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I Ionization and Nodes in the Hydrogen Wave Function

  1. Nov 6, 2016 #1
    14GAePg.png
    As you can see from figure 4.4 from Griffiths book on QM, the radial wave function of the hydrogen atom has clear points where ## |R_{nl} (r)|^2 = 0 ##. My question is three fold:
    First, how is the electron able to traverse this region? My intuition is that with the uncertainty principle, the electron will only ever be observed at ##r \pm \Delta r## for which r satisfies ## |R_{nl} (r)|^2 = 0 ## though this is not as satisfying as I would like. Surely it would be possible (even if extraordinarily unlikely) to observe an electron at this specific point.
    Second, how fast does the electron move during ionization? My initial guess was c since it is emitting/absorbing a photon, however ## v = \sqrt{\frac{2 E_k}{m_e}}=\sqrt{\frac{2 \times 13.6eV}{510eV/c^2}} \cong .008c ## which seems reasonable.
    Third, is there a wave function for a transitioning electron? Perhaps ## P = \left< \psi' | Q | \psi \right>## where ##\psi'## and ##\psi## are the overall initial and final wave functions?
    Thanks
     
  2. jcsd
  3. Nov 6, 2016 #2

    mfb

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    Staff: Mentor

    In those eigenstates, the electron does not move - it does not traverse anything.

    Those eigenstates are only exact with a single point-like charge, an electron, and nothing else in the universe. Every external influence will disturb them and break the symmetry and the general zero-crossing.

    The speed of an electron cannot be expressed as number. Changes in the wave function in the process of ionizing hydrogen might have something like 0.01 c as typical propagation speed, but I'm not sure how meaningful that is either.

    You can calculate the wave function of an electron in the combination of the field from the nucleus and an external electromagnetic field. It will get a component that is still at the nucleus and a component that leaves the nucleus.
     
  4. Nov 6, 2016 #3
    Could you expand a little more on this? Is this like the electron is a "cloud" and that the electron is somewhere there but not known until observed and the act of observing it alters where the node is and then the electron goes back to being a "cloud"?
    The rest is clear. Thank you.
     
  5. Nov 6, 2016 #4

    mfb

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    No. Without a position measurement, "the position of the electron" is a meaningless concept (apart from "it is in this atom").

    If you observe the position, you get a completely new wavefunction, localized at some random point. If you then stop interacting with it, the wavefunction will spread out again, but this time in a superposition of many energy eigenstates.
     
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