The transition in ${\mathrm{He}}^{\oplus }$ ion that would have the same wavelength as the first Lyman line in hydrogen spectrum is

$\begin{array}{l}{\overline{\mathrm{v}}}_{{\mathrm{He}}^{\oplus }}=\frac{1}{\mathrm{\lambda }}=\mathrm{R}\times 2^2\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right)=\mathrm{R}\left(\frac{4}{{\mathrm{n}}_1^2}-\frac{4}{{\mathrm{n}}_2^2}\right)....\left(\mathrm{i}\right)\\ {\overline{\mathrm{v}}}_{{\mathrm{H}}_2\oplus }=\frac{1}{\mathrm{\lambda }}=\mathrm{R}\times 1^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)....\left(\mathrm{i}\right)\end{array}$

Compare equations (i) and (ii), we get

$$\begin{gathered} \mathrm{R}\times 1^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\mathrm{R}\times 4^2\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right) \\ {\mathrm{n}}_1^2=4, {\mathrm{n}}_1=2 \\ {\mathrm{n}}_2^2=16, {\mathrm{n}}_2=4 \end{gathered}$$

$(4\to 2)$