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Energy levels of helium/equation

  1. Jan 12, 2009 #1
    the equation for the energy levels of a hydrogen-like atom is:

    [​IMG]

    Note that aμ, is approximately equal to a0, (the Bohr radius). If the mass of the nucleus is infinite then μ = me, and aμ = a0

    [​IMG]

    but what is the equation for the energy levels of a helium or helium-like atom? I've heard that heliums spectrum is simply 2 hydrogen spectrums superimposed so it should be quite simple.
     
    Last edited: Jan 12, 2009
  2. jcsd
  3. Jan 12, 2009 #2

    malawi_glenn

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    as far as I know, you can't solve those system exactly. Only two-body systems can be
     
  4. Jan 12, 2009 #3
    I'm not asking for a 'solution'. I'm asking what equation fits the empirically observed spectrum.
     
  5. Jan 12, 2009 #4

    Vanadium 50

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    But that's what a solution is.
     
  6. Jan 12, 2009 #5
    I'm not even going to touch that.
     
  7. Jan 12, 2009 #6

    f95toli

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    So what ARE you asking for then?
    Are you asking if there is e.g. an interpolating polynomial (or more realistically; an expansion using some other bases; e.g. Lorentzians) that fits the shape of an experimental spectrum?

    I doubt such a thing exist; it is of course possible to create but it would need to contain so many terms that it would be useless; it is much easier to look up the data in a table or just run a computer simulation.
     
  8. Jan 16, 2009 #7

    Redbelly98

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    In practice, as far as I know, people either look up the energy levels from a table or chart, OR they run computer simulations to calculate them.

    For the simulations, you might do a search on Gordon W. F. Drake. He practically made a living from accurate calculations of helium, at least in the 1990's. Three references to his work are given here:

    http://physics.nist.gov/PhysRefData/Handbook/Tables/heliumtable7.htm

    EDIT: Understanding Drake's calculations in any detail pretty much requires grad-school level quantum mechanics.
     
    Last edited: Jan 16, 2009
  9. Jan 16, 2009 #8
    thank you. that was extremely helpful.
     
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