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

$\begin{array}{l}\mathrm{For}{\mathrm{He}}^{+}\mathrm{ion},\\ \frac{1}{\mathrm{\lambda }}={\mathrm{Z}}^2\mathrm{R}\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]\\ ={\left(2\right)}^2\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=\frac{3\mathrm{R}}{4}\\ \mathrm{For}\mathrm{hydrogen}\mathrm{atom},\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]\\ \mathrm{So},\frac{3\mathrm{R}}{4}=\mathrm{R}\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]\\ \mathrm{or}\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}=\frac{3}{4}\\ \mathrm{i}.\mathrm{e}.,{\mathrm{n}}_1=1\mathrm{and}{\mathrm{n}}_2=2\end{array}$