The ratio of minimum wavelengths of Lyman and Balmer series will be

Wavelength, \(\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}^2}-\frac{1}{\mathrm{~m}^2}\right)\)
For shortest wavelength in Lyman series:
\(\begin{aligned}
& \mathrm{n}=1, \mathrm{~m}=\infty \\
& \frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)=\mathrm{R}...(i)
\end{aligned}\)
For shortest wavelength in Balmer series:
\(\begin{aligned}
& \mathrm{n}=2, \mathrm{~m}=\infty \\
& \frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)=\frac{\mathrm{R}}{4}...(ii)
\end{aligned}\)
Dividing equation (ii) by equation (i),
\(\frac{\lambda_{\mathrm{L}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{R} / 4}{\mathrm{R}}=\frac{1}{4}=0.25\)