$\frac{1}{\lambda}$ = $\overset{-}{v}$ = 1.097 × $10^{7} Z^{2} \left[ \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right] \mathrm{m}^{-1}$ where, $n_{1}$ = number of the lower energy level, $n_{2}$ = number of the higher energy level, Z = atomic number, λ = wavelength, $\overset{-}{v}$ = wavenumber For $\mathrm{He}^{+}$, $\frac{1}{\lambda}$ = 1.097 × $10^{7}$ × $(2)^{2} \left[ \frac{1}{(2)^{2}} - \frac{1}{(4)^{2}} \right] \mathrm{m}^{-1}$ = 1.097 × $10^{7} \times 4 \left( \frac{3}{16} \right)$ = 1.097 × $10^{7} \times \frac{3}{4}$ For H atom, $\frac{1}{\lambda}$ = 1.097 × $10^{7} \times (1)^{2} \times \frac{3}{4}$ = 1.097 × $10^{7} \left[ \frac{1}{n_{L}^{2}} - \frac{1}{n_{H}^{2}} \right]$ $\left[ \frac{1}{n_{L}^{2}} - \frac{1}{n_{H}^{2}} \right]$ = $\frac{3}{4}$. Thig gives $n_{L}$ = 1 , $n_{H}$ = 2 The transition $n_{L}$ = 1 to $n_{H}$ = 2 in H-atom would have the same wavelength as Balmer transition n = 4 to n = 2 of $\mathrm{He}^{+}$.