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In a diatomic gas, there are three degrees of freedom of rotation, of which the frozen mode (the rotation around the bond axis) is ruled out because the energy spacing of frozen rotational energies is about 100 000 times as great as the other two, non-frozen rotations (around the axes perpendicular to the bond axes).

But what if the temperature of the gas is very high (just beneath the temperature at which the molecular bonds start breaking up)? The two "normal" rotations possess an energy which grows quadratically with n, the number of angular momentum quanta ($L=\frac{nh}{2\pi}$): $T=\frac{n^2h^2}{8\pi^2I}$ (T is the kinetic energy, I the moment of inertia and h is Planck's constant). So if the molecule possesses 100 quanta (n=100) of angular momentum (for the "normal" rotations), the energy gets bigger 100 000 times as in the case n=1 (of course, the same is true for the frozen rotation).

Are the non-frozen rotational energies high enough before the molecular breakup to convey quanta of angular momentum to the two atoms in a "target" molecule, and give this molecule an angular momentum around the bond axis (n=1), in which case the equipartition theorem holds?

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  • $\begingroup$ What is the "frozen mode". I'm not familiar with that term, so perhaps others aren't as well. $\endgroup$
    – garyp
    Commented Jan 18, 2017 at 2:24
  • $\begingroup$ You are at 100 kph on this topic, the rest of us are at 50. :) I am not sure your first sentence makes sense, sorry.........slow down, slow down.....it was only this morning we talked (OK, I told you) about the lack of rotational freedom in diatomic molcules and I think you need to get the basics nailed first, no offence. Then your more advanced questions, and your own answers to them, which could well be correct and are interesting, can be dealt with. You are too clever for your own good here, really.. $\endgroup$
    – user140606
    Commented Jan 18, 2017 at 3:10
  • $\begingroup$ I will answer to the title , because the text under is not understandable. The answer is yes. You could excite the electronic structure of the atoms and it can be seen as rotational excitation. $\endgroup$
    – ceillac
    Commented Jan 18, 2017 at 4:07
  • $\begingroup$ @Countto10-You're right (fifth time today!). But if I'm in the water, I always want to touch the bottom (hoping not to drown...). By the way, the number 100 000 I read in an answer to another question. :-) $\endgroup$
    – Deschele Schilder
    Commented Jan 18, 2017 at 11:28

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Let us consider, e.g., a hydrogen molecule. The bond dissociation energy is about 5 eV (https://en.wikipedia.org/wiki/Bond-dissociation_energy). On the other hand, to initiate rotation of the diatomic hydrogen molecule around its axis you need to drive the electrons in hydrogen atoms from the ground state to higher levels (the moment of inertia of the nucleus is negligible), which requires at least 10 eV (http://astro.unl.edu/naap/hydrogen/levels.html) (I guess this energy is not dramatically different in the molecule compared to the atom). Thus, temperature increase can only partially "unfroze" this degree of freedom before the molecule dissociates.

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