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HUP inside the atom

  1. Jul 25, 2008 #1
    I am kinda confused on the whole electrons falling into the nucleus thing (I know you guys have probably seen this question a million times but when I searched the forum, I could hardly find a satisfying answer )

    So I have heard that the reason electrons don't fall into the nuclues is because of the HUP - if they did fell into the nucleus then their uncertainty in the position would very small which would imply that their momentum would be very large and hence making them fly off elsewhere.

    But then what about energy levels? Can't you say that an electron can't fall into the nucleus because it can't go below the lowest energy level? (Bohr's explanation I believe).

    Also, what do we do about the Coulomb force of attraction? It should technically be cancelled out by some other force right?


    Also, what keeps the protons inside the nuclueus? Is it also because of HUP? What about the strong nuclear force? Isn't it responsible for the stability of the nucleus? Are these two explanations self-consistent?
     
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  3. Jul 25, 2008 #2

    mathman

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    The strong nuclear force keeps the nucleus together. Quantum mechanics determines the electron energy levels. Essentially electrons won't be captured by nuclei unless the result is at a lower energy level after the capture. HUP doesn't have any direct bearing on these things.
     
  4. Jul 25, 2008 #3
    Calling for Heisenber's principle about the stability of atoms is not irrelevant, it's just utterly confusing. I've always considered that innapropriate, but maybe I'm missing the physics behind.

    The way I see it, it can be justified in the following manner. Radiation reaction will cause canonical commutators for a harmonic oscillator to decay to zero unless coupling to the fluctuating vacuum is included. Commutator in that case are consistently preserved indeed. In a sort of fluctuation-dissipation relation, one can impose that "spontaneous emission" is exactly cancelled by "spontaneous absorption" in the ground state, thus cancelling out radiation reaction. If you do that, you find that the cancellation gives you Bohr's quantization condition for the ground state (for excited states, you have different conditions on the "spontanenous emission vs absorption", corresponding to the necessity of vacuum fluctuations for transitions to occur).

    Anyway, I don't think anybody takes those very seriously. This is how some physicists have interpreted Bohr's condition back in the early XXth century. But it's out of proportion to introduce that when calculating say Hydrogen's level, at the very beginning of QM.
     
  5. Jul 25, 2008 #4

    jtbell

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    The HUP is part of the reason we know that there can't be any electrons confined inside the nucleus. They'd have to have very large range of possible momenta, and occasionally we'd see one tunnel out with a high momentum and energy... but we don't. The electrons produced in beta decay have much lower energies. That's how we figure those electrons are created in the beta-decay process, and weren't in the nucleus beforehand.

    As for atomic electrons, they do sometimes "fall into" the nucleus, because the probability distribution is not zero at the origin (the nucleus), in general. The nucleus can capture the electron in a process that is related to beta decay, if the total mass comes out less afterwards than before. This is called (surprise!) electron capture.
     
  6. Jul 25, 2008 #5
    To be slightly more specific ("Quantum mechanics" is a big topic :biggrin:):
    You might be familiar with the problem of a "particle in a box", or the simple introductory QM problem in which you work out the allowed energies of a paricle in some potential well. The physics of electron energy levels is essentially the same, but with harder maths- your 1D square well potential is replaced by a spherically symmetric coulomb potential, and the energy levels are determined by the allowed solutions to the Schroedinger eqn.

    Also, the HUP has a discernible effect in the case of the square well- as a particle in a well defined position with precisely zero energy would have a precisely defined momentum, your 'zero-point energy' is shifted up to the lowest allowed non-zero energy state. This is why people say the HUP is responsible for the stability of atoms- electrons don't collapse into the centre of the atom because they would be localised and have precisely zero momenta. The analogy isn't perfect, however; an electron at the very centre of an atom would theoretically have infinitely negative energy, rather than zero. Really, of course, the HUP is an emergent property of quantum mechanics rather than an additional rule to which particles must conform. In the case of the electron in the atom, the reason for the fact that the "zeroth energy level" doesn't exist is complicated, and relates to the fact that the radial dependence of the energy eigenfunctions is constrained by an integer (the principal quantum number!) which must be greater than the quantum number for orbital angular momentum (which is greater than or equal to zero).
     
  7. Jul 25, 2008 #6
    What about Heisenberg's principle for light quarks inside the proton ? They don't have such a larger mass than the electron !
     
  8. Jul 26, 2008 #7

    malawi_glenn

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    Electron capture occurs in certain nuclei, so in one sense, they have fallen into the nucleus.
     
  9. Jul 26, 2008 #8
    Do you mean that HUP has been violated?
     
  10. Jul 26, 2008 #9
    No. In electron capture, it ceases to be an electron! What happens is the process
    p + e ---> n + v_e
    i.e. a proton and an electron combine to form a neutron, with the emission of an electron neutrino. The end result has neither zero size nor zero momentum, so there's plenty of room for the HUP to manouvre.
     
  11. Jul 27, 2008 #10

    malawi_glenn

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    The nucleus also have a certain spread in space and the interaction is not point-like.
     
  12. Jul 27, 2008 #11
    I thought I provided some material for discussion, after all the HUP has the commutator in the r.h.s. and I thought it was relevant that, once the vacuum was included, the condition for the commutator not to decay to zero was a detailed balance fluctuation-dissipation relation between spontaneous absorption and emission, giving exactly the Bohr levels. The details can be found in any textbook on the QED vacuum.

    But let's get to more elementary considerations. Say we have a spherically symmetric electron distribution, with potential energy like [tex]<V>=-\frac{e^2}{r_0}[/tex]. The kinetic energy will look something like [tex]T\approx \frac{1}{2m}(\Delta p)^2=\frac{\hbar^2}{2mr_0^{2}}[/tex]. Now minimize the sum :
    [tex]E_{min}=\frac{\hbar^2}{2mr_0^{2}}-\frac{e^2}{r_0}[/tex]
    achieved for
    [tex]r_0=\frac{\hbar^2}{me^2}[/tex]
    [tex]E_0=-\frac{me^4}{2\hbar^2}[/tex]
    which is the same as Bohr's model.

    See e.g. Cohen-Tannoudji, Diu, Laloe CI.

    Personally, I think this is not very profound.
     
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