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Electron Proton Repulsion

  1. Sep 7, 2011 #1
    What accounts for the repulsion of the Electron & Proton say for instance in the hydrogen atom.

    The coulomb force "classically" is counteracted by the centrifugal force allowing for a stable orbit but this picture isn't correct in quantum mechanics, for a charge undergoing acceleration should emit radiation and decay towards into the nucleus.

    I've never seen in any text book what counteracts the coulombic atraction. Its as if the coulomb force only acts in certain regimes.

    Thank you
  2. jcsd
  3. Sep 7, 2011 #2


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    It's not so clear to me what you are after.

    It starts with "the Coulomb force is counteracted by the centrifugal force". First of all this is a very classical picture, and I don't know, how to transfer it to quantum mechanics at all. In classical mechanics it's a very complicated idea, because you would have to sit on the electron, which is moving around the proton, i.e., in a non-inertial frame to make sense of your idea. It's much easier (and, btw, that's what makes Kepler's idea of the sun being the center of planetary motion so much easier than the older earth-centered point of view) to work in the inertial reference frame, where the center of mass of the proton and Neutron is at rest. In classical mechanics the mutual attraction together with a non-zero angular momentum (i.e., inial conditions, where the two particles are not in collinear head-on collision) leads to the Keplerian motion of both bodies, namely in conic sections (i.e., ellipses, parabolae, or hyperbolae) with the center of mass as a focus. In terms of forces, the Coulomb force is the centripetal force which makes the particles go around the center of mass in curved trajektories and not straight lines.

    As is well known, this classical picture of an atom is a complete failure to begin with. Such a system would very quickly lose energy through the radiation of electromagnetic waves since accelerated charges radiate off em. waves due to Maxwell's equations.

    Leaving out the also not satisfactory Bohr-Sommerfeld model, this problem has been cured properly only with modern quantum mechanics (invented by Heisenberg, Born, and Jordan; the hydrogen problem within modern qm solved first by Pauli). The solution is simply that quantum mechanics admits stationary solutions, given by the energy-eigen states of the Hamiltonian.

    Of course, looking a bit more precisely only the ground state is really stable since any excited state will also radiate off photons due to the coupling to the em. quantum field.
  4. Sep 7, 2011 #3
    Electrons and protons attract each other, not repel. Nothing accounts for them repelling because they do not repel. The attraction is caused by the electromagnetic force (an exchange of virtual photons).

    So why don't electrons spiral into the atomic nucleus? Because they don't fit. Electrons are not little balls, they are particle/waves. The more you try to squeeze an electron into a tight spot, the more it spreads out like a wave. So an electron in an atom is confined tightly enough that it acts like a wave and the wave takes up a certain space that only fits a certain distances from the nucleus. If an electron were to try to spiral into the nucleus, the closer it got, the more confined it would be, the more like a wave it would act, and the less this wave would fit.
  5. Sep 7, 2011 #4
    I know that they attract.... however if they stop approaching each other at some point there must be a repulsion to the coulomb force.

    Maybe I can redirect you to where I recently encountered this question/problem

    [crackpot link deleted]

    "That’s a really terrible answer, even to give to a 16 year old! In fact, it is just gibberish. The problem is, it really does represent the full answer, the one given to graduate students when they insist upon one (which they rarely do). The full answer has more gibberish, but not more content or logic. QM tries to answer the question by making the electron a cloud or probability, but we must imagine that no matter how probabilistic the electron is, it still must have a negative charge. It cannot have a negative charge far away from the nucleus, acting like a particle, then approach the nucleus and begin acting like a cloud with a positive charge. All this talk of momentum and kinetic energy and HUP is just misdirection. No matter how you represent the kinetic energy or momentum of the electron, you cannot create a repulsion. The dispersion of momentum or kinetic energy into a cloud or probability cannot switch the charge or create a repulsion. And the HUP simply has nothing to say about switching charges or creating repulsions from attractions. This “physicist” should be ashamed to be saying such things, especially to the young. "

    I'm not saying hes correct...
    Last edited by a moderator: Sep 8, 2011
  6. Sep 7, 2011 #5


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    There is no such point. The probability distribution for the electron in the ground state of a hydrogen atom is in fact maximum at the nucleus! See Fig. 3-4 on the following page:


    So there is a definite probability for the electron to "materialize" inside the nucleus. With hydrogen, nothing happens as a result, because there's not enough energy to make anything happen. A proton and electron can in principle combine to produce a neutron and a neutrino, but the mass of a neutron is bigger than the sum of the masses of the proton and electron, so this reaction can't "go" unless enough energy is added from an outside source to produce the extra mass.

