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Electron Configuration of Excited State

  1. Feb 24, 2016 #1
    1. The problem statement, all variables and given/known data

    An exercise examining the tin atom (Sn). Tin has a ground state electron configuration of ##[Kr]4d^{10}5s^25p^2##.

    a) Write down the electronic configuration of the first excited state.

    b) Illustrate with a vector diagram the allowed total angular momentum ##J## values for this excited state.

    2. Relevant equations

    N/a

    3. The attempt at a solution

    Hi all,

    For a) As I understand it, excited states for atoms are achieved when an atom in the outermost shell makes a transition to an atomic orbital with greater energy, according to the selection rule of ##ΔL=±1##. However I'm unsure as to what would constitute the first excited state.

    I think it would be one of the two following:

    (1) ##[Kr]4d^{10}5s^25p^16s^1##

    (2) ##[Kr]4d^{10}5s^25p^15d^1##

    The ##5d## level has a higher energy than that of the ##6s##, so perhaps (1) is the first excited state?

    For b) I'm not entirely sure about it, but I presume that the correct answer for a) must be determined first.
     
    Last edited: Feb 24, 2016
  2. jcsd
  3. Feb 25, 2016 #2

    DrClaude

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    Staff: Mentor

    Careful here. When considering the excited states by themselves, do not refer to selection rules. The first excited state of an atom may not be reachable by a dipole transition from the ground state. The best example is hydrogen, where the first excited state is 2s.

    The rule that works most of the time is to take a valence electron and put it in the next highest energy orbital.

    Yes.

    Yes.
     
  4. Feb 28, 2016 #3
    I see. Then the electron configuration of the first excited state would be: ##[Kr]4d^{10}5s^25p^16s^1##.

    Then for part b), since the total angular momentum of an atom is given by that of its outer electron, the total ##J## will be defined by the single outer electron in the ##6s## orbital. Therefore, we have ##l=5## and ##s=\frac{1}{2}##. And therefore the total angular momentum values we have are ##j=s±l=\frac{11}{2}## and ##\frac{9}{2}##. Therefore ##J=\sqrt{j(j+1)}ħ=\frac{143}{4}ħ## and ##\frac{121}{4}ħ##. I have a feeling this isn't right..
     
  5. Feb 28, 2016 #4

    DrClaude

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    This is not correct. It is only filled (sub)shells that have an overall contribution of 0 to the total orbital angular momentum and spin. Here, there are two subshells that are not filled.

    ???
     
  6. Feb 28, 2016 #5
    Ah OK, so both the ##5p## and ##6s## shells have contribution to the total angular momentum. And I went completely off track with ##l##. For the ##6s## orbital ##l=0## and ##s=\frac{1}{2}##. And so would that mean that ##j=s+l=\frac{1}{2}##, and so ##J=\sqrt{j(j+1)}=\frac{3}{4}ħ##?

    And for ##5p##, ##l=1## therefore ##j=\frac{3}{2}## and ##\frac{1}{2}##. So ##J=\sqrt{j(j+1)}=\frac{15}{4}ħ## and ##\frac{3}{4}ħ##. Am I on the right track?
     
  7. Feb 28, 2016 #6

    DrClaude

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    Staff: Mentor

    For Sn, LS-coupling should apply, so this is not the correct way to go about it.
     
  8. Feb 28, 2016 #7
    Ok, then I'll implement LS coupling as found on this page. And so let ##L_1## be the orbital angular momentum of the ##5p## electron, and ##L_2## that of the ##6s## electron. Then ##L=L_1+L_2=1+0=1##. And similarly for the spins: ##S=\frac{1}{2}±\frac{1}{2}=1## or ##0##. Therefore, ##J=L+S=1## or ##2##.

    I wonder if the atomic nucleus might be perhaps too large for this type of coupling?
     
  9. Feb 28, 2016 #8

    DrClaude

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    Staff: Mentor

    That is correct, but
    is not complete. You have to find all possible J's for each combination of L and S.

    Tin is at the limit where LS coupling applies, and it could be argued that jj-coupling should be used. I've checked the NIST website, and they use LS term symbols for Sn.
     
  10. Feb 28, 2016 #9
    So including all possible ##Js## gives ##J=0,1,2##?
     
  11. Feb 28, 2016 #10

    DrClaude

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    Staff: Mentor

    You found that L = 1 and S = 0 or 1. That gives you two terms, 1P and 3P. Each term has its own possible set of J values.
     
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