How to tell whether a transition state is metastable?

  • Thread starter Thread starter kratos
  • Start date Start date
  • Tags Tags
    State Transition
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
kratos
Messages
5
Reaction score
0

Homework Statement



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

Homework 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

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:
Physics news on Phys.org
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.