The half-life of radioisotope bromine - 82 is 36 hours. The fraction which remains after one day is ________ $\times 10^{-2}$.

(Given antilog $0.2006=1.587$)

<p>Half life of bromine $-82=36$ hours</p> <p>$$\begin{aligned} & \mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \\ & \mathrm{~K}=\frac{0.693}{36}=0.01925 \mathrm{~hr}^{-1} \\ & 1^{\text {st }} \text { order rxn kinetic equation } \\ & \mathrm{t}=\frac{2.303}{\mathrm{~K}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \\ & \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=\frac{\mathrm{t} \times \mathrm{K}}{2.303}(\mathrm{t}=1 \text { day }=24 \mathrm{~hr}) \\ & \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=\frac{24 \mathrm{hr} \times 0.01925 \mathrm{hr}^{-1}}{2.303} \\ & \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=0.2006 \\ & \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=\mathrm{anti} \log (0.2006) \\ & \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}=1.587 \\ & \text { If }=1 \\ & \frac{1}{1-\mathrm{x}}=1.587 \Rightarrow 1-\mathrm{x}=0.6301=\text { Fraction remain } \\ & \mathrm{after} \text { one day } \end{aligned}$$</p>