Explanation

(d) Radius of orbit of electron in $n$th excited state of hydrogen
$\ \ \ \ \ \ \ \ \ \ \ \ \ r\ = \frac{\varepsilon_{0}h^{2}n^{2}}{{\pi mZe}^{2}}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ r \propto \frac{n^{2}}{Z}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
$\therefore\ \ \ \ \ \ \ \ \ \ \frac{r_{1}}{r_{2}} = \frac{n_{1}^{2}}{n_{2}^{2}} \times \frac{Z_{2}}{Z_{1}}$
But $r_{1} = r_{2}$
So, $n_{2}^{2} = n_{1}^{2}\ \times \frac{Z_{2}}{Z_{1}}$
Here,
$n$$_{1}$ = 1(ground state of hydrogen),
$\ Z$$_{1}$= 1(atomic number of hydrogen),
$Z$$_{2}$= 4(atomic number of beryllium)
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt{n_{2}^{2}}$ = (1)$^{2}$$\times \frac{4}{1}$
or${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n}_{2}^{2}$ $= 4$
or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n$$_{2}$$= 2$