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Finding total angular momentum qnumber for 3d4d

  1. Apr 13, 2012 #1
    Hey,

    I've been able to do most of these problems but at this point I stopped because it gives 5 values for L and I just wanted to double check it is correct.

    L = 2+2,....,2-2 = 4,3,2,1,0

    S = 1,0

    then you have to evaluate J for each L and S which will give 10 possibilities.

    It's a pretty small question but, is it ok to have 5 values for L?
     
  2. jcsd
  3. Apr 13, 2012 #2

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    Hi linda300! :smile:

    Can you provide some more information?

    What do you mean by a "qnumber"?
    What do you mean by "3d4d"?
    What do you mean by an L-number? Do you mean an angular momentum?
    What do you mean by an S-number?
    What do you mean by J?

    I guess I didn't learn from the same book you're doing now and I may be used to different symbols and definitions.
    I'll adjust to the symbols you are using, but I will need their definitions.
     
  4. Apr 13, 2012 #3
    oh sorry,

    i've read a few books which used wither L or l' notation for angular momentum,

    qnumber i meant quantum number but i thought it might not fit in the heading,

    3d4d means one electron at n=3, l=2 and one at n=4,l=2

    S total spin= s1+s2, s1+s2-1,...,|s1-s2|

    J total angular momentum= L+S, L+S-1,...,|L-S|

    I think it's fine to have 10 different combinations they all look good,

    However I do have another question related to this,

    Spectroscopic notation is 2S+1XJ
    Where X is the angular momentum according to below
    If X= S, P, D, F, G, H....
    Then L= 0 , 1, 2, 3, 4, 5....

    Take 5P0 for example. so S=2, L=1 and J = 0

    But according to S=2, L=1, J=3,2,1. so this state is impossible right?


    Also consider 3D1 S=1, L=2, J=1 which is possible,

    Since S=1 does this mean that there must be 2 optically active electrons?

    I think that's right but i just wanted to double check,

    Thanks for answering
     
    Last edited: Apr 13, 2012
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