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Energy levels hydrogenic atoms

  1. Feb 14, 2012 #1
    I don't get why for hydrogenic atoms the 2s and 2p orbitals have the same energy. i do get it mathematically, but im thinking that the fact that there are angular nodes in 2p and not in 2s MUST affect the energy!!!!
     
  2. jcsd
  3. Feb 14, 2012 #2

    Bill_K

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    It's an "accidental" degeneracy of course, but qualitatively here's why: the energy depends only on the principal quantum number n = ℓ + nr + 1 where nr is the radial quantum number, i.e. the number of radial nodes. And so more nodes in the angular direction tends to increase the energy, but it is accompanied by fewer nodes in the radial direction which tends to decrease it.
     
  4. Feb 14, 2012 #3

    Uhm okay, but I still don't get why for the 2s being more core-like than 2p, for the hydrogenic atom this isn't taken into account and hence both orbitals have the same energy.

    why does it only take it into account when we talk about multi electron atoms??
     
  5. Feb 14, 2012 #4

    jtbell

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    This holds only for the Coulomb potential, V ≈ -1/r, IIRC.

    In a multi-electron atom, an individual electron "feels" not only the attraction of the nucleus,but also the repulsion of the other electrons. The "effective" potential is not -1/r as with a one-electron atom.
     
  6. Feb 14, 2012 #5

    phyzguy

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    Your intuition is correct. In reality they don't have the same energy. The relativistic corrections and the spin-orbit coupling breaks the degeneracy.
     
  7. Feb 14, 2012 #6
    Right... so then why in the H atom the 2s has the same energy as the 2p ?

    also... how does spin coupling affect the energies. as far as I knew, spin coupling arises due to the interaction of the orbital angular momentum and spin angular momentum. So....?
     
  8. Feb 14, 2012 #7

    Vanadium 50

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    This question has several answers, depending on the exact details of the question asked, the mathematical sophistication involved, and to some extent the starting point.

    Potentials of the form rk have an extra symmetry for the cases k = 2 (harmonic oscillator) and k = -1 (inverse square). In the inverse square case, this extra symmetry appears in three places: the n-l degeneracy, the fact that a classical orbit does not precess, and the fact that quantum mechanically the variables separate in two coordinate systems: spherical and parabolic. (Indeed, in parabolic coordinates, the n-l degeneracy makes more sense. The price you pay is that it requires a little more mathematical expertise to do it this way)
     
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