Chemistry - Atomic Structure Question with Solution | TestHub

Accordinq to Rydberg's equation

$\overline{\mathrm{v}}=\frac{1}{\mathrm{\lambda }}={\mathrm{RZ}}^2\left[\frac{1}{{\mathrm{n}}_{\mathrm{L}}^2}-\frac{1}{{\mathrm{n}}_{\mathrm{H}}^2}\right]$

where,

$\overline{\mathrm{v}}=$ wave number

$\mathrm{\lambda }=$ wavelength

$\mathrm{R}=$ Rydberg constant $\left(=109677 {\mathrm{cm}}^{-1}\right)$

$\mathrm{Z}=$ atomic number

${\mathrm{n}}_{\mathrm{L}}=$ lower energy state

${\mathrm{n}}_{\mathrm{H}}=$ higher energy state.

i.e. $\frac{1}{\mathrm{\lambda }}\propto {\mathrm{Z}}^2$

Thus, $\frac{\mathrm{\lambda }\left({\mathrm{He}}^{+}\right)}{\mathrm{\lambda }\left(\mathrm{H}\right)}=\frac{{\mathrm{Z}}^2\left(\mathrm{H}\right)}{{\mathrm{Z}}^2\left({\mathrm{He}}^{+}\right)}$

To find $\mathrm{\lambda }\left({\mathrm{He}}^{+}\right)=?$

Given, $\mathrm{\lambda }\left(\mathrm{H}\right)=91.2 \mathrm{nm}$

$\mathrm{Z}\left(\mathrm{H}\right)=1$

$\mathrm{Z}\left(\mathrm{He}\right)=2$

$\therefore \mathrm{\lambda }\left({\mathrm{He}}^{+}\right)=\frac{1\times 91.2}{4}=22.8 \mathrm{nm}$