Physics - Modern Physics Question with Solution | TestHub

Wavelength for Lyman series $${\lambda }_L=\frac{1}{R\left(1-\frac{1}{4}\right)}=\frac{4}{3R}$$ and wavelength for Balmer series $$\begin{array}{rl}& {\lambda }_B=\frac{1}{R\left(\frac{1}{4}-\frac{1}{9}\right)}=\frac{1}{R\left(\frac{5}{36}\right)}=\frac{36}{5R}\\ & \therefore \frac{{\lambda }_L}{{\lambda }_B}=\frac{4}{3R}\times \frac{5R}{36}=\frac{5}{27}\\ & \Rightarrow {\lambda }_L:{\lambda }_B=5:27\\ & \end{array}$$