$\text{ Given, }\mathrm{\Delta E}={\overline{\mathrm{R}}}_{\mathrm{H}}{\mathrm{Z}}^2\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]$

${\overline{\mathrm{R}}}_{\mathrm{H}}$ (Rydberg constant) for H-atom is x m^-1. Corresponding value for this constant for He⁺ ion is

$\begin{array}{l}\overline{\mathrm{v}}={\overline{\mathrm{R}}}_{\mathrm{H}}{\mathrm{Z}}^2\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]\\ {\overline{\mathrm{R}}}_{\mathrm{H}}=\frac{2{\mathrm{\pi }}^2{\mathrm{me}}^4{\mathrm{k}}^2}{{\mathrm{ch}}^3}\\ =\mathrm{x} {\mathrm{m}}^{-1} \mathrm{for} \mathrm{H}-\mathrm{atom}\\ {\overline{\mathrm{R}}}_{\mathrm{Z}}=\frac{2{\mathrm{\pi }}^2{\mathrm{me}}^4{\mathrm{k}}^2}{{\mathrm{ch}}^3}\left({\mathrm{Z}}^2\right)\end{array}$

For a species of atomic number Z (with one electron only)

$\begin{array}{l}\frac{{\overline{\mathrm{R}}}_{\mathrm{Z}}}{{\overline{\mathrm{R}}}_{\mathrm{H}}}={\mathrm{Z}}^2\\ \therefore {\overline{\mathrm{R}}}_{\mathrm{Z}}=(2{)}^2{\overline{\mathrm{R}}}_{\mathrm{H}}=4\times {\mathrm{m}}^{-1}=\frac{4\times }{100}\left({\mathrm{cm}}^{-1}\right)=0.04\times {\mathrm{cm}}^{-1}\end{array}$