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Mössbauer Effect

  1. Jul 10, 2006 #1
    Hello

    I’m currently reading up on Mössbauer spectroscopy. Now I understand that you can have a low energy gamma emission become absorbed in an atom of the same radionuclide due to resonance (either from a recoilless event or oscillating a source/absorber). While I don’t need to know the details of "how/why" for what I’m doing per se, but it’s been bugging me.

    What I don’t understand is the absorption. Is this like a compton scatter? Or does it directly excite the nucleus? Or is there something else I’m missing? What’s going on here?

    (edit: ruled out one possibility I proposed)
     
    Last edited: Jul 11, 2006
  2. jcsd
  3. Jul 10, 2006 #2
    also, heres another question. Say a ground state atom absorbs the gamma. Wouldnt it just spit the energy back out right away to reach the ground state? So then in Mössbauer spectroscopy why do you even get those resonance valleys, same isotope, same energy...
     
  4. Jul 11, 2006 #3

    Astronuc

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    Compton scattering and photo-electric effect would be a factor, but it is the absorption of the emitted photon by a nucleus the same radionuclide, which is the characteristic of the Mössbauer effect.

    It is explained quite well on the Hyperphysics site -
    http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/mossb.html#c2

    http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/mossfe.html

    http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/ironze.html#c1
     
    Last edited: Jul 12, 2006
  5. Jul 11, 2006 #4

    Gokul43201

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    You can think of it as a combination of Compton scattering accompanied by a nuclear excitation (and relaxation) process. If you used an X-ray photon, you'd probably have a pure (nuclear) Compton effect. With higher energies - in the gamma ray range - you have that and more.

    To excite a nucleus to its first excited state (~MeV, I think), you need to supply the excitation energy E(ex) plus the recoil energy E(r) needed to conserve momentum. So, E(incident photon) needs to be E(ex) + E(r), to induce a nuclear excitation. The nucleus goes to the excited state and possesses the necessary recoil momentum before it radiatively relaxes to the ground state with another recoil. Again, writing the energy-momentum conservation equations, the energy of the emitted photon turns out to be E(ex) - E(r). The difference, 2E(r) resides in the KE of the nucleus (or atom).

    For a system of isolated atoms, this difference (given by E(ex)2/Mc2) in energy between absorbed and emitted photons is too large to sustain resonance. (ie: the spectral width of the photons is not enough to make up for the recoil energy loss and so, an emitted photon does not have enough energy to cause an excitation in another nucleus of the same element).

    In a solid, you don't make single atoms recoil. Instead you must excite a phonon mode (ie: you make the entire crystal recoil). Think of this, for simplicity, as increasing the effective mass (M, in the above expression for the recoil energy) from that of a single atom to that of the entire sample. This makes the recoil energy miniscule - and hence the probability of an emitted photon producing a subsequent excitation is no longer vanishingly small.
     
  6. Jul 11, 2006 #5
    Thank you Astronuc & Gokul43201
     
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