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In a multielectron atom

  1. May 6, 2009 #1
    In a multielectron atom, the lowest-l state for each n (2s, 3s, 4s, etc.) is significantly lower in energy than the hydrogen state having the same n. But the highest-l state for each n (2p, 3d, 4f, etc.) is very nearly equal in energy to the hydrogen state with the same n. Explain?
  2. jcsd
  3. May 7, 2009 #2


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    Explain why you think this is the case.
  4. May 7, 2009 #3
    Never heard about that. Can you specify a little bit more or give numbers?
  5. May 7, 2009 #4
    In multi-electron atoms, the inner electrons hide (shield, screen) the full effect of the nuclear charge from the outer electron(s). This is especially true in atoms that have filled shells +1 electron, like lithium (2 + 1 =3 electrons), sodium (10 + 1 =11 electrons), and potassium (18 + 1 =19 electrons). The outer electron in these atoms effectively "sees" only one nuclear charge.
    Last edited: May 7, 2009
  6. May 7, 2009 #5
    does the electron in a hydrogen atom ever go into the s subshell except for n=1. I thought (based on the zeeman effect) that the electron in a hydrogen atom only (or at least normally) transitioned between states where l=n-1.
    Last edited: May 7, 2009
  7. May 7, 2009 #6
    Electrons are usually captured into high n states, most probably with high l (probability proportional to 2l+1), and as they cascade down, most electrons eventually end up cascading through l=n-1 states. This is why 4f-3d, and 3d-2p transitions are so bright. An electron could find itself stuck in the 2s state, and cannot get to the 1s state by the standard delta-l = +/- 1 rule. It does eventually get there by higher order effects, which allows it to get to the 2p state.
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