We just finished studying Gauss’ law and were puzzled by this thought. If I look at a copper atom and focus on the 29th electron in the 4th shell, according to Gauss’ law, I can draw a Gaussian surface that completely encloses the first three shells plus the nucleus. It is my understanding that the 29th electron will not contribute to the E-field inside this Gaussian because this charge is outside the Gaussian. However, it does not mean that the 29th electron doesn’t apply an electric force on the inner electrons, especially the 3rd shell electrons. So here is my dilemma: wouldn’t this effect the E-field of the inner shell electrons?

Please, if possible, answer this question without quantum if possible.

  • $\begingroup$ Good question. Short answer, to zeroth order: no from filled shells; yes (but likely less that you expect) from un-filled shells. It is important to recall that electrons in an orbital don't have a position they have a position distributions. For filled shell the total distributions is spherical (to zeroth order). Lots of QM involved. $\endgroup$ – dmckee Sep 15 '13 at 0:30
  • $\begingroup$ @dmckee: I can imagine bringing an electron very close to filled-shell electrons in an atom. How can (to zeroth order) it be that these electrons will not feel an electric force from the outer electron? $\endgroup$ – Smith Sep 15 '13 at 1:25
  • $\begingroup$ @Smith Sorry. The inner electrons don't feel a net force from filled outer shells. You can, impose a external field and affect them, but the other bound electrons don't have positions: they only have orbitals and those tend to average out. They average out nearly completely for filled shells (exactly completely before perturbations). $\endgroup$ – dmckee Sep 15 '13 at 2:08

It is a misconception to think that just because the 29th electron is outside the gaussian surface, it will not have an effect on the electric field inside it. The total flux through the surface is indeed zero, but that doesn't mean there is no influence: imagine a point charge and draw up a sphere next to it. The electric field goes in on one face and out the other to make a zero flux, but there's definitely a field inside.

The situation for copper is slightly more complicated because the electron is in a spherically symmetric s orbital. This does reduce its influence on the inner electrons, not because of Gauss's law but because of a theorem of Newton:

the electric field due to a spherical shell of charge vanishes inside it.

However, the 4s orbital is not simply a spherical shell of charge around the [Ar] 3d10 core. Even in the most simplistic, hydrogen-based models, the higher s orbitals still have a sizable probability of being inside the core, and have multiple shells of positive probability:

Hydrogen 6s orbital

Source: Wikipedia. Note that this is a 6s hydrogen orbital with no actual relation to copper; a 4s orbital has four such spherical shells.

Because of this, the 4s electron does create a nonzero electric field inside the atomic core, and this does affect the inner electrons.

As you have guessed, of course, in real life this is much, much more complicated, and quantum chemists will send you packing if you tell them you need an accurate ab initio description of the whole copper atom all the way to the 1s2 core. The electrons are highly correlated, and all of them interact strongly with each other, including such lovely job-simplifying features like exchange interactions. Quantum mechanically, the bottom line to your question is that it's meaningless to talk about "the 29th electron," because all the electrons are indistinguishable and therefore you must consider them all as a whole. (Specifically, "the charge density due to the electron in the 4s orbital" is not a meaningful physical quantity.)

However, the quantumness actually helps. The reason for this is clear if you try to come up with a classical model of a copper atom. Even if you preserved the shell structure with the lone 4s electron orbiting well outside them, the inner electrons will be a hugely complicated jumble of electrons whizzing around, and the system will be nowhere near spherically symmetric. While you can still draw a gaussian sphere and get information about the total flux, this does not help at all in describing the electric field at any particular point.

In the quantum case, on the other hand, everything is much simpler. You have a bunch of closed shells and a single $s$ electron, and this means that the whole atom is spherically symmetric. Electric fields must be radial and uniform, and drawing gaussian spheres does help you calculate the electric field at a given radius.

  • $\begingroup$ great answer! I've never heard this in my undergraduate quantum course. Where (courses) does one get this exposure to this information? $\endgroup$ – Carlos Sep 15 '13 at 1:42
  • $\begingroup$ @Carlos It depends which bit of information you're talking about. The first bit is mostly electrostatics, but if you want the whole thing then you should be taking a (good) atomic physics course. $\endgroup$ – Emilio Pisanty Sep 15 '13 at 1:50
  • $\begingroup$ I am familiar with electrostatics but not with atomic physics. Would you recommended any good books along these lines? I want the whole picture as you described. $\endgroup$ – Carlos Sep 15 '13 at 2:21
  • $\begingroup$ To be honest my atomic physics course was not all that good and I've picked it up from here and there along the way. I understand Haken and Wolf is pretty good, though I don't have it with me and haven't seen it in a while. $\endgroup$ – Emilio Pisanty Sep 15 '13 at 2:28
  • $\begingroup$ @EmilioPisanty: after your answer, I can see how naive I was about asking it classically without quantum physics. So let me see if I understand you: The outer electrons do affect the inner electrons because they come inside the inner orbitals of the atom. However, if you have a lone 4s electron for a brief moment, it does "see" a spherical symmetric shell of charge. At this time, can I say that Gauss law is valid? $\endgroup$ – Lisa Lee Sep 17 '13 at 22:17

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