    With certain kinds of heavier nuclei, on the other hand, it does happen, because the resulting nucleus ends up with less mass than the original nucleus. This is called electron capture and is a well-known method of nuclear decay.
  7. Sep 7, 2011 #6

    I'm not sure what you mean by "at the nucleus". Who says that the Bohr radius is where the nucleus starts.

    I know that the most probable location of the electron is the Bohr radius, and the average value is 1.5 Bohr radii. But why, yes I can do the quantum mechanics that tells me so.

    I want to say that the electron is orbiting the proton but that is a classical argument, so there is not centripetal motion, or centrifugal force in the non inertial frame. Also there would be energy loss due to radiation from acceleration.

    I mean the coulomb force is still there, The negative gradient of the potential energy yields a force.

    All that i can think of to rationalize it, is the allowed energy levels are flat areas of the potential energy, and the gradient of a scalar is zero, thus there is not force acting on the electron in the allowed energy levels?

    Thank you all for comments
  8. Sep 7, 2011 #7
    Crackpot blogs are probably not the best place to learn anything about physics.
  9. Sep 8, 2011 #8


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    It's not at a maximum at the Bohr radius, it's at a maximum at r=0, which is truly "at the nucleus".

    No, that's the most probable radius. The probability of being at a radius r is the sum over the probabilities of every point at that radius; so it's 4*π*r2|Ψ|2 which is not the same thing as the probability for an single location in space, which is |Ψ|2.

    The reason an atom has a finite extent is due to basic quantum mechanics. The momentum of a particle is related to the derivative of its (spatial) wave function and so, location-probabilities. The more sharply located the electron (or any particle) is located in space, the higher its momentum and kinetic energy. Alternately, you can use the uncertainty principle to make a heuristic argument along the same lines.
    Or you can make a classical analogy to a standing wave, and consider how its fundamental frequency (and so energies) will increase as the wave-length decreases, and chalk it down to 'wave-like' behavior on the part of the electron. Or you can take the simple particle-in-the-box model and trivially see how its energy changes with box size. The coulomb potential energy, which 'wants' the electron near the nucleus is counteracted by the kinetic energy, which would be minimized by dispersing the electron as much as possible. The points where those 'balance' are the stationary states of the system. (This is essentially the time-independent Schrödinger equation expressed in words)

    And as JeffKoch says, you're not going to learn anything from reading crackpot blogs like that one. There's nothing wrong with our theoretical understanding of electrons in atoms. It's one of the most accurately-measured and well-verified things around. (specifically, the Rydberg constant is the most accurately-measured fundamental constant) All that quote seems to say is "this doesn't make sense to me, and any explanation is just nonsense". The fact that things work this way is fundamental to quantum mechanics, and QM is a spectacularly successful theory at explaining how things work, especially electrons in atoms. If someone thinks it's wrong, they'll need to point to some objective experimental fact that the theory doesn't predict. It's not an argument that it doesn't fit your preconceptions of how reality "should" work.

    Nobody said quantum physics was intuitive or "made sense" from a classical POV. But nature isn't under any obligation to "make sense" to us, either.
  10. Sep 9, 2011 #9
    First thank you.

    secondly, your arguments make sense but I still haven't been able to rationalize the complete picture (I think its because I'm trying to combine classical and quantum ideas into the same picture).

    The electron's kinetic energy would be maximal near the center of the nucleus, and the "probability" is highest there?

    One of the issues i have trouble with is the coulomb force creating an acceleration of the electron and releasing energy (through radiation), obviously this doesn't happen for an isolated hydrogen atom. The electron will stay in its groundstate.

    So its groundstate must correspond to some condition where it does not undergoe acceleration (as it exists & moves in its "orbit" about the nucleus), thus it must not feel the coulomb force?

    When we obtain that the energy levels are quantized, the energy should be the sum of kinetic plus potential, the potential energy is due to the coulomb potential. Does this mean that the coulomb potential is quantized?

