we know \(\frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^{t/{T_{1/2}}}}\)

For \(20\%\) decay \(\frac{N}{{{N_0}}} = \frac{{80}}{{100}} = {\left( {\frac{1}{2}} \right)^{{t_1}/20}}\)..... \((i)\)

For \(80\%\) decay \(\frac{N}{{{N_0}}} = \frac{{20}}{{100}} = {\left( {\frac{1}{2}} \right)^{{t_2}/20}}\)..... \((ii)\)

Dividing \((ii)\) by \((i)\)

\(\frac{1}{4} = {\left( {\frac{1}{2}} \right)^{\frac{{({t_2} - {t_1})}}{{20}}}};\) on solving we get \({t_2} - {t_1} = 40\,min.\)