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Why did Compton use X-rays in his famous experiment? Can it be done using other types of electromagnetic waves?

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2 Answers 2

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The Compton effect is the inelastic scattering of photons by electrons.

Compton's initial experiment used electrons in a graphite crystal to act as scatterers. These electrons are not free, they are bound, but the X-ray energies (17 keV) were large compared with the binding energies, so they approximated to free electrons.

Photons of lower energies (UV to a few keV) may liberate electrons in the graphite via photoelectric absorption, rather than scatter. Lower energy photons would also interact with electrons pseudo-classically - Thomson scattering; any scattering would be very close to elastic. Very high energy photons, above 1.02 MeV are capable of interacting with nuclei and create electron/positron pairs, instead of scattering.

So, I think your answer is that Compton needed a "thick" target of free electrons, but in the absence of such, improvised using the electrons in a graphite crystal. At low photon energies, the interactions are dominated by photoelectric absorption or elastic scattering. At very high energies, pair-production dominates. It is only at intermediate photon energies (roughly 10-1000 keV for carbon) that Compton scattering dominates (see cross-section plot below, where $\sigma_{inel}$ refers to Compton scattering, $\sigma_{pe}$ is photoelectric absorption, $\sigma_{pp}$ is pair production and $\sigma_{el}$ is Thomson scattering).

X-ray cross-section for Carbon

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  • $\begingroup$ a very clear idea indeed $\endgroup$ Aug 24, 2015 at 12:45
  • $\begingroup$ How are electrons considered free if they are still in graphite crystals ? How is it possible for an electron to not fully absorb the photon? $\endgroup$ Apr 2, 2019 at 0:01
  • $\begingroup$ @adosar I refer you to paragraph 2. Regarding your second point, actually it is impossible for a free electron to either fully absorb a photon or to scatter it perfectly elastically. Both are forbidden by conservation of energy/momentum. $\endgroup$
    – ProfRob
    Apr 2, 2019 at 5:51
  • $\begingroup$ Thanks for the answer. What bothers me is how the electrons are considered free in graphite. $\endgroup$ Apr 2, 2019 at 23:32
  • $\begingroup$ Thanks for the answer. What bothers me is how the electrons are considered free in graphite. I mean they are binded so we must provide them energy to leave the graphite. But why then dont the energy of the photon gets fully converted into kinetic of electrons as in photoelectric effect. Thanks in advance. $\endgroup$ Apr 2, 2019 at 23:32
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In theory this effect is always present, but it may be less present compared to the photoelectric effect. But it is noticeable only if the frequency of the incoming photon has (at least) a magnitude of 10^20-10^21 Hz. If the frequency used is less, the variation of the wavelength is negligible and so I could not identify it.

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