# How does the inter-atomic distance affect the energy of an electron?

In the Born-Oppenheimer approximation, we are able to separate the electron solutions from the nuclei solutions. This gives an eigen value equation for the nucleus which has a potential given by the electron energies (which are obtained for each configuration of the nuclei). Let's say we are dealing with a hydrogen molecule with one electron (no electron-electron interaction). How will the electron energies vary with each inter-atomic spacing? I have been told the electron energy will diverge for close inter atomic spacing and go to zero for large inter-atomic spacing, having a minimum around intermediate inter-atomic spacing. Is it correct? if it is, what can be the physical interpretation of that?

• you mean a single hydrogen atom? the Born-Oppenheimer approximation makes sense only in the context of some structure that the nuclei have to assume, in response to the effective electronic potential. If you have a single atom it cannot really apply. What exactly is the setup you are considering?
– user245141
Nov 5, 2019 at 15:05
• The situation is a hydrogen molecule with just one electron. Nov 5, 2019 at 15:11
• Oh a molecule! missed that. thanks for the clarification
– user245141
Nov 5, 2019 at 15:17
• You are welcome! So, how can this be explained? Nov 5, 2019 at 15:20
• see nyu.edu/classes/tuckerman/adv.chem/lectures/lecture_13/… you do not need approximations Nov 5, 2019 at 15:38

## 2 Answers

The Hamiltonian of a your $$H_2^+$$ molecule is given by $$\hat{H} = \hat{T}_\text{electron} + \hat{T}_\text{nuclei} + \hat{V}_\text{electron-nucleus} + \hat{V}_\text{nucleus-nucleus}.$$ If you want to get an energy for the electron from this, it is kind of arbitrary how you divide the contribution $$\hat{V}_\text{electron-nucleus}$$ between the nuclei and the electron. In the Born-Oppenheimer Approximation, we "freeze" the nuclei, so we have $$\hat{H}_e = \hat{T}_\text{electron} + \hat{V}_\text{electron-nucleus} + \hat{V}_\text{nucleus-nucleus}.$$ The eigenvalue of this Hamiltonian is called the electronic energy and is not actually the energy of the electron (which would be arbitrary anyway). This diverges when the nuclei get close to each other because of the repulsion between the nuclei $$\hat{V}_\text{nucleus-nucleus}$$.

As for a behavior when going to infinite separations: By convention one sets the potential to be zero if all particles (electrons and nuclei) are infinitely far apart. In your case however, only the nuclei get separated ($$H_2^+ \rightarrow H^+ + H$$). So the electronic energy at infinite separation is the formation energy of a hydrogen atom from an electron and a proton, which is non-zero. The minimum in between is due to the attraction between nuclei and electrons. It is actually energetically favourable for the two protons to share one electron than for one to be "naked" and the other have the electron by itself. This is what we call a chemical bond and you can find more details about it on the Chemistry Stack Exchange.

The electron energy will go down but the proton-proton repulsion will increase more. However your question was about electron energy. This will go down from 1 rydberg from infinite to 4 rydberg for zero proton separation. If you throw in one or two neutrons you will able to create a stable He nucleus from 2 protons but that takes us to the topic of hydrogen fusion.