In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

The wavelength of a spectral line in the Lyman series  is

$\frac{1}{{\lambda }_L}=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right),n=2,3,4,\dots \dots$

and that in the Balmer series is 

$\frac{1}{{\lambda }_B}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right),n=3,4,5$

$\text{ For the longest wavelength in the Lyman series, }n=2$

$\therefore \frac{1}{{\lambda }_L}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=R\left(\frac{1}{1}-\frac{1}{4}\right)=R\left(\frac{4-1}{4}\right)=\frac{3R}{4}$

$\text{ or }{\lambda }_L=\frac{4}{3R}$

For the longest wavelength in the Balmer series, n = 3

$\therefore \frac{1}{{\lambda }_B}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=R\left(\frac{1}{4}-\frac{1}{9}\right)=R\left(\frac{9-4}{36}\right)=\frac{5R}{36}$

$\text{ or }{\lambda }_B=\frac{36}{5R}$

$\text{ Thus, }\frac{{\lambda }_L}{{\lambda }_B}=\frac{\frac{4}{3R}}{\frac{36}{5R}}=\frac{4}{3R}\times \frac{5R}{36}=\frac{5}{27}$