The half-life of radioisotopic bromine - 82 is 36 hours. The fraction which remains after one day is…

Half life of bromine $-82=36$ hours

${\mathrm{t}}_{1/2}=\frac{0.693}{\mathrm{K}}$

$\mathrm{K}=\frac{0.693}{36}=0.01925{\mathrm{hr}}^{-1}$

$\mathrm{K}=\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{N}}_0}{{\mathrm{N}}_{\mathrm{t}}}$

 $\left(\mathrm{t}=1 \mathrm{day} = 24 \mathrm{hr}\right)$

$\log \frac{{\mathrm{N}}_0}{{\mathrm{N}}_{\mathrm{t}}}=\frac{24\mathrm{hr}\times 0.01925{\mathrm{hr}}^{-1}}{2.303}$

$=0.2006$

$Or \frac{{\mathrm{N}}_0}{{\mathrm{N}}_{\mathrm{t}}}=$ anti log $\left(0.2006\right)$

$\frac{N_0}{{\mathrm{N}}_{\mathrm{t}}}=1.587$

Or $\frac{{\mathrm{N}}_{\mathrm{t}}}{{\mathrm{N}}_0}=\frac{1}{1.587}=0.630$

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