In the helium atom, if you drop the shielding approximation for the remaining electron in the ionized state, as common sense would suggest, the ionization energy would become negative! This illustrates the dangers of mixing models at random. This problem might also be why the discussion in [25] is based on the zero shielding approximation, rather than the full shielding approximation used here.
But zero shielding does make the base energy levels of the critical outer electrons of heavy atoms very large, proportional to the square of the atomic number. And that might then suggest the question: if the energy levels explode like that, why doesn't the ionization energy or the electronegativity? And it makes the explanation why helium would not want another electron more difficult. Full shielding puts you in the obviously more desirable starting position of the additional electron not being attracted, and the already present electrons being shielded from the nucleus by the new electron. And how about the size of the atoms imploding in zero shielding?
Overall, this book prefers the full shielding approach. Zero shielding would predict the helium ionization energy to be 54.4 eV, which really seems worse than 13.6 eV when compared to the exact value of 24.6 eV. On the other hand, zero shielding does give a fair approximation of the actual total energy of the atom; 109 eV instead of an exact value of 79. Full shielding produces a poor value of 27 eV for the total energy; the total energy is proportional to the square of the effective nucleus strength, so a lack of full shielding will increase the total energy very strongly. But also importantly, full shielding avoids the reader’s distraction of having to rescale the wave functions to account for the nonunit nuclear strength.
If eventually X-ray spectra need to be covered in this book, a
description of hot
relativistic inner electrons would
presumably fix any problem well.