Physics - Modern Physics Question with Solution | TestHub

For longest wavelength in Paschen's series: $\frac{1}{\lambda }=\mathrm{R}\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]$ For longest ${\mathrm{n}}_1=3$ ${\mathrm{n}}_2=4$ $\begin{array}{rl}& \frac{1}{\lambda }=\mathrm{R}\left[\frac{1}{(3{)}^2}-\frac{1}{(4{)}^2}\right]\\ & \frac{1}{\lambda }=\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right]\\ & \frac{1}{\lambda }=\mathrm{R}\left[\frac{16-9}{16\times 9}\right]\\ & \Rightarrow \lambda =\frac{16\times 9}{7\mathrm{R}}=\frac{16\times 9}{7\times 1.097\times 10^7}\\ & \lambda =1.876\times 10^{-6}\mathrm{m}\end{array}$