It's cheating, but it's the same cheat that everyone else uses. (Except that I happened to do an especially nice job of streamlining the math by using some very cool scaling arguments!) You get a pretty good approximation to the ground state energy level, and you can't really improve on it unless you go the whole hog and calculate the 6-dimensional wave function of both electrons.
Then I did something that was very very cool. I generalized the solution method to apply to the whole isoelectronic series of Helium: that is, every possible ion consisting of a nucleus with two electrons. That would be H(-) (yes, there is a stable species of Hydrogen with an extra electron), He(neutral), Li(+) and Be(2+) etc. The funny thing about the series is that the higher up you go, the more accurate it gets. I actually worked out a table of values which I promised to post, where I compared the calculated energies to the experimentally determined values. Then I got distracted and I never did post my table. Well, here it is now:
You can see there are three terms in the formula: the first is the potential energy, the second is the kinetic energy, and the third is the interaction energy of the two electrons, due to their mutual repulsion. The k factor tells you how much the wave function spreads out to accomodate this repulsion, and you calculate it just by using basic first-year calculus to minimize the energy. For helium you get a relaxation factor of 27/32. (Other people get 27/16 but they are defining their terms a little differently from me.) It turns out the relaxation factor approaches one as z approaches infinity. This means that the interaction energy becomes less significant as the atoms get bigger. For U(90+) the electrons just basically share the scaled-down hydrogen ground-state orbital. (That's uranium stripped down to its last two electrons, in case you didn't figure that out.)
My formula is very accurate at the high end, but starts to diverge at the low end. In fact, for the hydrogen negative ion, it does something quite bad. My calculated value is -12.86 eV, and the actual energy is -14.35 eV, which is more than 10% off. That might not seem so bad, but in fact it makes a huge difference in the physics. The binding energy of ordinary hydrogen is 13.6 eV, or as it is known, one Rydberg. My calculated binding energy is just under one Rydberg, so it is actually preferable for the hydrogen to eject the extra unwanted electron. In fact, the true binding energy is just slightly greater than one Rydberg, and this makes the configuration stable with two bound electrons. In fact, H(-) is a stable species, which you wouldn't have known from my calculations.
Why doesn't my formula work so well for the lightest atoms? Because the electrons have a more intricate way of minimizing their energy, which goes beyond what I can account for with my simple model. You have to treat the system in 6-dimensional phase space in order to do the optimisation. Not about to happen on this blog. (Not without some tricks anyhow).
But that's not exactly where I wanted to go with this. There's something very funny about the binding energies, and I'm not sure anyone has actually solved this problem. As the atoms get lighter, the binding energy gets less and less. What is the lightest atom which is stable with two electrons? Obviously, hydrogen. But what if you could shave a bit of charge off the proton? Wouldn't it still be stable, up to a point? I'm interested in that very special atom consisting of something like 0.92 protons and two electrons, where the binding energy is...exactly zero. Has anyone calculated this atom? Because I think I know how to write down the wave function for it, and I mean the real wave function, not just an approximation.