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Do vibrational energy levels decay?

  1. Sep 22, 2010 #1
    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?
     
  2. jcsd
  3. Sep 22, 2010 #2
    It will decay to ground state by emitting photon!
     
  4. Sep 22, 2010 #3

    alxm

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    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.
     
  5. Sep 22, 2010 #4
    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.
     
  6. Sep 22, 2010 #5

    alxm

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    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".
     
  7. Sep 22, 2010 #6
    Thanks for clearing that up for me :)
     
  8. Sep 23, 2010 #7
    Hi,
    some people also say:
    A molecule absorbs photon: Here the molecule is excited
    When decay: Molecule is de-excited.
     
  9. Sep 23, 2010 #8
    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.
     
  10. Sep 23, 2010 #9
    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.
     
  11. Sep 24, 2010 #10
    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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