Calculate the energy of ionization of the dihydrogen [itex]H_2[/itex], knowing that the binding energy of [itex]H_2[/itex] and [itex]H_2^+[/itex] are worth -4.48 eV and -2.65 eV respectively and that the ionization energy of the hydrogen atom is 13.6 eV.
2. Relevant thoughts
Energy of ionization of [itex]H_2[/itex] is the energy required to remove 1 electron from the system "2 protons+1 electron".
Binding energy of [itex]H_2[/itex] is the energy that the system :2 electrons+2 protons have. Since it's negative it means it's stable (when totally dissociated the molecule has 0 energy).
The Attempt at a Solution
I don't really know how to use the given data.
A try: The binding energy increases by 1.83 eV when one removes an electron from [itex]H_2[/itex]. The ion molecule is thus less stable (but still is). So I'm guessing that it's easier by an amount of 1.83 eV to remove an electron in the [itex]H_2[/itex] molecule compared to the hydrogen atom. This would make the answer to the problem 11.77 eV.
However in wikipedia I find 15.603 eV. I have absolutely no idea how to reach such a number (or 15.6 for that matter).
Any tip is welcome.