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HCl molecule, rotational and vibrational energy question

  1. Apr 16, 2012 #1

    fluidistic

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    Since the HCl molecule has a non vanishing dipole moment, it is allowed to have pure rotational spectrum. If I understand this correctly then it's possible to excite the molecule in such a way that its vibrational energy stays the same while its rotational energy increases by some amount. This can basically be found there: http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/rotrig.html#c4.

    However in http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/vibrot.html#c2, one reads that the quantum number v must change in a transition according to [itex]\Delta v =\pm 1[/itex] (and [itex]\Delta j = \pm 1[/itex]). So that it seems that the molecule cannot (due to the [itex]\Delta v =\pm 1[/itex] rule) have a purely rotational spectrum. To me this contradicts the first assertion but obviously I'm missing something.
    What is going on?
     
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  3. Apr 16, 2012 #2

    fzero

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    There's no problem with the rotational spectrum. This is an issue of allowed transitions, which have to obey conservation laws. The transitions in question involve emission or absorption of one or more photons, which have angular momentum (magnetic quantum number) [itex]\pm 1[/itex]. Since angular momentum must be conserved, the angular momentum of the molecule must change by [itex]\Delta j= \pm 1[/itex] for each photon involved in a transition.

    A transition like [itex] j=0, \upsilon =0[/itex] to [itex] j=0, \upsilon =1[/itex] can't occur due to single photon absorption. It can however proceed by absorption of two photons (with opposite angular momentum). Such a 2nd order process is suppressed relative to the 1st order processes, so it's much less visible in data.
     
  4. Apr 16, 2012 #3

    fluidistic

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    Ok thanks fzero. I'm still confused, I think I understand all what you mean.
    My understanding is that for any molecule (no need to have a non zero dipole moment), transitions like j=1 v=0 to j=0 v=1 are allowed and transitions like j=0 v=0 to j=1 v=0 are allowed only for non vanishing dipole moment molecules.
    However hyperphysics states that [itex]\Delta v[/itex] must either be 1 or -1 for the HCl molecule. To my understanding this excludes any pure rotational spectrum.
    In other words hyperphysics says that j=0 v=0 to j=1 v=0 transition is not allowed for the HCl. Why not? It looks like a pure rotational transition and should be allowed according to hyperphysics (other link).
     
  5. Apr 16, 2012 #4

    fzero

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    I don't think that they're saying that [itex]\Delta v = 0[/itex] pure rotational transitions don't occur. They just occur at lower frequencies than the vibrational transitions. Look at the figure on http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/vibrot.html#c2 where the level spacings are (roughly) compared. j=0 to j=1 with [itex]\Delta v = 0[/itex] corresponds to a different part of the EM spectrum compared to, for example, j=0, v=0 to j=1,v=1.

    The HCl vib-rot spectrum occurs around [itex]8.6\cdot 10^{13} ~\mathrm{Hz}[/itex]. I estimated from the rotational wavelengths on http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/rotrig.html#c4 that the pure rotational transitions are around [itex]3\cdot 10^{12} ~\mathrm{Hz}[/itex]. So if you zoom into the vibrational spectrum, you won't see the pure rotational spectrum.
     
  6. Apr 16, 2012 #5

    fluidistic

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    Thank you very much. By the way now I cannot find where they say that delta v must obbey [itex]\Delta v =\pm 1[/itex]. I must be crazy... :smile: I think I misread and was misleaded by the figures.
    Problem solved.
     
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