 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2022/paper-1/a2/c/i/solution

The atomic nuclei are much heavier than electrons, so nuclear motion is slower. In the Born-Oppenheimer approximation, first hold the nuclear positions $R$ fixed and solve
$\hat{H}_{e} (r; R) ψ_{n} (r; R) = U_{n} (R) ψ_{n} (r; R) .$
(1)
Approximate the complete wavefunction by $Ψ (r, R) = ψ_{n} (r; R) χ_{n} (R)$ and neglect the nuclear-derivative couplings involving $\nabla_{R} ψ_{n}$ and $\nabla_{R}^{2} ψ_{n}$ .
Nuclear motion then occurs on one electronic potential energy surface:
$[\hat{T}_{N} + U_{n} (R) + V_{NN} (R)] χ_{n} = E χ_{n} .$
(2)
Here $U_{n}$ excludes the nuclear repulsion $V_{NN}$ , which must be included once.
The mass separation makes the discarded coupling terms small when electronic energy gaps are sufficiently large; near a degeneracy or avoided crossing, a single-surface description can fail.

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