Find the value of wave number ($\overline{\mathrm{v}}$) in terms of Rydberg’s constant, when transition of electron takes place between two levels of He⁺ ion whose sum is ‘4’ and difference is ‘2’.

one Energy level 'n₁' & other energy level 'n₂'

n₁+n₂ =4 ----------(1)

n_1-n₂ =2------------(2)

By adding Eq 1 and eq 2

2n₁ = 6

n₁ = 3

n₁ value is substituting in equation (1)

n₁+n₂ =4 

3+n₂ =4 

n₂ = 1

$$\begin{gathered} \stackrel{-}{\mathrm{v}} =\frac{1}{\mathrm{\lambda }} ={\mathrm{R}}_{\mathrm{H}}{\mathrm{Z}}^2[\frac{1}{1^2}-\frac{1}{3^2}] \\ \stackrel{-}{\mathrm{v}} =\frac{1}{\mathrm{\lambda }} =\mathrm{R}{2}^2[\frac{1}{1}-\frac{1}{9}] \\ \stackrel{-}{\mathrm{v}} =\mathrm{R} 4\left[\frac{8}{9}\right] \\ \stackrel{\mathit{-}}{\mathit{v}}\mathit{ }\mathit{=}\frac{\mathit{32}\mathit{R}}{\mathit{9}} \end{gathered}$$