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Allowed combinations of remaining quantum numbers when n = 3

  1. Dec 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Hey everyone

    I couldnt really fit a good description into the title. Basically here's what it is.

    Consider an atom in which all the states with principal quantum number n = 3 and angular momentum quantum number l = 2 are occupied by electrons (this is called a closed subshell). List the allowed combinations of the remaining quantum numbers [itex]m,m_{s}[/itex] which are related to the eigenvalues of [itex]\hat{L}_{z},S_{z}[/itex]. Hence, how many electrons are in the n = 3, l = 2 closed subshell?


    2. Relevant equations

    Dont know of any

    3. The attempt at a solution

    I dont have a clue...need some help T_T
     
  2. jcsd
  3. Dec 4, 2013 #2
    What's the spin of an electron? That determines the allowed values for ms. What's the thing that tells you what values m can take?
     
  4. Dec 4, 2013 #3
    Spin of an electron is either +1/2 or -1/2...that's [itex]m_{s}[/itex] right? as far as [itex]m[/itex] goes, is that the one that ranges from -l to +l...so it has 2l+1 values right? I just dont get what all this has to do with the eigenvalues of [itex]\hat{L}[/itex] and [itex]S_{z}[/itex]...
     
  5. Dec 4, 2013 #4
    So m is the eigenvalue corresponding to operator [itex] \hat{L}_z [/itex], and ms corresponds to [itex] \hat{S}_z [/itex].

    You basically got it, now you just need to figure out which values can m and ms take together, and how many combinations there are.
     
  6. Dec 4, 2013 #5
    for m = 1/2:
    m = -2, -1, 0, 1, 2

    for m = -1/2
    m= -2, -1, 0, 1, 2

    So total is 10 electrons...?
     
  7. Dec 4, 2013 #6
    Yap that looks correct
     
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