How Can the First Ionization Energy of Helium be Accurately Calculated Using Basic Quantum Mechanics?

There is a book which summarizes the fairly accurate calculation of the First Ionization Energy of Helium, but it would be nice to see if there is a more direct answer than the one calculated, that might have better accuracy even.

The experimental ionization energies for Hydrogen and Helium are given at this link: Experimental Values of Ionization Energies and in Table 3 in the linked Chemistry book here:

Table 3: Ionization Energies (kJ/mol)

Element 1 2 3 4 5 6 7 8
H 1312
He 2372 5250
Li 520 7297 11810
Be 899 1757 14845 21000
B 800 2426 3659 25020 32820
C 1086 2352 4619 6221 37820 47260
N 1402 2855 4576 7473 9442 53250 64340
O 1314 3388 5296 7467 10987 13320 71320 84070
F 1680 3375 6045 8408 11020 15160 17860 92010
Ne 2080 3963 6130 9361 12180 15240
Na 496 4563 6913 9541 13350 16600 20113 25666
Mg 737 1450 7731 10545 13627 17995 21700 25662

Are these the most currently-accepted experimentally-confirmed values?

Now how can these energies be calculated using radially-symmetric basic Quantum Mechanics Theory accurately?

I have made fairly accurate Helium ionization calculations at this link. What are other accurate approaches that use just the radially-symmetric Schrödinger wave equation, with a few small simplifications?

  • 2
    $\begingroup$ You should define what you mean by "accurate" in context of your post. As any quantum calculations of a three-body system are inaccurate, unless we put the inaccuracy threshold above their inaccuracies. $\endgroup$
    – Poutnik
    Jun 21, 2023 at 9:18
  • 1
    $\begingroup$ Note that using photos/screenshots of text instead of typing text itself is highly discouraged. The image text content cannot be indexed nor searched for, nor can be reused in answers. Specifically handwritten scripts can be difficult to decipher. Consider copy/pasting or rewriting of at least essential parts. // Optionally, here are formatting guides for texts and formulas/equations/expressions. $\endgroup$
    – Poutnik
    Jun 21, 2023 at 9:19
  • $\begingroup$ @Poutnik I try to reference the picture as a link and a picture only in the case where the copy link option **produces an error and does not work **. If the link to the picture works, I always use the link, whenever possible. In these cases where I got an error message with the link option, I took a screenshot of the picture and posted a separate html link reference. Also, the accuracy calculation is given in the online book link reference. $\endgroup$ Jun 21, 2023 at 10:46
  • 1
    $\begingroup$ The point is, the question content policy focuses on questions being self-contained, with quoted or reproduced essential content. Links are provided rather just as the content origin reference, or for further reading. By other words, the question should make perfect sense without following any link. // Instead of table images, there are preferred tables created by SE Markdown or by MathJAx/LaTex code. See e.g. SE meta - tables // I think this your case is acceptable, just for future. $\endgroup$
    – Poutnik
    Jun 21, 2023 at 10:59
  • 4
    $\begingroup$ Is this an ad for your book? $\endgroup$
    – Buck Thorn
    Jun 21, 2023 at 16:35

1 Answer 1


There is no need to reach the experimental results by basic methods, the wave function corresponding to $He$ atom has no analytical expression. The numerical methods used do not often bare a significant scientific meaning. Trying to reach this experimental result is more or less a computational challenge. The expansion of the ground-state function in a minimal basis-set leads to an accurate description depending on a right choice of the basis functions for this expansion. We are more interested in the suitable basis functions for reducing the computational cost.

The current version of the book that I downloaded, my calculated ionization energy for Helium was 22.1433 eV (2136.6417 kJ/Mol ) conversion at unitsconverters.com/en/Ev/Particle-To-Kj/Mol/Utu-6180-6179 . Then the accuracy is (2372−2136.6417)/2372∗100= +9.9% accurate

Assuming that your second ionization energy is given by the analytical formula of hydrogenic atoms $E_2 = \frac{Z^2}{2n^2}$ a.u. , your total energy, related to the first ionization energy $E_1=22.14$ eV will be $E_{tot} = -(E_1+E_2) = -76.54$ eV. This ground state energy is not accurate at all for a numerical method. A variational method applied considering $V_{ee}$ as a perturbation gives an analytical solution of $E_{tot} = - 77.7$ eV more accurate that your numerical result.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.