The first member of the Paschen series in hydrogen spectrum is of wavelength 18,800 $\stackrel{0}{\mathrm{A}}$. The short wavelengths limit of Paschen series is

For paschen series $\overline{\mathrm{v}}=\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left(\frac{1}{3^2}-\frac{1}{{\mathrm{n}}^2}\right)\mathrm{here}\mathrm{n}=4,5,6...$

For first member of paschen series n = 4

$\begin{array}{l}\frac{1}{{\mathrm{\lambda }}_1}=\mathrm{R}\left[\frac{1}{3^2}-\frac{1}{4^2}\right]\Rightarrow \frac{1}{{\mathrm{\lambda }}_1}=\frac{7\mathrm{R}}{144}\\ \Rightarrow \mathrm{R}=\frac{144}{7{\mathrm{\lambda }}_1}=\frac{144}{7\times 18800\times {10}^{-10}}=1.1\times {10}^{-7}\end{array}$

For shortest wave length n = $\infty$

$\begin{array}{l}\mathrm{So}\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left[\frac{1}{3^2}-\frac{1}{{\mathrm{\infty }}^2}\right]=\frac{\mathrm{R}}{9}\\ \Rightarrow \mathrm{\lambda }=\frac{9}{\mathrm{R}}=\frac{9}{1.1\times {10}^{-7}}=8.225\times {10}^{-7}\mathrm{m}=8225\stackrel{0}{\mathrm{A}}\end{array}$