Explanation

(d) From Rutherford-Soddy law, the number of atoms left after $n$ half-lives is given by
$\ \ \ \ \ \ \ \ \ N = N_{0}\left( \frac{1}{2} \right)^{n}$
Where,$N_{0}$ is original number of atoms.
The number of half-life
$\ \ \ \ \ \ \ \ \ \ n = \frac{time\ of\ decay}{effective\ half - life}$
Relation between effective disintegration constant ($\lambda$) and half-life ($T)$ is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = \frac{\ln 2}{T}$
$\therefore\ \ \lambda_{1} + \lambda_{2} = \frac{\ln 2}{T_{1}} + \frac{\ln 2}{T_{2}}$
Effective half –life
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{T} = \frac{1}{T_{1}} + \frac{1}{T_{2}} = \frac{1}{1620} + \frac{1}{810}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{T} = \frac{1 + 2}{1620} \Rightarrow T \Rightarrow 540\ yr$
$\therefore\ \ \ \ \ \ \ \ \ \ \ n = \ \frac{t}{540}$
$\therefore\ \ \ \ \ \ \ \ \ \ N = N_{0}\left( \frac{1}{2} \right)^{t/540}$
$\Rightarrow \ \ \ \ \ \frac{N}{N_{0}} = \left( \frac{1}{2} \right)^{2} = \left( \frac{1}{2} \right)^{t/540}$
$\Rightarrow$ $\frac{t}{540} = 2$
$\Rightarrow \ \ \ \ \ \ t = \ 2\ \times \ 540\ = \ 1080\ yr$