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Inelastic collisions

  1. Apr 27, 2009 #1
    I wonder how inelastic a collision can be? I am thinking about a gas at room temperature; for the sake of argument let it be helium, where the first excited state is well above the available collision energies. Even in a low-velocity collision, it seems there ought to be some distortion of the orbitals, which to me suggests that the atoms are at least partially driven into a higher state. The question is: must this energy of distortion be completely recovered in the kinetic recoil of the atoms, or is there some residual distortion which is left over to be radiated away electromagnetically?

    I can put the question another way. Even at room temperature, according to Planck's Law of radiation, there is a very small amount of equilibrium radiation in the ultraviolet band corresponding to the first excititation level of helium. We can explain this by saying that the velocity distribution of helium atoms contains a very tiny percentage of atoms with enough kinetic energy to drive this transition, which explains where the radiation comes from. Or we can say that every single collision drives this transition to a very small extent, and the equilibrium of radiation results from the relaxation of these very small distortions.

    Are these two explanations equivalent or do the facts require us to choose between them?
     
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  3. Apr 27, 2009 #2

    alxm

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    Okay, let's consider Helium. Since it's not a molecule you only have two significant energy states here, electronic states, and kinetic energy of the atom (which is a continuum). Collisions at room temperature (or any fairly 'normal' temperature, is well in that, kinetic energy, range.

    Well, that's a bad assumption (c.f. actual values of Zeeman splitting) - Orbitals wouldn't deform enough to significantly alter their total energy and with it, the transition probabilities. Another evidence of that is the relative tinyness of chemical energies with overall electronic energies. Nuclear motion is to a large extent decoupled from electronic transitions, hence the Born-Oppenheimer approximation and Franck-Condon principle.

    Only at very high temperatures. The opposite process (electronic transitions followed by non-radiative relaxation) happens all the time though.
     
  4. Apr 27, 2009 #3
    I don't believe you can be correct in stating that orbitals wouldn't significantly deform in an ordinary room-temperature collision. At the moment of collision, we ought to have two stationary helium nuclei separated by a distance of one "atomic diamter". If we add four electorns to this system and look for a solution to its Schroedinger equation, surely we will get a very different solution than the case of two isolated helium atoms.

    My question wasn't really as to whether the orbitals would deform: unless I'm completely wrong they must surely deform quite a bit. My question was whether the energy of deformation goes back into the elastic collision, or whether some of it remains behind in the partially excited atoms, to be subsequently radiated away?
     
  5. Apr 27, 2009 #4

    alxm

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    No, you don't get a 'very' different solution, which is why LCAO remains a fairly decent descriptor of molecules. What's more important is what happens in terms of energy. Consider moving two hydrogen atoms very close to each other - they form a "chemical bond". That's what happens to the Schrödinger equation. The total electronic energy is about 1 Hartree. The bonding energy (difference relative the free atoms) is 0.16 Hartrees, and much less that what would be required to cause an electronic excitation.

    Chemical reactions in general, which are caused by atomic and molecular collisions, tend to proceed along the potential energy surface of the electronic ground state. (With some exceptions like spin-state transitions in transition-metal reactions, and photochemical reactions)

    Again, deformation of orbitals has little to do with it. Two atoms or molecules will collide inelastically only if there's somewhere for that energy to go, and at normal temperatures, that does not mean an excited electronic state. It will often, at ordinary temperatures, mean an excited vibrational/rotational state (if a molecule).

    And when we're talking vibrational/rotational states, you're talking about kinetic energy of the nuclei. And talking about the 'deformation of the orbitals' is simply not much of a factor there (again, Born-Oppenheimer and Franck-Condon). The kinetic energy is transferred directly from nucleus to nucleus, without the electrons acting as a mediator. From the nuclear 'time scale', the electrons adapt instantly to an approaching atom/ion/electrical field, and so there is no 'recoil' - no kinetic energy (within the approximations, in reality: little energy) is transferred to the nuclei from the electrons. The nucleus weighs thousands of times more than the electrons, and moving them over to one side of the atom or the other isn't going to make it budge much. They're too fast for it.
     
  6. Apr 27, 2009 #5
    That's what I'd call a "very different solution".

    But it's still very different from the isolated atoms, which was my point.


    Why not a partially excited state? That was my original question. I know there isn't enough energy in the collision to drive the transition to completion. But if the orbitals can be distorted, that means the transition would be driven at least partially. It is a reasonable mechanism whereby energy might be stored during the collision.

    I excluded these cases when I suggested we consider the case of helium atoms.

    Again, isn't that what I've been saying? If the electrons "adapt" to the approaching ion, isn't that exactly the same as saying the orbitals are distorted during the collision?
     
  7. Apr 28, 2009 #6

    alxm

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    Because there's no such thing as a 'partially excited state'.


    You're not reading what I wrote. I didn't say the orbitals didn't 'distort', I said that that 'distortion' is not sufficient, energy-wise to cause an transition between electronic states, and nor does it, to any significant degree, transfer kinetic energy to the nuclei. As far as electronic orbitals are concerned, atomic and molecular collisions are perfectly elastic at ordinary temperatures.
     
  8. Apr 28, 2009 #7

    edguy99

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    I would think the collision. Is it not correct to say that although this is mostly an inelastic collision regarding the 2 very heavy helium nucleus, it may well involve the emission of a photon or two and a corresponding slight energy difference after the collision. Or put another way, the collision may drive an electron out of its energy level and a photon may be emitted when it falls back into the same or perhaps a slightly different energy level that may exist because of the new situation (post collision).
     
  9. Apr 29, 2009 #8
    Sure there is. Yes, if the molecules are very slow, the orbitals distort quasi-statically, so the distorted orbitals are at all times in their ground state. But as you look at collisions with greater and greater energies, you get to the point where the orbital cannot readjust to the changing configuration fast enough to stay in the ground state. This becomes significant well before the point where the collision energy becomes equal to the minimum excitation energy. So you get partially excited states.
     
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