Ratio between longest wavelength of H atom in Lyman series to the shortest wavelength in Balmer series of ${\mathrm{He}}^{+}\mathrm{is}$

Longest wavelength in Lyman Series of hydrogen atom arises from transition between n = 2 to n = 1 whose number is given by

${\overline{\mathrm{v}}}_{\mathrm{H}\left(2\to 1\right)}=\mathrm{R}\times 1^2\left(1\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3}{4}\mathrm{R}$

Shortest wavelength in Balmer series of $H{e}^{+}$ arises from transition between n = $\infty$ to n = 2 whose wave number is given by

$\begin{array}{l}{\overline{\mathrm{v}}}_{{\mathrm{He}}^{+}\left(\infty -2\right)}=\mathrm{R}\times 2^2\left(\frac{1}{2^2}-\frac{1}{{\text{∞}}^2}\right)=\mathrm{R}\\ \frac{{\mathrm{v}}_{{\mathrm{He}}^{+}}}{{\mathrm{v}}_{\mathrm{H}}}=\frac{{\mathrm{\lambda }}_{\mathrm{H}}}{{\mathrm{\lambda }}_{{\mathrm{He}}^{+}}}\\ \therefore \frac{{\mathrm{\lambda }}_{\mathrm{H}}}{{\mathrm{\lambda }}_{{\mathrm{He}}^{+}}}=\frac{4\mathrm{R}}{3\mathrm{R}}=\frac{4}{3}\end{array}$