In class we showed that for Compton Scattering (incoherent scattering) the energ
ID: 1688749 • Letter: I
Question
In class we showed that for Compton Scattering(incoherent scattering) the energy of a scattered photon can be written as:
h? 'scattered = h?incident
1+? (1- cos? )
where ? = h?
mec2
a) (10 points) Using energy and momentum conservation derive the
expression for the energy of the scattered photon in terms of the
incident photon energy and the scattering angle. Hints: 1) momentum
in the x- and y- direction must be conserved, 2) E = h? = pc for
relativistic particles (i.e. photons) and 3) do not forget your
Trigonometry.
Explanation / Answer
Energy of incident photon E = h v Energy of scattered photon E' = h v' angle between scattered photon and incident photon is f angle between recoil electron and incident photon is ? conservation of momentum in original photon direction h v / c + 0 = h v' / c cosf + p cos? ...(1) in perpendicular direction 0 = h v' / c sin f - p sin? ....(2) from equations (1) and (2) p cos ? = h v / c - h v' / c cosf P sin? = h v' / c sin f squaring the adding the equations p^2 c^2 =(hv)^2 - 2(hv) ( hv') cosf + ( h v')^2 ......(3 total energy of the particle E = k.E + mc^2 p^2 c^2 = kE^2 + 2mc^2 KE =(hv)^2 - 2(hv) ( hv') + ( h v')^2 + 2 mc^2 ( h v - hv')........(4) equating equation (3) and (4) (hv)^2 - 2(hv) ( hv') cosf + ( h v')^2 +2 mc^2 ( h v - hv' ) = (hv)^2 - 2(hv) ( hv') cosf + ( h v')^2 2(hv) ( hv') - 2(hv) ( hv') cosf =2 mc^2 ( h v - hv' ) 2 E E' - 2 EE' cosf = 2 mc^2 ( E - E ' ) EE ' = mc^2 ( E - E ' ) / ( 1- cos f ) E ' = m c^2 ( k.E ) / E (1- cos f )Related Questions
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