Mössbauer Effect: Learn About Its Absorption Process

  • Thread starter tehfrr
  • Start date
In summary, the Mossbauer effect is when an excited nucleus emits a gamma ray, and it must recoil in order to conserve momentum. The sharpness of an energy state in a potential target nucleus has a natural line width on the order of 10-5 eV, so that the shift in the photon energy prevents the target nucleus from absorbing the gamma photon. This difference (given by E(ex)2/Mc2) in energy between absorbed and emitted photons is too large to sustain resonance.
  • #1
tehfrr
52
0
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:
Physics news on Phys.org
  • #2
also, here's 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...
 
  • #3
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 -
The Mossbauer effect involves the emission and absorption of gamma rays from the excited states of a nucleus. When an excited nucleus emits a gamma ray, it must recoil in order to conserve momentum since the gamma ray photon has momentum. But this takes energy, and the gamma photon has less energy by about 1 eV for a 100 keV photon. The sharpness of an energy state in a potential target nucleus has a natural line width on the order of 10-5 eV, so that the shift in the photon energy prevents the target nucleus from absorbing the gamma photon.
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:
  • #4
tehfrr said:
What I don’t understand is the absorption. Is this like a compton scatter? Or does it directly excite the nucleus?
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.
 
  • #5
Thank you Astronuc & Gokul43201
 

Related to Mössbauer Effect: Learn About Its Absorption Process

1. What is the Mössbauer Effect?

The Mössbauer Effect is a nuclear phenomenon in which gamma rays are emitted from a radioactive nucleus and are absorbed by a target material without any change in their energy. This effect is named after its discoverer, Rudolf Mössbauer, and has applications in various fields of science, including physics, chemistry, and materials science.

2. How does the Mössbauer Effect work?

The Mössbauer Effect works by using a Mössbauer source, which is a radioactive material that emits gamma rays. These gamma rays interact with a target material, causing a small fraction of them to be absorbed. The absorbed gamma rays then re-emit at the same energy level and direction as the original source, allowing for precise measurements of the target material's properties.

3. What is the absorption process in the Mössbauer Effect?

The absorption process in the Mössbauer Effect occurs when a gamma ray is absorbed by an atom in the target material. This absorption causes the atom's nucleus to be excited to a higher energy state. The nucleus then quickly returns to its ground state, emitting a gamma ray in the process. This re-emitted gamma ray has the same energy as the original source, allowing for precise measurements.

4. What are the applications of the Mössbauer Effect?

The Mössbauer Effect has numerous applications in different fields of science. In physics, it is used to study the properties of nuclei and their interactions. In chemistry, it is used to study chemical bonding and molecular structure. In materials science, it is used to study the properties of materials, such as magnetism and crystal structure.

5. What are the benefits of using the Mössbauer Effect?

The Mössbauer Effect has several benefits, including high sensitivity, allowing for precise measurements of small changes in the target material. It also has a high resolution, allowing for the detection of subtle differences in the energy levels of the gamma rays. Additionally, it is a non-destructive technique, meaning the target material can be studied without altering its properties.

Similar threads

Back
Top