Do vibrational energy levels decay?

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Discussion Overview

The discussion revolves around the behavior of vibrational energy levels in molecules, specifically whether these levels decay by emitting photons or through other mechanisms. It explores concepts related to vibrational excitation, relaxation, and the implications of these processes in different contexts, such as laser interactions and macroscopic systems.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • Some participants propose that vibrational energy levels decay by emitting photons, similar to electronic levels, but with different selection rules.
  • Others argue that vibrational energy levels can also lose energy non-radiatively through collisions with other molecules.
  • One participant suggests that the term "decay" could be replaced with "cool," indicating a broader macroscopic perspective on energy loss.
  • Another participant challenges this terminology, emphasizing that "cool" refers to temperature, a macroscopic property, while "decay" pertains to a change in a specific state.
  • There is a discussion about the concept of "vibrational relaxation" as a more appropriate term for the process of returning to lower energy states.
  • Some participants mention that in an isolated system, rapid excitation and relaxation of vibrational states can occur without a measurable cooling effect, governed by the Boltzmann distribution.
  • Concerns are raised about the conditions under which excitation and relaxation rates can be equal, particularly in isolated systems.

Areas of Agreement / Disagreement

Participants express differing views on the terminology used to describe vibrational energy level transitions and the mechanisms involved. There is no consensus on whether "decay" or "cool" is the more appropriate term, and the discussion remains unresolved regarding the implications of these processes in different contexts.

Contextual Notes

Limitations include the dependence on definitions of terms like "decay" and "cool," as well as the assumptions regarding isolated systems and the applicability of the Boltzmann distribution in various scenarios.

The-Exiled
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Hi,

I was just wondering if the vibrational energy levels in molecules decay, emitting a photon? or by some other mechanism?

For example, if a laser is used to excite a vibrational mode (Like that in Stokes-Raman Scattering) from say v = 0 to v = 1 state. Will the molecule remain in the v = 1 state or will it return to the ground state?
 
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It will decay to ground state by emitting photon!
 
Vibrational levels will emit photons and decay just as electronic levels do, they have different selection rules though. They can also lose energy non-radiatively in collisions with other molecules.
 
I think that at these energies, we can practically replace the word "decay" by "cool".

The laser crystal will simple cool, like any other substance.
 
I think that'd be too loose terminology. "Cool" implies temperature, which is a macroscopic or ensemble property, whereas a decay is a change of a single state.

I believe the most used term is "vibrational relaxation".
 
Thanks for clearing that up for me :)
 
Hi,
some people also say:
A molecule absorbs photon: Here the molecule is excited
When decay: Molecule is de-excited.
 
alxm said:
I think that'd be too loose terminology. "Cool" implies temperature, which is a macroscopic or ensemble property, whereas a decay is a change of a single state.

I believe the most used term is "vibrational relaxation".

Ok, but for a macroscopic crystal, the only real difference is in the number of states that are considered: 1 (or a few) for a "relaxing" laser crystal vs "many" for a "cooling" random object.
 
Dr Lots-o'watts said:
Ok, but for a macroscopic crystal, the only real difference is in the number of states that are considered: 1 (or a few) for a "relaxing" laser crystal vs "many" for a "cooling" random object.

If the crystal is an isolated system (no energy or mass is allowed to escape) you would still have rapid excitation and relaxation of vibarational states inside the crystal without any cooling effect. The Boltzmann distribution law would then let you approximate how many molecules/atoms are in each vibarational state at any given time.
 
  • #10
espen180 said:
If the crystal is an isolated system (no energy or mass is allowed to escape) you would still have rapid excitation and relaxation of vibrational states inside the crystal without any cooling effect. The Boltzmann distribution law would then let you approximate how many molecules/atoms are in each vibrational state at any given time.

Sure, the system has to very well isolated though for the excitation rate to be = to the relaxation rate i.e. no loss. With only one state involved, a temperature difference may not be measurable.
 

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