8
$\begingroup$

I asked my physics teacher why two electrons come in pairs and not push each other away as you would expect from negative charges. He said that according to the Pauli exclusion principle, there are a limited amount of states that can be occupied in certain energy levels, so you know that when there are two electrons in the same shell, one of them has spin 'up' and the other 'down'.

How does spin counteract the repulsive electric force?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Why do you think the repulsion shouldn't be present? $\endgroup$
    – BioPhysicist
    Commented Apr 19 at 15:33
  • $\begingroup$ You have a couple of answers already. I just wanted to confirm: you are right, Pauli principle is of different nature and has nothing to do with Coulomb force of repulsion. Check out planetary model of an atom for confirmation and change your physics teacher if you can. $\endgroup$
    – MsTais
    Commented Apr 19 at 15:54
  • $\begingroup$ The planetary model is so deeply flawed there is no good reason to suggest that a student look into it. Better that they learn the truth than hang on a lie. $\endgroup$
    – Matt Hanson
    Commented Apr 22 at 11:06

3 Answers 3

Reset to default
20
$\begingroup$

First of all, the electrons do exert a repulsive force on one another. This is inherently accounted for in the electrostatic repulsion terms in the atomic Hamiltonian that is used to solve the Schrödinger equation numerically. The fact that electrons will pair up is just a result of the fact that these states are the most optimal ones possible, despite this repulsion. I will treat this question at a rather basic level by hand waving away the fact that changing the electron configuration changes the energies of the electron levels. Instead, I will act as if there were orbital energy levels "already there" and we are just trying to decide how to fill them by considering effects like pair repulsion.

For a rather obvious example, take the He atom. Despite the fact that there is obvious pair repulsion associated to both electrons being in the $1s$ spatial orbital, the cost of promoting an electron to the $2s$ orbital to avoid this repulsion is unreasonably high, and so the spin paired electrons are the most stable configuration. It is also noteworthy that despite the fact that the electron repulsion is fairly strong in more spatially compressed orbitals, the energy levels of the orbitals are also highly stable in the first place.

To see a practical example of how electron repulsion can affect the relative stability of different electron configurations, one can look at the transition metals of the periodic table. Despite the fact that the $4s$ state is nominally lower in energy than the neighboring $3d$ state, the valence electrons in a Cr atom adopt a $4s^13d^5$ configuration. This is because spreading out the electrons in distinct spatial orbitals lowers the overall energy of the atom more than simply putting the electrons in the lowest energy orbitals. But the energy savings gained by avoiding pair repulsion is relatively small, and the associated levels need to be rather close in energy for this to outweigh the cost associated to promoting an electron to a higher shell or subshell. In fact, as we go to higher energy shells, the energy associated to pair repulsion generally decreases due to the larger spatial spread of these electron orbitals.

The issues of spin are many and highly complicated, so I will not attempt to flesh that out here. I will just point out that your final question is a bit confused, as the spin does not "counteract" the repulsive force in any way. The spin is just the spin, and for complex reasons it has to do with what states are allowed in the first place (i.e. electrons in the same spatial orbital must have opposite spins).

Improve this answer
$\endgroup$
16
  • 1
    $\begingroup$ Thank you for your answer. It makes sense. Another question: how can we consider electrons 'close to each other' (as in Lewis structures), in the Schrodinger model? $\endgroup$
    – Koen de Jong
    Commented Apr 19 at 16:52
  • 1
    $\begingroup$ It is based on the degree of spatial delocalization of the orbital itself. One potential way of calculating this would be the average $\langle r\rangle$ or standard deviation $\sigma_r$ in the radial direction to tell you roughly how “big” the orbital is. $\endgroup$
    – Matt Hanson
    Commented Apr 19 at 17:09
  • 2
    $\begingroup$ @hyportnex much appreciated. The simple answer that I would give is that it isn’t! It follows very neatly from the Boltzmann distribution of statistical thermodynamics when applied to the energies of a quantum system. But in the case of degeneracies, it is possible for a system to probabilistically favor states that aren’t the minimum of energy. But this typically applies to the rotational states where the energy gap is often small compared to $k_BT$, not to electronic states where the gap is far larger. $\endgroup$
    – Matt Hanson
    Commented Apr 19 at 23:57
  • 1
    $\begingroup$ To be honest, whenever I see the measurement problem come up I have to just throw up my hands and say “it’s magic.” I truly do not know in what sense quantum has any merit as an ontological theory of nature. But it sure does a nice job of predicting the outcomes of physical measurements. I even teach my students that the safest perspective on quantum theory is that it is a theory of measurements. I personally think there are missing pieces (like thermodynamics, though it could be argued that this isn’t missing so much as just a different emphasis), but I have no good argument for this! $\endgroup$
    – Matt Hanson
    Commented Apr 20 at 0:50
  • 1
    $\begingroup$ @MattHanson Great explanation! I now understand this far better than I ever did. Many thanks. $\endgroup$
    – Judith Jones
    Commented Apr 20 at 5:48
11
$\begingroup$

They do repel each other, but the nucleus of the atom attracts them so they keep hanging around. If you take helium, with 2 electrons, it is clear that the nucleus with charge +2 will dominate over the other electron with charge -1. But even in $H^-$, ionized hydrogen with a second electron present, the attraction of the +1 charge of the nucleus is apparently stronger than the -1 of the other electron (not unreasonable, the electron positions can have a distribution where they are on average on opposite sides).

The role of spin is only understandable if you first know about the exchange force, or "Pauli repulsion", which is not an electric force but an additional (effective) repulsion generated by QM for identical particles. This extra repulsion would of course not help (especially in cases like the $H^-$ mentioned above where there is not much margin!) But if the electrons have opposite spin they do not count as identical, so the exchange interaction will then not be present.

So in conclusion: electrically they definitely do push each other away. A positive nucleus is needed to keep them around. Exchange force adds even more repulsion. Spin does not counteract the repulsive electric force, but it does circumvent this exchange force.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ While the other answers are more technical, I do like this more layman's argument. It helped the more technical arguments click in my head. $\endgroup$
    – trlkly
    Commented Apr 21 at 0:17
5
$\begingroup$

Technically you would account for interactions between elections in multi-electron atoms, and this interaction would be repulsive, but there is no issue with there being a repulsion. This is where the shielding effect comes from essentially. Spin doesn't "cancel" the repulsion.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.