stylistic:tone, quality:correction
Screening effect

The energy that we obtained is too low because the repulsion term between the electrons was ignored, whose effect is to raise the energy levels. As Z gets bigger, our approach should yield better results, since the electron-electron repulsion term will get smaller. H ′ ¯ = k e 2 r 12 − k Z e 2 r 1 − V (r 1) − k Z e 2 r 2 − V (r 2) {\displaystyle {\bar {H'}}={\frac {ke^{2}}{r_{12}}}-{\frac {kZe^{2}}{r_{1}}}-V(r_{1})-{\frac {kZe^{2}}{r_{2}}}-V(r_{2})} V (r) is a central potential which is chosen so that the effect of the perturbation H ′ ¯ {\displaystyle {\bar {H'}}} is small. The net effect of each electron on the motion of the other one is to screen somewhat the charge of the nucleus, so a simple guess for V (r) is V (r) = − k (Z − S) e 2 r = − k Z e e 2 r {\displaystyle V(r)=-{\frac {k(Z-S)e^{2}}{r}}=-{\frac {kZ_{e}e^{2}}{r}}} where S is a screening constant and the quantity Z is the effective charge. The potential is a Coulomb interaction, so the corresponding individual electron energies are given by E 0 = − k (Z − S) 2 e 2 2 a 0 n 2 = − k Z e 2 e 2 2 a 0 n 2 {\displaystyle E_{0}=-{\frac {k(Z-S)^{2}e^{2}}{2a_{0}n^{2}}}=-{\frac {kZ_{e}^{2}e^{2}}{2a_{0}n^{2}}}} and the corresponding spatial wave function is given by ψ 0 (r 1, r 2) = Z e 3 π a 0 3 e − Z e (r 1 + r 2) {\displaystyle \psi _{0}(r_{1}\,,r_{2})={\sqrt {\frac {Z_{e}^{3}}{\pi a_{0}^{3}}}}e^{-Z_{e}(r_{1}+r_{2})}} If Z was 1.70, that would make the expression above for the ground state energy agree with the experimental value E = −2.903 a.u. of the ground state energy of helium. Since Z = 2 in this case, the screening constant is S = 0.30. For the ground state of helium, for the average shielding approximation, the screening effect of each electron on the other one is equivalent to about 1 3 {\textstyle {\frac {1}{3}}} of the electric charge.