Compton effect

Graphical representation of the dispersion of a γ photon (wallowed red line), by an electron. The scattered photon has a higher wavelength (or a smaller frequency) than before interacting with the electron.

The Compton effect (or Compton scattering) consists of the increase in the wavelength of a photon when it collides with a free electron and loses part of its energy. The frequency or wavelength of the scattered radiation depends solely on the angle of scattering.

Discovery and historical relevance

The Compton effect was studied by physicist Arthur Compton in 1923, who was able to explain it using the quantum notion of electromagnetic radiation as quanta of energy and Einstein's relativistic mechanics. The Compton effect was the final demonstration of the quantum nature of light after Planck's studies of the black body and Albert Einstein's explanation of the photoelectric effect.

Compton discovered this effect by experimenting with X-rays, which were directed against one side of a block of carbon. When the X-rays collided with the block, they spread out in various directions; as the angle of the scattered rays increased, their wavelength also increased. Based on quantum theory, Compton stated that the effect was due to the fact that the X-ray quantum acts as a material particle when colliding with the electron, for which the kinetic energy, which the quantum imparts to the electron, represents a loss in its original energy.

As a consequence of these studies, Compton won the Nobel Prize in Physics in 1927.

This effect is of special scientific relevance, since it cannot be explained through the wave nature of light. This must behave like a particle to be able to explain these observations, which is why it acquires a wave-corpuscle duality characteristic of quantum mechanics.

Mathematical formulation

The wavelength variation of scattered photons, Δ Δ λ λ {displaystyle Delta lambda }, can be calculated through the Compton ratio:

Δ Δ λ λ =hmec(1− − # θ θ ),{displaystyle Delta lambda ={frac {h}{m_{e}c}left(1-cos theta right),}

where:

• h{displaystyle h} It's Planck's constant,
• me{displaystyle m_{e} is the mass of the electron,
• c{displaystyle c} It's the speed of light.
• θ θ {displaystyle theta } the angle between incident and scattered photons.

This expression comes from the analysis of the interaction as if it were an elastic collision and its deduction requires only the use of the principles of energy conservation and timing. The quantity h/mec{displaystyle h/m_{e}c} = 0.0243 Å, is called Compton wavelength. For photons dispersed to 90°, the wavelength of the scattered X-rays is just 0.0243 Å greater than the primary emission line.

Mathematical deduction

The deduction of expression for Δ Δ λ λ {displaystyle Delta lambda } (called sometimes Compton corrimiento) can be considered the corpuscular nature of the radiation and the relations of the relativist mechanics. Consider a wavelength photon λ λ {displaystyle lambda } and momentum h/λ λ {displaystyle h/lambda } heading towards an electron at rest (mass at rest of the electron me{displaystyle m_{e}). The theory of special relativity imposes the conservation of the quadruple pμ μ =(E/c,p→ → ){displaystyle p^{mu }=. Yeah. λ λ ♫{displaystyle lambda} is the wavelength of the scattered photon and p→ → {displaystyle {vec {p}}} is the momentum of the dispersed electron is obtained:

hλ λ ♫without θ θ =pwithout φ φ {displaystyle {frac {h}{lambda'}}sin theta =psin phi}

hλ λ =p# φ φ +hλ λ ♫# θ θ {displaystyle {frac {h}{lambda }}}=pcos phi +{frac {h}{lambda '}}cos theta }

where θ θ {displaystyle theta } and φ φ {displaystyle phi } are, respectively, the dispersion angles of the photon and the electron (measures regarding the direction of the incident photon). The first of the above equations ensures the conservation of the component of the moment perpendicular to the incident direction, the second does the same for the parallel direction. Energy conservation gives:

hcλ λ +mec2=hcλ λ ♫+me2c4+c2p2{displaystyle {frac {hc}{lambda }}} +m_{e},c^{2}={frac}{lambda '}}} +{sqrt {m_{e}{2}{2}{,c^{4} +c^{2},p^{2}}}}}}

What follows is an elementary algebra work. Of momentum conservation equations it is easy to eliminate φ φ {displaystyle phi } to get:

p2=h2(1λ λ 2+1λ λ ♫2− − 2λ λ λ λ ♫# θ θ ){displaystyle p^{2}=h^{2}left({frac {1}{lambda ^{2}}}} +{frac {1}{lambda '^{2}}}}{frac {2}{lambda ,lambda '}}cos theta right)

In the expression for the conservation of energy we do:

[chuckles]hc(1λ λ − − 1λ λ ♫)+mec2]2=me2c4+c2p2{displaystyle left[hcleft({frac {1}{lambda }-}{frac {1}{lambda '}}{right)+m_{e},c^{2}{2}{2}{2}=m_{e}^{2}c^{4}+c^{2p^{2}{2}{2}{2}{2}}}}}}{2}}{2}{2}}{2}}{2}}{2}}}}{2}{2}}}

Animations and simulations

• Compton effect (in German)