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Moessbauer Effect

  1. Sep 5, 2014 #1


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    I was looking at this article, trying to prepare myself for experimenting this effect. What I understood is that Moessbauer effect explains the reason why you can have gamma absorbtion/emission in solids while you can't for gases.
    However at the description it says:
    From this I don't understand what's the problem of the recoiling...
    You send some gamma on a nucleus, it gets excited and then it emits the extra energy in another gamma and recoiling. Why should we care about the last gamma's energy?

    Secondly, what is the reverse transition? If it means that they can be reabsorbed by another nucleus of the same matterial, then it should excite it in the same energy level and not less. Suppose you have a 3MeV ray emitted from the 1st nucleus, it should excite the 2nd to 3 MeV again...and so on...

    Finally I don't understand why nuclei have fixed positions in a crystal and the recoil is taking place for the crystal as a whole. Couldn't they be recoiled but nothing happen to the crystal? like for example cause some oscillations within it. As far as I know the states [gas to solid] is an atomic thing, so why would it concern the nuclei?
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  3. Sep 5, 2014 #2

    Simon Bridge

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    You need to read beyond wikipedia - which is not, on the whole, very good for learning physics or preparing for a physics course. What other sources have you tried?

    Did you see what the phenomena of the mossbauer effect is - think about it in terms of an experimental setup: what gets detected?
  4. Sep 10, 2014 #3

    I think the transition in question has a relatively long life (compared with its energy), so that gammas emitted would, if the nucleus could be "bolted down", have a very narrow energy spectrum, Δ, the recoil can however "smear out" the narrow gamma spectrum.

    I think that under the right conditions the gravitational energy difference of 3 or 4 stories is enough to shift the energy between the emitter and absorber so that they are off resonance, the energy spectrum is that narrow. Also, at relative speeds between the emitter and absorber of order mm/ sec is enough to doppler shift out of resonance.

    At room temperature, after a gamma emission, the nucleus has two options, produce a quantum of sound, a phonon, or give its recoil energy to the the crystal as a whole. At room temp with many phonons around you might say that the existing phonons stimulate the emission of another phonon of the same type? This smears out the energy of the emitted gammas.

    At low temps with fewer phonons they don't do as much stimulation so we get more phonon-less gamma emission.

    Hopefully better late then never.

    See also,

    Last edited: Sep 10, 2014
  5. Sep 11, 2014 #4


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    How do you get the initial photon with "exactly" the right energy? It has to be a bit higher than the transition energy as the excited atom will move (due to the momentum of the photon).

    But without the Moessbauer effect, the photon won't have 3 MeV, it just has 2.9999 MeV (or something similar).

    The electrons follow the nucleus. A nuclear recoil is always a recoil of the whole atom. How do you imagine "nothing happens" if you start moving an atom?
  6. Sep 11, 2014 #5


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    for the last- Suppose that the recoil energy will be of the order of 1 MeV (and I'm exaggerating, eg for the Fe57 it's ~[itex]\frac{196keV^2}{2m_p}[/itex])... the mass of the atom will be at least of 1GeV ([itex]\sim m_{p}[/itex].. I didn't thinnk its velocity will be so large to change anything within the lattice
  7. Sep 11, 2014 #6


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    Exciting a phonon ("sound") needs tiny energies, some meV (milli, not Mega!) are sufficient. Displacing atoms in crystal lattices is possible with typical energies of 10 to 100 eV.
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