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How to tell whether a transition state is metastable?

  1. Apr 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Explain why the 1s2s (3P) excited state of helium is metastable?


    2. Relevant equations

    Spin multiplicity = 2S+1

    J = |L-S|, |L-S|+1,...,L+S, L+S-1,..

    Selection Rules:

    ΔL = +/-1
    Δm = 0,+/-1
    Δs = 0
    state must change parity
    Δj = 0,+/-1
    j = 0 -> j' = 0 NOT ALLOWED

    3. The attempt at a solution

    2S+1 = 3 so S = 1
    L = 0
    J = |L-S| = 1 (only one value I think)

    How do I tell whether 1s2s (3P) excited state of helium is metastable?

    Note: (The 3 is superscript in front of P)

    Any help would be great, thanks!
     
    Last edited: Apr 20, 2012
  2. jcsd
  3. Aug 8, 2012 #2
    I know this is kind of old, but you may want the help anyway.

    A metastable state occurs when an excited state cannot make the transition back to the ground state because one of the selection rules forbid it. I haven't worked it out exactly, but it is likely that the transition from your excited state to the ground state would disobey one of the selection rules.

    To solve this problem, work out the quantum numbers for the excited state, then get the numbers for the ground state. Then look at the differences (Δ's) to get the values for the transition. Compare that with the selection rules. If it is forbidden, it is a metastable state.

    To get the idea better, you could look at fluorescence and phosphorescence (wiki). Fluorescence obeys ΔS=0 while phosphorescence breaks this rule creating a long lifetime because of its triplet metastable state.
     
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