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→Observation
== Observation ==
== Observation ==
While indeed many of the scattered photons have an unchanged wavelength, there are some which lost energy, i.e., their wavelength lengthened. This lengthening depends on the angle of the deflection only.
While indeed many of the scattered photons have an unchanged wavelength, there are some which lost energy, i.e., their wavelength lengthened. This lengthening depends on the angle of the deflection only.
==Explanation==
The lengthening of the wave-length happens when a photon interacts with a free electron in the material. The effect can be calculated along the following lines.
1) '''Conservation of energy''': We assume that before the collision, the electron is nearly at rest, so its energy is entirely its rest-energy <math>E_e=m_e c^2</math>. Here, <math>m_e</math> is the rest mass of the electron. The photon has an energy of <math>E_{\gamma}= h f</math>, where <math>h</math> is Planck's constant, and <math>f</math> is its initial frequency. After the interaction, the photon's frequency changed to <math>f'</math>, so its energy is now <math>E'_{\gamma} = h f'</math>. The energy of the electron after the collision is <math>E'_e = \sqrt{p^2_e c^2 + m^2_e c^4}</math>, and we get the equation;
<center><math> h f + m_e c^2 = h f' +\sqrt{p^2_e c^2 + m^2_e c^4} </math></center>
(here, <math>\vec{p_e}</math> is the momentum of the electron after the event)
2) '''Conversation of momentum''': As we assume that the electron is nearly at rest at first, its initial momentum is zero. The momentum of the photon is at first <math>\vec{p_p}</math> and then <math>\vec{p'_p}</math>. The seize of the momentum of a photon can be calculated via <math>p = E/c = {h f}/c</math>.
Looking at the picture, we see that
<math>p^2_e = (\vec{p_p}-\vec{p'_p})\cdot (\vec{p_p}-\vec{p'_p}) = </math>
<math>p^2 + p'^2 - 2 p p' \cos \theta</math>, thus
'''I''': <math>(p_e c)^2 = (h f)^2 + (h f')^2 - 2 h^2 f f' \cos \theta</math>
Squaring <math> h (f-f') + m_e c^2 = \sqrt{p^2_e c^2 + m^2_e c^4} </math> and rearranging leads to:
'''II''': <math>(p_e c)^2 = (h f)^2 + (h f')^2-2 h^2 f f' + 2m_e c^2 (h f - h f')</math>
We equate our '''I''' and '''II''' and get:
<math>-2 h^2 f f' \cos \theta = -2 h^2 f f' + 2 m_e c^2(h f - h f')</math>
This we rearrange to:
<math> \frac{1}{h f'} - \frac{1}{h f'} = \frac{1}{m_e c^2} (1 - \cos \theta)</math>
As for a photon <math>f \lambda = c</math>, we can bring this is Compton's form:
<math>\lambda' - \lambda = \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta)</math>
This describes exactly the effect observed by Compton! Compton himself used a similar derivation, also including the relativistic energy expression.
== Probability of Compton Scattering ==
The probability for Compton scattering is approximately proportional to the [[atomic number]] Z, and for energies greater than 500 [[keV]] approximately proportional to <math>\frac{1}{E^\gamma}</math>,<ref>{{cite web|url=http://ie.lbl.gov/education/glossary/glossaryf.htm|title=Glossary of Nuclear Science Terms|accessdate=January 10, 2013}}</ref> the energy of the [[gamma ray]] [[photon]].
The probability for Compton scattering is approximately proportional to the [[atomic number]] Z, and for energies greater than 500 [[keV]] approximately proportional to <math>\frac{1}{E^\gamma}</math>,<ref>{{cite web|url=http://ie.lbl.gov/education/glossary/glossaryf.htm|title=Glossary of Nuclear Science Terms|accessdate=January 10, 2013}}</ref> the energy of the [[gamma ray]] [[photon]].