right)$

$\dfrac{{N}_{0}}{{N}_{t}} = 1 . 587$

Or $\dfrac{{N}_{t}}{{N}_{0}} = \dfrac{1}{1 . 587} = 0 . 630$

Hence, the fraction remain after day $= 63 \times {10}^{- 2}$