    And if the coulomb potential is quantized, then it will have locations where the force will be zero.

    OR is the problem that I'm trying to mix forces and accelerations from the classical picture with potentials, energy, and probability. When an electron moves from spot to spot (alternate probable locations) does it not accelerate. Maybe the electron just appears there when we measure it so it never accelerated?

    Is the electron under a force?, if so isn't it accelerating, if so, why doesn't it radiate

    Thank you
    Last edited: Sep 9, 2011
  11. Sep 9, 2011 #10
    If you are so determined to combine classical and quantum pictures, you could think of it this way: The electron is orbiting because the Coulomb force is causing it to accelerate into an orbit. In general, accelerating charges should emit electromagnetic radiation. But in this case, if an an electron in the ground state were to emit em radiation, it would loose kinetic energy in the process and thus spiral into the nucleus. The electron's state cannot "fit" more compactly than the ground state, so the kinetic energy cannot change, so the accelerating orbiting electron is forbidden from emitting em radiation. The electron cannot find a stable orbit lower in energy than its ground state. It's as if the electron tries to emit em radiation because it is accelerating, but it cannot, because there is no where for it to step down to when it looses energy. This picture may be clunky, but that may be the best you can get describing quantum effects with classical pictures.
  12. Sep 9, 2011 #11


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    You are trying to mix classical mechanics and quantum mechanics. It will not work. Classical mechanics has been shown to be inaccurate at the atomic level for almost 100 years.
  13. Sep 10, 2011 #12


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    The maximum energy needed to seperate an electron from a proton in hydrogen is 13.6 electron volts. This just happens to be the same as the coulomb force between the electron and proton at a distance of 53 picometers (the bohr radius). If you imagine a hydrogen proton as this size (as is often shown in chemistry text books for the purposes of illustrating bonding) and assume that that electron feels no force from the proton once inside this shell (ie. the coulomb attraction force does not grow past 13.6ev), you wont need the centrifugal force to counteract the coulomb force and the electron no longer needs to be accelerating or emitting radiation. In fact, there is no need for the electron to be moving around at all once it is inside the proton shell - its just there.. somewhere... but where...
  14. Sep 10, 2011 #13
    Yes.. doesn't that agree with my explanation for a quantized coulomb force or rather a quantized coulomb potential.

    Since the energy is quantized and the force is related to the potential energy by negative gradient of the potential energy, the gradient of constant yields a zero force in a certain level.

    Obviously its unnecessary to even talk about the force at all, but if someone where to ask where is the coulomb force, your answer seems to justify whats going on. The electron "floats" around, no centrifugal or coulomb force, no acceleration, no radiation.

    When the electron changes energy levels we can get a brief glimmer of the force / radiation released or absorbed.

    Thank you.
  15. Sep 10, 2011 #14


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    There is no reason to believe the force itself is quantized. Not to my knowledge at least.
  16. Nov 10, 2012 #15
    RE: Electron Proton Repulsion
    I wonder if, by changing assumptions of exclusion, that the free neutron state might be considered to be equivalent to an unstable entrapment of a dense electron within a diffuse proton. If that state required that the electron and proton to be counter-rotating (perhaps to prevent radiation) in the classical sense, then their classical magnetic dipoles would be of the same sense and repulsive.

    Also, and additional 'empty' proton nearby might supply the additional retentivity to make the state stable.

    Perhaps in such a construction the 'Strong Force' might be equivalent to the unbalanced classical electrical field of the captured electron and perhaps the 'Weak Force' would be equivalent classical magnetic dipole forces.

    This is just a question about whether such a construction is precluded by established data or if a change in basic assumptions, such that of the electron being unable exist as an independent entity inside a proton, might allow an equivalent alternative theory.
  17. Nov 10, 2012 #16


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    Current scientific observations have confirmed that 3 particles make up protons and neutrons. They are called Quarks and the Standard Model of Particle Physics is heavily dependent on them. So no, current evidence does not allow an alternative theory unless it makes the same predictions and explains the same observations. I advise you to stop speculating before this thread gets locked, since personal theories and speculation are not allowed as per PF rules.
  18. Nov 10, 2012 #17


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    MLBar, I suggest you click the Rules link at the top of the page and review our Discussion Guidelines.
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