Find the ratio of longest wavelength and the shortest wavelength observed in the five spectral series of emission spectrum of hydrogen.

$\begin{array}{l}{\text{ Shortest wavelength comes from n}}_1=\infty {\text{ to n}}_2=1\text{ and longest wavelength }\\ \text{ comes from }{n}_1=6\text{ to }{n}_2=5\text{ in the given case }\\ \text{ Hence }\frac{1}{{\lambda }_{\min }}=\text{R}(\frac{1}{1^2}-\frac{1}{{\infty }^2})=\text{R}\\ \frac{1}{{\lambda }_{\max }}=\text{R}(\frac{1}{5^2}-\frac{1}{6^2})=\text{R}\left(\frac{36-25}{25\times 36}\right)=\frac{11}{900}\text{R}\\ \therefore \frac{{\lambda }_{\max }}{{\lambda }_{\min }}=\frac{900}{11}\end{array}$