Explanation

(d) In time $t = T\ ,\ N = \frac{N_{0}}{2}$
In another half-life,(ie, after 2 half-lives)
$N = \frac{1}{2}\frac{N_{0}}{2}\ = \ \frac{N_{0}}{4}\ = \ N_{0}\left( \frac{1}{2} \right)^{2}$
After yet another half-life ,($ie$, after 3 half-lives)
$\ \ N = \frac{1}{2}\left( \frac{N_{0}}{4} \right)\ = \frac{N_{0}}{8}\ = \ N_{0}\left( \frac{1}{2} \right)^{3}$and so on. Hence, after $n$
half-lives
$N = N_{0}\left( \frac{1}{2} \right)^{n}$
$\ \ \ \ \ \ \ = N_{0}\left( \frac{1}{2} \right)^{t/T}$
where $t = \ n \times T\ =$ total time of $n$ half-lives.
Here$,\ \ \ \ \ \ \ n\ = \ \frac{t}{T}\ = \ \frac{19}{3.8}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5$
$\therefore$ The fraction left
$\frac{N}{N_{0}}\ = \ \left( \frac{1}{2} \right)^{n} = \left( \frac{1}{2} \right)^{5} = \frac{1}{32}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.031$