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N00b question: Why don't Atoms Collapse?

  1. Jul 5, 2004 #1
    This question is probably pretty silly looking at the complicated discussions going on here, but I'll ask none the less.

    Why don't Atoms spontaneously collapse? If there is a proton and there is an electron, shouldn't they attract each other and form a neutron?

    In trying to understand why this happens I was told that it's because of the quantization of energy. For the electron to move in towards the proton it would have to travel in such a way that it would violate the quantization, and have less than a quanta of energy.

    However, I later learned about Electron Capture, which clearly states that it is able to collapse.

    So my question is why doesn't it USUALLY collapse, and why does it collapse under certain circumstances? What keeps protons and electrons seperate 90% of the time, and allows them to unite at other points in their lives?

    Thanks, and try to use small words and talk slowly when answering.
    -- Physics n00b.
  2. jcsd
  3. Jul 6, 2004 #2
    Actually, this question appears again and again on this forum ! There are several possible answers. I will give my favorite (^_^)

    The electron is a particle, and also a wave. Therefore, there is a wave length associated to the electron. Now for a physical trajectory of the electron around the proton, the total length of the trajectory must be an integer of time the wavelength. If the total length equal 1.5 the wavelength, the electron would interfere destructively with itself.
    This is very old fashioned quantum mechanics. But it's cute.
    You can picture it this way : the wavelength is inversely proportional to the mass. The mass of the proton is big : it just sits in the center without moving. The electron is really light, therefore, its wavelength is large : in order to respect Heisenberg principle, it cannot be well localized around the nucleus. It is blurried somewhere around.

    There is another reason by the way : the free neutron actually decays into proton + electron + antineutrino. You see, it is not energetically favorable for the Hydrogen atom to collapse ! Now, of course, if you stack a lot of protons and electrons together without any neutron around, and you compress them (as in neutron stars !) then as the pressure grows, electrons and protons will indeed create neutrons, but they do so in order to reduce pressure energy.

    Simple questions are often the most difficult. Why is it dark at night ?
    Last edited: Jul 6, 2004
  4. Jul 6, 2004 #3
    wow, that's a nice explanation.
  5. Jul 7, 2004 #4
    Thanks Humanino, that makes sense. It makes sense to me that the wavelength has to be maintained, and that it can't travel in a path that isn't a multiple of the wavelength, but then I don't understand how it can occasionally collapse (during electron capture).

    If you could explain that a little bit more (specifically why the non-integer multiple of the wavelength ceases to matter) I would really appreciate it.
  6. Jul 7, 2004 #5
    You should not take what I wrote too seriously. It is inspired from the old-fashionned quantum mechanics. It is actually possible to explain nowadays why this kind of picture works. But it might become technical.

    First, let me try to elaborate on
    >>I don't understand how it can occasionally collapse (during electron capture).
    The electron is blurried around the nucleus : i.e. it is described by a wave function, and in position variables, the modulus squared of the wavefunction is the density of probability for the electron to be at the position x. Likewise, in momentum variable, the wavefunction depends on the variable p, and the modulus squared of the wavefunction is the density of probability for the electron to have momentum p. I am sorry if I am going to simple here.
    So the wavefunction in position space might or might not vanish very close to the nucleus. For electrons in outer shells, the wavefunction actually vanishes around the nuclei. But for electrons in the inner shell, the least energetic ones, the wavefunction is certainly non-zero around the nucleus.

    Consider for example a gaussian distribution function. Imagine that
    |psi(2)|^2 = exp(-R^2/r0^2)
    psi is the wavefunction, R is the distance of the electron to the center of the nuclei (since we consider electrons in the innerest shell, the wavefunction is actually spherically symmetric : it really depends only on R, not the angles). OK, now the size of the nucleus is very small compared to the caracteristic r0 of the electron. Even though the gaussian is peaked at zero where the nucleus is, the nucleus is so small that it will take a real while before the electron kicks the nucleus. But in the end, it might happen. This can lead to electron capture.

    Now I am trying to argue that paths taken by the electron have a length which is a multiple of the wavelength of the electron. This is not gonna be very convincing. Sorry.

    First, you can take a glance at :
    http://www.colorado.edu/physics/2000/quantumzone/debroglie.html [Broken]
    this is very "naive"
    To motivate the assumption, and get a feeling of what happens, you can think of a particle in a potential well : the wavefunction must vanish on the edges, so you must have a integer number of wavelength in the well.
    Now a stronger argument : maybe you are already familiar with Feynman's formulation of the amplitude : the amplitude for a particle to go from A to B is :
    A(A->B) = Sum[ exp(- i * S(path)/hbar) ]
    The sum is performed over all paths from A to B. The phase in the exponential is the action S. The classical action for this path ! note that h =2pi hbar.
    So consider a double cycle of the electron. The amplitude for the electron to come back twice at the point O can be devided as the sum of the two amplitudes : come back once, and then come back once again. The two phases of the amplitude must be close to each other (say equal for definiteness, or differ by a multiple of 2pi) in order to have constructive interferences. If the phases differ by an angle pi, the interferences are destructive, and the electron cannot even exist !
    So consider a closed path of length D, and a second one slightly modified, whose length is D+d. We assume d<<D. S = E*t = 1/2 m v^2 t and v = D/ t, so S = 1/2 m D^2 / t. Now we use d<<D to calculate the phase difference : it is approximately
    (S1 - S2)/hbar = m v d /hbar = 2pi Pd/h = 2pi * d / wavelength.
    We get that two paths interfering constructively have length which differ by an integer of times the wavelength. So i failed. I did my best.

    You can find many good informations on the web. Check also
    Last edited by a moderator: May 1, 2017
  7. Jul 11, 2004 #6
    In a Hydrogen atom the electron has two kinds of energy, potential energy which depends on the relative position of the electron and proton, and kinetic energy, which depends on the momentum of the electron. If the electron (wavefunction) is closer to the proton the potential energy is less, but the kinetic energy increases. The ground state of the atom is the one for which the total energy is minimum.
  8. Aug 6, 2004 #7
    How come in positronium, the electron does collapse onto the positron ending in a pair annihilation?
  9. Aug 7, 2004 #8
    The electron and the positron have same mass, which is not the case for Hydrogen atom. In the case of Positronium, the situation is symmetric between the two "orbiting" particles, up the electric charge. So the wavefunction for the positron, like for the electron, tends to smear the particle around. The probability of presence of the two particles is blurried, and that implies better chance for them to meet.

    I imagine that the converse should be : since the nuclei have a large mass, their wavefunction is well-localized at the scale of their movement under Electromagnetic orbiting. Therefore, if one builds an anti-nucleus composed of anti-proton and anti-neutron, and puts it into orbit around a regular nucleus, one sould obtain some kind of stable "Nucleitronium" :confused:
    Last edited: Aug 7, 2004
  10. Aug 20, 2004 #9
    I'm new at this, but doesn't this got to do with the uncertainty relation? When electrons are attracted to the nucleus, their position becomes highly localized, which means that it has to have a high momentum, but since, an electron has very small mass, it is impossible for it to have high momentum, enough to make it that localized. Also, that's why electrons don't fall to the bottom of a potential well.

    Am I right to say that?
  11. Aug 20, 2004 #10
    Yes, I would agree with you.
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