Explanation

(b) The radius of the orbit of the electron in the $n$th excited state
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r_{e}\ =$ $\frac{n^{2}4\pi\varepsilon_{0}h^{2}}{{4\pi}^{2}{mZe}^{2}}$
For the first excited state
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = 2\ ,\ Z = 1$
$\because\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r’\ =$ $\frac{{4\varepsilon}_{0}h^{2}}{{\pi me}^{2}}$
For the ground state of hydrogen atom
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = 1\ ,\ Z = 1$
$\because\ \ \ \ \ \ \ \ \ \ \ \ r’’\ =$ $\frac{h^{2}\varepsilon_{0}}{{\pi me}^{2}}$
The ratio of radius
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{r^{'}}{r^{''}} = \frac{4}{1}$
The ratio of area of the electron orbit for hydrogen atom
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{A^{'}}{A^{''}} = \frac{4\pi(r')^{2}}{4\pi(r'')^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{A^{'}}{A^{''}} = \frac{16}{1}$