Compton Scattering When light encounters charged particles, the particles will interact with the light and cause some of the light to be scattered. incident photon scattered photon light wave electron motion of electron after hit electron motion of electron

Compton Scattering From the wave theory, we can understand that charged particles would interact with the light since the light is an electromagnetic wave! But the actual predictions of how the light scatters from the charged particles does not fit our simple wave model. If we consider the photon idea of light, some of the photons would “hit” the charged particles and “bounce off”.The laws of conservation of energy and momentum should then predict the scattering.

Compton Scattering As we will see in this course, photons do have momentum as well as energy. The scattered photons will have less energy and less momentum after collision with electrons, and so should have a larger wavelength according to the formula:  = scattered - incident = (h/mc)[1-cos()]

Compton scattering Einstein has proved that the energy of light is quantized E = h Compton’ s scattering demonstrated light (photon) also has momentum p = h/ Moreover, the scattering experiment can ONLY be successfully explained by the particulate nature of light

i --wavelength of incident X-ray f – wavelength of the scattered X-ray -- deflection angle of scattered X-ray -- angle of recoiled electron Experimental result Modern Physics Chapter One

Derivation Relation between photon momentum and wavelength Energy expression (in relativistic case) When v c Total energy of a particle of rest mass mo where is called the relativistic mass The total energy can also be expressed by The second term is called rest energy

Conservation of momentum Along x -axis (2) Momentum of photon of f at an angle  Momentum of electron at an angle  Momentum of photon of i Total momentum before scattering Total momentum after scattering

Conservation of energy Scattered photon energy Total energy of recoiled electron of momentum pe Initial photon energy Rest energy of stationary electron (1)

(3) From (2) (4) From (3) (5) (4)2+(5)2

(6) From (1) (7)

Equating (6) and (7)

Compton Scattering  = scattered - incident = (h/mc)[1-cos()] Note that the maximum change in wavelength is (for scattering from an electron) 2h/mc = 2(6.63 x 10-34 J-s) / (9.1 x 10-31 kg * 3 x 108 m/s) = 4.86 x 10-12 m =  which would be insignificant for visible light (with l of 10-7m) but NOT for x-ray and -ray light (with l of 10-10 m or smaller) .

Problem An x-ray beam of wavelength 0.01 nm strikes a target containing free electrons. Consider the xrays scattered back at 1800 Determine (a) change in wavelength of the xrays (b) change in photon energy between incident and scattered beams (c) the kinetic energy transferred to the electron (d) the electron’s direction of motion

Solution X-ray beam has =.01 nm = 10 pm =1800 ` -  =(h/mec) (1 - cos) = c (1 - cos) c is Compton wavelength of the electron (a) =(h/cme)(1-cos(180))= 2h/cme =2(6.63x10-34)/[(3x108)(9.11x10-31)] =2(2.43 pm)= 4.86 pm (b) E={ hc/ ` -hc/} =(6.63x10-34)(3x108){1/14.86 -1/10}/(10-12) =-.65x10-14 J = -.41x105 eV = -41 keV

Solution (c) K (electron) = 41 keV (d) direction of electron? Momentum conserved => electron moves forward p