Explanation

(c) After $n$ half-lives($ie,\ at\ t = nT$) the number of nuclides left undecayed,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ N = N_{0}\left( \frac{1}{2} \right)^{n}$
Given, $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{N}{N_{0}} = \frac{1}{16}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{16} = \left( \frac{1}{2} \right)^{n}$
or$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \frac{1}{2} \right)^{4} = \left( \frac{1}{2} \right)^{n}$
Equating the powers, we obtain
$n = 4$
$ie,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{t}{T}\ = \ 4$
or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t = 4T$
Or $t = 4\ \times \ 5730\ = \ 22929\ yr$ ($\because T = 5730\ yr)$