1. 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

2. 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.

3. 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()]

4. 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

5. 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

6. 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

7. 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

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

9. (3)
From (2)
(4)
From (3)
(5)
(4)2+(5)2

10. (6)
From (1)
(7)

11. Equating (6) and (7)

12. 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 J-s) / (9.1 x kg * 3 x 108 m/s)
= x m = 
which would be insignificant for visible light (with l of 10-7m) but NOT for x-ray and -ray light (with l of m or smaller) .

13. 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

14. 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/ /10}/(10-12) =-.65x10-14 J = -.41x105 eV = -41 keV

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