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In Compton scattering, all theoretical derivations I've seen consider the case of a photon interacting with a free electron. However, in experiments of Compton scattering I see that photons are directed at an aluminum target (such as here) or a graphite target (such as here).

This makes it seem that the experiments are not dealing with free electrons but rather bound electrons. So then I would ask, how come experiments are dealing with bound electrons if the effect is meant to apply to free electrons? Is there anything I'm missing or misunderstanding?


Edit: Apparently my question has been marked as duplicate of How can a photon collide with an electron?, but I don't understand what possible similarity that question has with mine. I'm just trying to clarify the distinction/contrast between the idealized treatments of Compton scattering vs tests of Compton scattering in the real world.

The justification (which others in the answers have adequately provided) for the difference between theory and experiment is usually absent from many online sources, so I thought it would be interesting to have something here that clarifies this.

My question has nothing to do with the cited post, which asks how photons can collide to begin with, nor do I think any responses there even answer my question. (If I'm mistaken, you can point out a specific quote , and I will stand corrected). I'm not asking how they can collide.

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You are right. The atomic electrons are not free, but bound to their atom. And the Compton scattering is usually explained as if the photon hits a free electron. This contradiction is resolved as follows.

  • The outermost electrons of atoms have binding energies of only a few eV. And even the innermost electrons af carbon and aluminium atoms have binding energies less than $200$ eV.
  • On the other hand the Compton experiment is done with X-rays (i.e. the photon energy is larger than $10$ keV) or with gamma rays (i.e. photon energy even larger).
  • So in any case the photon energy is much larger than the binding energy of the electron hit by the photon. Therefore it is a valid approximation to completely neglect the binding energy of the electron, and treat the electron as nearly free and at rest (very small potential and kinetic energy). The photon kicks the electron out of the atom, nearly unaffected by its binding energy.
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  • $\begingroup$ 1 MeV as a lower limit for gamma seems to be way too high. Above 1 MeV (1022 keV) you get pair production, but gamma is usually taken well below that. Many typical gamma emitters are below 1 MeV. $\endgroup$ Commented Nov 15, 2022 at 13:50
  • $\begingroup$ @VladimirFГероямслава You are right. I have updated the answer. $\endgroup$ Commented Nov 15, 2022 at 14:21
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In practice, it's difficult to keep a sufficiently dense collection of free electrons together to do the experiment. So, if you want to approximate pure Compton scattering, you use energetic gamma rays with a target that contains no elements with tightly bound electrons. The electrons are thus almost free, effectively.

In other situations, Compton-like inelastic scattering of less energetic photons from tightly bound electrons occurs, but the kinematics and cross section are modified.

As for Compton versus photoelectric interactions, they can be difficult to distinguish. They differ in that for the photoelectric effect the photon is completely absorbed, for the Compton effect it isn't. The easy thing to detect is the recoiling electron, but whether a scattered photon emerges or not can be difficult to determine.

We sometimes surround photoelectric x-ray detectors with "active shields". An active shield is itself a gamma ray detector. A photon that scatters in the detector has a high probability of scattering into the shield and being detected. So, we reject apparent x-ray detections that are accompanied by a detection in the shield. This also rejects photons that leak through the shield, as long as they scatter at least once on their way through.

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So then I would ask, how come experiments are dealing with bound electrons if the effect is meant to apply to free electrons? Is there anything I'm missing or misunderstanding?

Compton's scattering is a term for one simple model of scattering of X-ray radiation, which explains some characteristics of the observed scattered radiation in real experiments. Interaction of the nuclei with the electrons is ignored in this model.

Compton introduced this photon - free electron billiard model to explain one major aspect of his experiments with scattering off graphite - scattered radiation wavelength depending on the angle of scattering - and possibly also to show the measured wavelength shift as evidence for the very photon idea.

Processes in X-ray scattering experiments are much more complicated than Compton's model would suggest - there is interaction between electrons, between electrons and nuclei, the scattered radiation is influenced by the scatterer chemical composition, crystalline structure, etc. and all this has some impact on the scattered radiation.

X-ray scattering off "really free" electrons (electrons in vacuum, not in atoms) is much harder to do than scattering of solid targets, but I think this should be possible in particle accelerators, where we can have zillions of electrons in macroscopic bunches in vacuum orbiting the accelerator ring. I am afraid measurement of the wavelength shift was never done in such a setting, as it is probably not easy (target it moving, its secondary radiation would be very weak) and nobody expects to discover anything there, except validating the Compton formula.

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