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

We know that, wavelength of hydrogen spectrum is given by $$ \frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) $$ For minimum wavelength of Lyman series, $$ \begin{aligned} & n_{1}=1 \text { and } n_{2}=\infty \\ \therefore \quad & \frac{1}{\left(\lambda_{L}\right)_{\min }}=R\left(\frac{1}{(1)^{2}}-\frac{1}{\infty}\right)=R \\ \Rightarrow \quad\left(\lambda_{2}\right)_{\min }=\frac{1}{R} \end{aligned} $$ For minimum wavelength of Balmer series, $n_{1}=2$ and $n_{2}=\infty$ $\therefore \quad \frac{1}{\left(\lambda_{B}\right)_{\min }}=R\left(\frac{1}{2^{2}}-\frac{1}{\infty}\right)$ $=R\left(\frac{1}{4}-0\right)=\frac{R}{4}$ $\Rightarrow \quad\left(\lambda_{B}\right)_{\min }=\frac{4}{R}$ $\therefore \quad \frac{\left(\lambda_{L}\right)_{\min }}{\left(\lambda_{B}\right)_{\min }}=\frac{\frac{1}{R}}{\frac{4}{R}}=\frac{1}{4}=0.25$