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Finding components of 2s2p configuration using LS coupling

  1. Mar 5, 2012 #1
    1. The problem statement, all variables and given/known data
    The lowest excited states of the Be atom correspond to a 2s2p configuration of optically active electrons. What 2s+1Lj components originate in this configuration? Assume that Hund's rules hold for the multiplet and deduce the ordering of the energy levels. Refer to a table of beryllium levels to see if the ordering occurs as predicted.


    2. Relevant equations
    l=|l1-l2|,|l1-l2|+1,...,l1+l2-1,l1+l2


    j=|l-s|,|l-s|+1,...,l+s-1,l+s

    3. The attempt at a solution
    spin-orbital assignments of 2s2p configuration:
    (l1=0,s1=1/2) and (l2=1,s2=1/2)

    So that,
    total spin states: s=0 and 1
    and
    l=0,1, and 2
    ...............................???
    I don't know what to do next.

    Answer at the back of the book:
    3P0,1,2,1P1
     
  2. jcsd
  3. Mar 6, 2012 #2
    You pretty much have it. Except, I don't believe you calculated the possible l values properly.

    |1-0|=1, so only l=1 works.

    Now, all you have to do is use Hund's rules to find the order.

    This will lead to the answer in the book.
     
    Last edited: Mar 6, 2012
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