An excited He⁺ ion emits two photons in succession with wavelength 108.5 and 30.4 nm respectively. The quantum number 'n' corresponding to its initial excited state is, $\mathrm{E}=\frac{1240}{\mathrm{\lambda }\left(\mathrm{nm}\right)}\mathrm{eV}$

$$\begin{gathered} \mathrm{\Delta E}={\mathrm{\Delta E}}_1+{\mathrm{\Delta E}}_2=\frac{1240}{{\mathrm{\lambda }}_1}+\frac{1240}{{\mathrm{\lambda }}_2} \\ =1240\left[\frac{1}{108.5}+\frac{1}{30.4}\right]=11.43+40.78=52.2 ev \end{gathered}$$

$\begin{array}{l}{\mathrm{E}}_1\text{ of }{\mathrm{He}}^{+}=-13.6\times 4=-54.4\mathrm{eV}\\ {\mathrm{E}}_2\text{ of }{\mathrm{He}}^{+}=\frac{-13.6\times 4}{4}=-13.6\mathrm{eV}\\ {\mathrm{E}}_3\text{ of }{\mathrm{He}}^{+}=\frac{-13.6\times 4}{9}=-6.04\mathrm{eV}\\ {\mathrm{E}}_4\text{ of }{\mathrm{He}}^{+}=\frac{-13.6\times 4}{16}=-3.4\mathrm{eV}\\ {\mathrm{E}}_5\text{ of }{\mathrm{He}}^{+}=\frac{-13.6\times 4}{25}=2.176\mathrm{eV}\\ \text{Thus, }{\mathrm{E}}_5-{\mathrm{E}}_1=54.4-2.17=52.2\mathrm{eV}\end{array}$