Explanation

(b) Here $T_{1/2} = 20\ minutes,$ 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 $t_{2} - t_{1} = 40\ min$