Questions about heat

Discussion in 'Physics & Math' started by Magical Realist, Oct 1, 2017.

  1. exchemist Valued Senior Member

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    James, I think to be accurate the physical motion of molecules is incapable of generating visible light. It requires excitation of electrons within the molecules or atoms to do this, does it not? The size of the energy gap needed to generate a photon in the visible region would exceed that available from molecular vibration, surely?
     
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  3. exchemist Valued Senior Member

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    You sound disappointed!

    (The fire ants idea is rather good - a missed opportunity.)
     
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  5. James R Just this guy, you know? Staff Member

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    I don't know.

    Maybe somebody else can help.
     
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  7. exchemist Valued Senior Member

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    OK, just to check I am not losing my memory, I've looked up some sources.

    I quote Wiki: " Visible light (and near-infrared light) is typically absorbed and emitted by electrons in molecules and atoms that move from one energy level to another." Taken from this link: https://en.wikipedia.org/wiki/Electromagnetic_spectrum

    However it is interesting that the energy range of visible spectrum photons goes down to around 1eV, whereas I found that a typical C-H bond dissociation energy is around 4eV: https://en.wikipedia.org/wiki/Bond-dissociation_energy So I suppose in theory one could have a C-H bond excited to a high vibrational level, a bit short of dissociation, which could radiate at a visible wavelength as it returned to its ground state.

    In reality, though, at normal temperature it is only the ground state, and to some extent the 1st vibrationally excited state, that are appreciably populated, due to Boltzmann distribution considerations (the energy gap between successive levels is >kT). Furthermore the behaviour of the bond oscillation at low values of the quantum number v is close to harmonic. For the harmonic oscillator, the selection rule for electric dipole-induced transitions is △v = +/-1. At high vibrational levels the bonds cease to behave like harmonic oscillators, so the selection rule would become weakened, so perhaps there could, maybe, be some intensity from direct transitions to the ground state.

    Thus in practice at normal temperatures one will not get significant transitions from states with high values of v direct to the ground state, which is what would be needed for a change in vibrational excitation to produce a photon in the visible region. However, from the foregoing, I suppose that in very hot conditions, close to molecular dissociation, one might get some visible photons from highly anharmonic stretched bonds dropping direct to the ground state. But I cannot find any information on this.

    Phew, the argument is more complex than I thought!
     
  8. arfa brane call me arf Valued Senior Member

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    Maybe someone can explain why a gas in a plasma ball emits visible light and why filaments appear? What would the temperature of these filaments need to be?
     
  9. ajanta Registered Senior Member

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    Thanks, few months ago; exchemist told me about electron that one electron can absorb minimum two photons.

    May be it(electron) can absorb two low energetic photons and one photon is re-emitted(that is more energetic than two incident photons.

    This process, can produce visible light from a hot body that I assumed...
     
  10. exchemist Valued Senior Member

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    Eh? Did I? That doesn't sound right at all. Can you refer me to what I said?
     
  11. ajanta Registered Senior Member

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    It was about a wiki link.
     
  12. exchemist Valued Senior Member

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    Can you find the post?

    Because I would not have intended to say electrons absorb a minimum of two photons. That's nuts. Almost all absorption processes involve single photons.

    So either I drafted my post badly or you have got the wrong end of the stick as to what I meant.
     
  13. DaveC426913 Valued Senior Member

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    Yep. Good ol' Two Photon Exchemist is what we call him.
     
  14. ajanta Registered Senior Member

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    I Will try to find it sir !
    But I have the link.....

    https://en.m.wikipedia.org/wiki/Two-photon_absorption
     
  15. exchemist Valued Senior Member

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    Ah OK but this is not a "minimum" requirement. Far from it. The article points out that 2 photon processes are far weaker than one photon processes.
     
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  16. Fednis48 Registered Senior Member

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    This is interesting stuff, but to a large extent you're overthinking it. At high temperatures, the translational degree of freedom gets excited enough to emit visible light; that is, the molecules are bouncing around fast enough to glow regardless of their internal energy structure. That's why blackbody radiation is independent of the blackbody's chemical composition. Of course, the vibrational and other degrees of freedom can come into play as well, for instance when you get different colored flames from burning different elements. But the red glow from a hot range stove is just a product of thermal motion.
     
  17. exchemist Valued Senior Member

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    Ah OK. Different translational states like the particle in a box, I suppose. What is the emission process, though? How do you get a transition dipole from changes in translational state? Or can you point me to a reference to read about it for myself?
     

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