Which transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of ${\text{He}}^{+}$ spectrum ?

$\frac{1}{\lambda }={\text{R}}_{\text{H}}{\text{Z}}^2\left(\frac{\text{1}}{{\text{n}}_{\text{1}}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{2}}^{\text{2}}}\right)$

$$\begin{gathered} {\text{R}}_{\text{H}}{{\text{Z}}_{\mathrm{H}}}^2\left(\frac{\text{1}}{{\text{n}}_{\text{1}}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{2}}^{\text{2}}}\right){\text{=R}}_{\text{H}}{\text{Z}}_{{\mathrm{He}}^{+}}^2\left(\frac{\text{1}}{{\text{2}}^2}-\frac{\text{1}}{4^2}\right) \\ \text{1}\left(\frac{\text{1}}{{\text{n}}_{\text{1}}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{2}}^{\text{2}}}\right)\text{=4}\left(\frac{\text{1}}{4}-\frac{\text{1}}{16}\right) \\ \left(\frac{\text{1}}{{\text{n}}_{\text{1}}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{2}}^{\text{2}}}\right)\text{=}\frac{3}{4} \end{gathered}$$

n₁=1, hence n₂=2,3,4,..

For first line, n₁=1 and n₂=2

$\left((\frac{\text{1}}{{\text{n}}_{\text{1}}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{2}}^{\text{2}}}\right)=\frac{1}{1}-\frac{1}{4}=\frac{3}{4}$