Explanation

(c) According to Bohr, the wavelength emitted when an electron jumps from $n_{1}$th to $n_{2}$th orbit is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ E\ = \ \frac{hc}{\lambda}\ = \ E_{2}\ –E_{1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda}$ $= \ R$ $\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$
For first line in Lyman series
$\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda_{L}}$ $= R\left( \frac{1}{1^{2}} - \frac{1}{2^{2}} \right)$ = $\frac{3\ R}{4}$ …(i)
For first line in Balmer series,
$\ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda_{B}}$ $= R\left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right)$ = $\frac{5\ R}{36}$ …(ii)
From Eqs. (i) and (ii)
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\lambda_{B}}{\lambda_{L}}$ $= \frac{3\ R}{4} \times \frac{36}{5R}$ $=$ $\frac{27}{5}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda_{B}$=$\frac{27}{5}\lambda\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because\ \lambda_{L} = \lambda)$