How to tell whether a transition state is metastable?

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SUMMARY

The 1s2s (3P) excited state of helium is classified as metastable due to its inability to transition back to the ground state, which is prohibited by selection rules. Specifically, the spin multiplicity is calculated as 3 (2S+1), with S equaling 1 and L equaling 0, leading to a total angular momentum J of 1. The transition from this excited state to the ground state violates one or more selection rules, confirming its metastable nature. Understanding this concept is crucial for analyzing quantum states and their transitions.

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  • Quantum mechanics fundamentals
  • Understanding of selection rules in quantum transitions
  • Knowledge of spin multiplicity and angular momentum
  • Familiarity with metastable states and their characteristics
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  • Research the concepts of fluorescence and phosphorescence
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Students and professionals in physics, particularly those focusing on quantum mechanics, atomic structure, and spectroscopy. This discussion is beneficial for anyone seeking to understand the behavior of excited states and their transitions.

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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:
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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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