Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is

The wavelength of different spectral lines of Lyman series is given by

$\frac{1}{{\lambda }_L}=R\left[\frac{1}{l^2}-\frac{1}{n^2}\right]wheren=2,3,4,\mathrm{....}$

where subscript L refers to Lyman. For longest wavelength, n = 2

$\therefore \frac{1}{{\lambda }_{L_{longest}}}=R\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{3}{4}R---\left(i\right)$

The wavelength of different spectral series of Balmer series is given by

$\frac{1}{{\lambda }_B}=R\left[\frac{1}{2^2}-\frac{1}{n^2}\right]wheren=3,4,5,\mathrm{....}$

where subscript B refers to Balmer. For longest wavelength, n= 3

$\therefore \frac{1}{{\lambda }_{B_{longest}}}=R\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=R\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5R}{36}---\left(ii\right)$

Divide (ii) by (i), we get

$\frac{{\lambda }_{L_{longest}}}{{\lambda }_{B_{longest}}}=\frac{5R}{36}\times \frac{4}{3R}=\frac{5}{27}$