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Strange behavior of 3d level of Helium

  1. Apr 11, 2009 #1
    Could someone explain to me the reason why the 3d level of Singlet Helium is lower in energy than the 3p, since it is supposed to be higher because of less penetration...? The energy diagram is here:
    Thank you!
  2. jcsd
  3. Apr 11, 2009 #2
    Could someone please help me...I need the answer desperately...please...
    Thank you very much, and I'm very sorry if I seems too demanding...but I have only 1 day left before the Science Fair...
    Thank you again.
  4. Apr 12, 2009 #3


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    Spin-orbit coupling probably.
  5. Apr 14, 2009 #4
    The experimental electron energy levels of ortho (triplet) and para (singlet) helium for all levels through n=4 are given in Bethe and Salpeter "Quantum mechanics of One and two Electron Atoms" (Springer-1957) on page 127. (The energy levels are all measured in wave numbers below the continuum). The table shows the 3D state as very very slightly more tightly bound than the 3P state for the para (singlet) helium, while the 3D is higher (less tightly bound) than the 3P in the triplet (ortho). The 4D and 4P singlet states are nearly identical.
  6. Apr 14, 2009 #5
    Oh! Thanks!
    Is there a good explanation for this that i can find on the internet?
  7. Apr 15, 2009 #6
    I read that the following book covers the 3-body problem of the un-ionized helium atom. It is a very difficult nucleus plus 2 electron system and can be calculated only approximately..

    "Intermediate Quantum Mechanics" by Hans Bethe and Roman Jackiw. Most
    books that discuss the Hartree-Fock method will have discussions on
    two-electron atoms.
  8. Apr 16, 2009 #7


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    Wheeler (1986):
    Bob S, I'd have to disagree with the statement it can "only be calculated approximately". It cannot be calculated analytically by any known method. But it can be calculated to arbitrary precision.

    There probably does exist an exact mathematical solution, at least for the nonrelativistic equation. They did find one for the classical three-body problem. It's unlikely though, that the mathematically exact solution is something which is itself easier to calculate than a numerical method.

    The nonrelativistic ground state energy is -2.903724 a.u. Relativistic: -2.903855 (a difference by ~0.3 J/mol.. not much by any standard)
    Just to compare some methods, since I have the numbers handy:
    Hartree-Fock: -2.86
    Thomas-Fermi model: -2.19
    1st order perturbation theory: -2.75

    The simplest, most exact method was the one Hylleraas used as early as 1929; a direct variational-principle approach, which is completely exact (nonrelativistic) for a given basis.
    A single parameter (analytically solvable): -2.85
    3 parameters: -2.847 (Hylleraas 1929)
    6 parameters: -2.902 (Hylleraas 1929)
    14 parameters: -2.90370 (Chandrasekhar 1955)
    10257 parameters: -2.90372437703411959831115924519440444 (Schwartz, 2002)

    Which is absurd, really, since relativistic effects come into play. Also, Helium is essentially a bit of a special case; Hylleraas method doesn't scale well to systems of more than two electrons. In other cases you'd probably use full-CI.
    Last edited: Apr 16, 2009
  9. Apr 16, 2009 #8
    If you are interested in studying 2-electron atoms, look at H-minus, a single proton with two attached electrons. The last (whichever one it is) has about 0.75 eV of binding energy, and can be stripped with visible light.
  10. Apr 17, 2009 #9


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    If anyone is interested, and has access to Physical Review journals, search for articles by G. W. F. Drake. He did a lot of work (mostly in the 1970's and 1980's ?) doing precision calculations of helium energy levels. For example:

    http://prola.aps.org/abstract/PRL/v59/i14/p1549_1" [Broken]
    "New variational techniques for the 1snd states of helium"

    Last edited by a moderator: May 4, 2017
  11. May 15, 2009 #10
    Thanks you guys a lot ^^
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