Chem 3240 · Lecture 7.1
\[\hat{H} = -\frac{\hbar^2}{2m_e}\left(\Delta_1 + \Delta_2\right) - \frac{1}{4\pi\epsilon_0}\left(\frac{Ze^2}{r_1} + \frac{Ze^2}{r_2} - \frac{e^2}{r_{12}}\right)\]
\[\hat{H} = \hat{H}_1 + \hat{H}_2\]
\[\hat{H}_i = -\frac{\hbar^2}{2m_e}\Delta_i - \frac{Ze^2}{4\pi\epsilon_0 r_i}\]
\[E = E_1 + E_2 = -RZ^2\left(\frac{1}{n_1^2} + \frac{1}{n_2^2}\right)\]
\[\psi(r_1,r_2) = \frac{1}{\pi}\left(\frac{Z}{a_0}\right)^3 e^{-Z(r_1 + r_2)/a_0}\]
\[E = \langle\psi|\hat{H}|\psi\rangle = \left[Z^2 - \frac{27Z}{8}\right]\frac{e^2}{4\pi\epsilon_0 a_0}\]
\[\frac{dE}{dZ} = \left(2Z - \frac{27}{8}\right)\frac{e^2}{4\pi\epsilon_0 a_0} = 0\]
\[Z = \frac{27}{16} \approx 1.7, \qquad E \approx -77.5 \text{ eV}\]
Helium has no exact solution because the \(1/r_{12}\) repulsion couples the electrons. We approximate it as two hydrogenlike orbitals, then sharpen the estimate with the variational principle: optimizing the nuclear charge gives \(Z \approx 1.7\), our first measure of shielding, and an energy of \(-77.5\) eV against the exact \(-79.0\) eV.