Wait... I did a mistake earlier >.< I think there was some number on my calculator from a previous work which got in the calculation.
I used the simple E = hf
Now, using:
Meaning that the ion gained 2.75 eV to get on to the n+1 state.
The ionisation energy from the n+1 state is thus = 7.65 - 2.75 = 4.90 eV
Okay, so the ionisation energy for:
n = 7.65 eV
n + 1 = 4.90 eV
Now, for this formula you gave, this is the first time I see it and I did some research.
Okay, so let's try He^+
Z = 2
If En and En+1 give satisfactory values of n, then He^+ is the one.
If not, we try with Li^+
Z = 3