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

The series end of Lyman series corresponds to transition from \(\mathrm{n}_{\mathrm{i}}=\infty\) to \(\mathrm{n}_{\mathrm{f}}=1\), corresponding to the wavelength
\[
\begin{aligned}
& \frac{1}{\left(\lambda_{\min }\right)_{\mathrm{L}}}=\mathrm{R}\left[\frac{1}{1}-\frac{1}{\infty}\right]=\mathrm{R} \\
\Rightarrow \quad\left(\lambda_{\min }\right)_{\mathrm{L}} &=\frac{1}{\mathrm{R}}=912 Å
\end{aligned}
\]
For last line of Balmer series
\[
\begin{aligned}
& \frac{1}{\left(\lambda_{\min }\right)_{B}}=R\left[\frac{1}{(2)^{2}}-\frac{1}{(\infty)^{2}}\right]=\frac{R}{4} \\
\Rightarrow \quad &\left(\lambda_{\min }\right)_{B}=\frac{4}{R}=3636 Å
\end{aligned}
\]
Dividing Eq. (i) by Eq. (ii), we get
\[
\frac{\left(\lambda_{\min }\right)_{\mathrm{L}}}{\left(\lambda_{\min }\right)_{\mathrm{B}}}=0.25
\]