Building the Periodic Table — Question 5

Question 5

A simple pairing model predicts when an orbital splitting overcomes Hund filling

Two electrons occupy two spatial orbitals of energies 0 and $\Delta>0$. In a supplied model, pairing both in the lower orbital costs interaction energy U; placing one in each with parallel spins costs $\Delta+U\prime-J$, where $J>0$ is an exchange lowering.

1. Derive the criterion for the parallel-spin state to be lower.

2. For $U=4.0\,\mathrm{eV}$, $U\prime=2.8\,\mathrm{eV}$ and $J=.6\,\mathrm{eV}$, identify the lower state at $\Delta=1.0$ and $2.5\,\mathrm{eV}$.

Question 5 - Solution

A simple pairing model predicts when an orbital splitting overcomes Hund filling

(a) Method and calculation

$$E_{\mathrm{pair}}=U,\quad E_{\mathrm{parallel}}=\Delta+U\prime-J.$$$$E_{\mathrm{parallel}}<E_{\mathrm{pair}}\Rightarrow\boxed{\Delta<U-U\prime+J}.$$

(b) Interpretation and further calculation

$$\Delta_c=4.0-2.8+.6=\boxed{1.8\,\mathrm{eV}}.$$$$\boxed{\Delta=1.0:\ E_{\mathrm{parallel}}=3.2<U;\qquad\Delta=2.5:\ E_{\mathrm{pair}}=4.0<4.7}.$$

Physical check

Both states obey exclusion. The competition between orbital splitting, direct repulsion, and exchange determines the spin state in this explicitly simplified model.