Explanation

(c) The wavelength of spectral line in Balmer series is given by
$\frac{1}{\lambda} = R\left\lbrack \frac{1}{2^{2}} - \frac{1}{n^{2}} \right\rbrack$
For first line of Balmer series, $n = 3$
$\Rightarrow \frac{1}{\lambda_{1}} = R\left\lbrack \frac{1}{2^{2}} - \frac{1}{3^{2}} \right\rbrack = \frac{5R}{36}$; For second line $n = 4$
$\Rightarrow \frac{1}{\lambda_{2}} = R\left\lbrack \frac{1}{2^{2}} - \frac{1}{4^{2}} \right\rbrack = \frac{3R}{16}$
$\therefore\frac{\lambda_{2}}{\lambda_{1}} = \frac{20}{27} \Rightarrow \lambda_{2} = \frac{20}{27} \times 6561 = 4860\mathring{\mathrm{A}}$