Deslocamento Compton (Δλ)

Compton shift: Δλ = (h/mc)(1 − cos θ)

Scatter a photon off a free electron and its wavelength grows by Δλ = (h/m_e·c)(1 − cos θ). Here θ is the scattering angle, and the prefactor h/(m_e·c) ≈ 2.43·10⁻¹² m is the Compton wavelength of the electron (2.43 pm). The shift peaks at θ = 180°, the backscatter case, where Δλ = 4.86 pm, and it drops to zero at θ = 0°. As an example, an X-ray scattered at θ = 90° picks up exactly one Compton wavelength, Δλ ≈ 2.43 pm. When Arthur Compton ran the experiment in 1923 (it earned him the Nobel Prize in 1927), the result was hard to argue with: light was acting like particles that carry momentum p = h/λ, something classical wave theory said should produce no wavelength change at all.

Applications

It shows up in radiology and CT, where Compton scatter is the dominant X-ray interaction in soft tissue between 100 keV and 10 MeV. You also meet it in nuclear medicine and PET imaging, in gamma-ray astrophysics through Compton telescopes such as COMPTEL and INTEGRAL, in the design of radiation shielding, and in Compton cameras built to pin down where a source is.

FAQ

Why doesn't the photon's wavelength change at θ = 0°? At that angle there is no scattering, so no momentum gets handed to the electron and the photon keeps the energy and wavelength it started with.

Does Compton scattering work with protons? It does, but the proton's Compton wavelength is h/(m_p·c) ≈ 1.32 fm, roughly 1836 times smaller than the electron's. At X-ray energies that makes the shift negligible.

Is Compton scattering elastic? In the relativistic sense it is, since energy and momentum are conserved across the whole interaction. For the photon by itself it's inelastic, because it hands some of its energy to the recoiling electron.