Explanation

(b) The wavelength of series for $n\ $is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda}$ =$\ R\left( \frac{1}{2^{2}} - \frac{1}{n^{2}} \right)$
were$\ R$ is Rydberg’s constant.
For Balmer series $n$=3 gives the first member of series and $n$=4 gives the second member of series. Hence,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda}$ =$\ R\left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda_{1}}$ =$\ R\left( \frac{5}{36} \right)$ …(i)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\lambda_{2}}$ =$\ R\left( \frac{1}{2^{2}} - \frac{1}{4^{2}} \right)$
= $R$ $\left( \frac{12}{16\ \times \ 4} \right) = \frac{3R}{16}$ …(ii)
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \frac{\lambda_{2}}{\lambda_{1}} = \frac{16}{3} \times \frac{5}{36} = \frac{20}{27}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda_{2} = \frac{20}{27}\ \lambda\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because\ \lambda_{1} = \lambda)$