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

Wavelength for Lyman series is,
\(\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{\mathrm{n}^2}\right]\)
For the longest wavelength, \(\lambda=\lambda_{\max }\) and \(\mathrm{n}=2\)
\(\frac{1}{\lambda_{\max (\mathrm{L})}}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{3}{4}\)
Wavelength for Balmer series is,
\(\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{\mathrm{n}^2}\right]\)
For the longest wavelength, \(\mathrm{n}=3\)
\(\begin{aligned}
& \frac{1}{\lambda_{\max (B)}}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{5}{36} \\
& \frac{\lambda_{\max (\mathrm{L})}}{\lambda_{\max (\mathrm{B})}}=\frac{4}{3} \times \frac{5}{36}=\frac{5}{27}
\end{aligned}\)