\(\frac{1}{\lambda_{L}}=R\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right), n=2,3,4, \ldots \ldots\)

and that in the Balmer series is

\(\frac{1}{\lambda_{B}}=R\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right), n=3,4,5, \ldots \ldots\)

For the longest wavelength in the Lyman series,

\(n=2\)

\(\therefore \quad \frac{1}{\lambda_{L}}=R\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=R\left(\frac{1}{1}-\frac{1}{4}\right)=R\left(\frac{4-1}{4}\right)=\frac{3 R}{4}\)

or \(\quad \lambda_{L}=\frac{4}{3 R}\)

For the longest wavelength in the Balmer series,

\(n=3\)

\(\therefore \quad \frac{1}{\lambda_{B}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R\left(\frac{1}{4}-\frac{1}{9}\right)=R\left(\frac{9-4}{36}\right)=\frac{5 R}{36}\)

or \(\quad \lambda_{B}=\frac{36}{5 R}\)

Thus, \(\frac{\lambda_{L}}{\lambda_{B}}=\frac{\frac{4}{3 R}}{\frac{36}{5 R}}=\frac{4}{3 R} \times \frac{5 R}{36}=\frac{5}{27}\)