The energy of a hydrogen atom in the nth excited state is given by - 13.6/n2 eV
Hence the energy of a ionised helium atom would be given by
for first excited state, the transition is from n = 1 to n = 2 state.
Now for n = 1
Hence the excitation energy of the H\(^+_e\) ion is –13.6 – (-54.4) = 40.8 eV
If V1 is the first excitation potential for H\(^+_e\)