If the shortest wavelength of spectral line of H - atom in Lyman series is x, Then match the following for ${\mathrm{Li}}^{+}$

Shortest wavelength of spectral line of H-atom in Lyman series is ‘X’ shortest wavelength in Lyman series for H-atom.

$\mathrm{A}\to \mathrm{B}$

$\stackrel{-}{\mathrm{\vartheta }}=\frac{1}{\lambda }=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]\times 2^2$

$\mathit{=}R\left[\frac{1}{1}-\frac{1}{{\infty }^2}\right]\times 2^2$

$=\mathrm{R}\left(1\right)\times {\left(1\right)}^2=\mathrm{R}$

$\mathrm{\lambda }=\frac{1}{\mathrm{R}}=\mathrm{X}$

Shortest wavelength in Lymon series for ${\mathrm{Li}}^{+2}$

$=\frac{\mathit{1}}{\lambda }R\left[\frac{1}{1^2}-\frac{1}{{\infty }^2}\right]\times \left(3^2\right)=9\mathrm{R}$

$\mathrm{\lambda }=\frac{1}{9\mathrm{R}}\Rightarrow \mathrm{\lambda }=\frac{\mathrm{x}}{9}$

$\mathrm{B}\to \mathrm{S}$

Longest wavelength in Lymon series for ${\mathrm{LI}}^{+2}$

$\frac{1}{\mathrm{\lambda }}=R\left[\frac{1}{1}-\frac{1}{2^2}\right]\times 3^2$

$\mathrm{R}\times \frac{3}{4}\times 9=\frac{27}{4}\mathrm{R}$

$\mathrm{but} \frac{1}{\mathrm{R}}=\mathrm{X}$

$\therefore \mathrm{\lambda }=\frac{4}{27}\frac{1}{\mathrm{R}}=\frac{4}{27}\mathrm{X}$

$\mathrm{C}\to \mathrm{Q}$

Shortest wavelength in Balmer series

$\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{{\infty }^2}\right]\times 9$

$\frac{9\mathrm{R}}{4}$

$\therefore \mathrm{\lambda }=\frac{4}{9}.\frac{1}{\mathrm{R}}=\frac{4\mathrm{x}}{9}$

$\mathrm{D}\to \mathrm{P}$

Longest wavelength in Balmer series

$\frac{1}{\mathrm{\lambda }}=R\left[\frac{1}{2^2}-\frac{1}{3^2}\right]\times 9$

$\frac{1}{\mathrm{\lambda }}=\frac{45}{36}\mathrm{R}$

$\frac{36}{45}.\frac{1}{R}=\frac{4}{5}X=\lambda$