Explanation

(b) The time required for the number of parent nuclei to fall to 50% is called half-life $T^{1/2}$ and may be related to $\lambda$ as follows.
Since $\ \ \ \ \ \ \ 0.5\ N_{0}\ = \ N_{0}e^{{- T}_{1/2}}$
We have, $\ \ \ \ \ \ \lambda T_{1/2}$= ln (2) = 0.693
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{1/2} = \frac{0.693}{\lambda}$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = \frac{0.693}{T_{1/2}}$
Given, $\ \ \ \ \ \ \ \ \ \ \ T_{1/2}$ = 77 days
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = \frac{0.693}{77\ } = 9 \times 10^{- 3}days^{- 1}$