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Total Angular Momentum and Binding Energy

  1. Jul 22, 2013 #1
    Why, for states with angular momentum l >0, do states with smaller total angular momentum J have a higher binding energy? For example, why does the 2p1/2 state have a higher binding energy than the 2p3/2 state? If the 2p orbital is filled (2p6), wouldn't Hund's third rule indicate that the highest J value would have the lowest energy (in other words, the highest binding energy)?
     
  2. jcsd
  3. Jul 24, 2013 #2
    There is an energetic interaction (typically amounting to a few eV for core levels) between the spin magnetic moment and the orbital magnetic moment. If an orbital is half-filled, the two magnetic moments prefer to line up (have lowest energy). If the orbital is filled (as in the 2p6 configuration) there is only one possible state. Thus in a photoelectron transition:

    initial state: 2p6
    final states: either 2p5 with spin up, or 2p5 with spin down (non-denegerate).
     
  4. Jul 26, 2013 #3
    So why would the j=1/2 state (spin -1/2) have a higher binding energy than the j=3/2 (spin 1/2)?
     
  5. Jul 26, 2013 #4
    Sorry, my mistake - lowest energy [final] state is when the spin and orbital moments are in opposite directions (j = 1/2).
     
  6. Jul 26, 2013 #5
    But from Hund's 3rd rule (http://en.wikipedia.org/wiki/Hund's_rules), would the higher J (3/2) have lower energy (higher binding energy)?
     
  7. Jul 27, 2013 #6
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