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Homework Help: Lande g-factor values

  1. Nov 13, 2006 #1

    neu

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    Im getting very confused about how to calculate the lande g-factor for the 3S1, 3P0, 3P1, and 3P2 states

    I know its equal to

    http://www.pha.jhu.edu/~rt19/hydro/img208.gif [Broken]

    but if i have state 3P0 where S=1 as 2S+1 = 3 and L=P=1 and J=0, but J=L+S which isn't =1?

    i've read myself into a hole can someone help us out?


    I should say the g-value is used in the zeeman effect. Gives the energy shift as ratio of bohr magneton

    http://www.pha.jhu.edu/~rt19/hydro/img207.gif [Broken]
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Nov 13, 2006 #2

    OlderDan

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    J = L±S yes?
     
    Last edited by a moderator: May 2, 2017
  4. Nov 13, 2006 #3
    Why would you need to calculate it if there is no electron at that energy level? Guess I'm missing something.
     
  5. Nov 13, 2006 #4

    nrqed

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    Just plug in the values of S,L and J.

    Your problem does not seem to be in finding g but in vector addition in QM. Recall that in QM, writing [itex] {\vec J } = {\vec L } + {\vec S} [/itex] means that J will range from |L-S| to |L+S| in steps of 1. So, if S=1 and L=1, J could take any value between |1-1| and |1+1| so J may be equal to 0, 1 or 2. Your 3P0, 3P1 and 3P2 states correspond to those three possible values of J.

    Hope this makes helps.

    Patrick
     
    Last edited by a moderator: May 2, 2017
  6. Nov 14, 2006 #5

    neu

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    Thats exactly the clarity i needed thankyou
     
  7. Mar 10, 2011 #6
    I'm stuck with calculating g and p for Eu(3+).

    The outtermost orbitals in Eu is 4f7 5s2 5p6 6s2. Eu(3+) has 4f6 as the last orbital.

    Thus, S = 3, L = 3 and J = 0 since J = L - S here.

    How do I calculate g (using the formula given above) and then p. (p = g[S(S+1)]).

    The experimental value for p = 3.4 and I read that g must be 2 in this case.

    I am at a loss how to arrive at this result.

    Can anyone help?
     
  8. Mar 12, 2011 #7
    :cry::confused::uhh:

    Someone please help . . .
     
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