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Chapter2: Compton Effect. Professor Mohammad Sajjad Alam University at Albany September 28, 2010 Adapted from Web. Adapted from the Web. Compton effect. Another experiment revealing the particle nature of X-ray (radiation, with wavelength ~ 10 -10 nm).

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chapter2 compton effect

Chapter2: Compton Effect

Professor Mohammad Sajjad Alam

University at Albany

September 28, 2010

Adapted from Web

Adapted from the Web

compton effect
Compton effect
  • Another experiment revealing the particle nature of X-ray (radiation, with wavelength ~ 10-10 nm)

Compton, Arthur Holly (1892-1962), American physicist and Nobel laureate whose studies of X rays led to his discovery in 1922 of the so-called Compton effect.

The Compton effect is the change in wavelength of high energy electromagnetic radiation when it scatters off electrons. The discovery of the Compton effect confirmed that electromagnetic radiation has both wave and particle properties, a central principle of quantum theory.

compton s experimental setup
A beam of x rays of wavelength 71.1 pm is directed onto a carbon target T. The x rays scattered from the target are observed at various angle q to the direction of the incident beam. The detector measures both the intensity of the scattered x rays and their wavelengthCompton’s experimental setup


experimental data
Although initially the

incident beam consists of

only a single well-defined

wavelength (l) the

scattered x-rays at a given

angle q have intensity

peaks at two wavelength

(l’ in addition), where l ‘>l

Experimental data


q =45 

q = 0 


q =135 

q =90 

compton shouldn t shift according to classical wave theory of light
Compton shouldn’t shift, according to classical wave theory of light
  • Unexplained by classical wave theory for radiation
  • No shift of wavelength is predicted in wave theory of light
modelling compton shift as particle particle collision
Modelling Compton shift as “particle-particle” collision
  • Compton (and independently by Debye) explain this in terms of collision between collections of (particle-like) photon, each with energy E = hn = pc, with the free electrons in the target graphite (imagine billard balls collision)
  • E2=(mc2)2+c2p2
  • Eg2=(mgc2)2+c2p2=c2p2
Part of a bubble chamber picture (Fermilab'15 foot Bubble Chamber', found at the University of Birmingham). An electron was knocked out of an atom by a high energy photon.

Scattered photon, E’=hc/l’, p’=h/l’



Initial photon, E=hc/l,


Initial electron, at rest, Eei=mec2, pei=0



1: Conservation of E:

cp + mec2= cp’ + Ee

Scattered electron, Ee,pe

2: Conservation of momentum:p = p’ + pe (vector sum)

conservation of momentum in 2 d

p’sinq= pesinf

p = p’cosq + pecosf

Conservation of momentum in 2-D
  • p = p’ + pe (vector sum) actually comprised of two equation for both conservation of momentum in x- and y- directions

Conservation of in y-direction

Conservation of in x-direction

some algebra
Some algebra…

Mom conservation in y : p’sinq = pesinf


Mom conservation in x : p - p’ cosq= pecosf


Conservation of total relativistic energy:

cp + mec2= cp’ + Ee


(PY)2+ (PX)2, substitute into (RE)2 to eliminate f, pe

and Ee(and using Ee2 = c2pe2 + me2c4 ):

Dl≡ l’- l= (h/mec)(1 – cosq)

compton wavelength
Compton wavelength

le = h/mec = 0.0243 Angstrom, is the Compton wavelength (for electron)

  • Note that the wavelength of the x-ray used in the scattering is of the similar length scale to the Compton wavelength of electron
  • The Compton scattering experiment can now be perfectly explained by the Compton shift relationship

Dl≡ l’- l= le(1 - cosq)

as a function of the photon scattered angle

  • Be reminded that the relationship is derived by assuming light behave like particle (photon)

For q = 00“grazing” collision =>Dl= 0

l l’


Dl≡ l’- l= (h/mec)(1 - cosq)

Notice that Dl depend on q only, not on the incident wavelength, l..

Consider some limiting behaviour of the Compton shift:

l’=0.1795 nm


for q 180 0 head on collision d l d l max
For q1800“head-on” collision =>Dl= Dlmax

q1800 photon being reversed in direction

Dlmax=lmax’- l=(h/mec)(1 – cos 180)

  • = 2le =2( 0.00243nm)

initially l

q =180o

After collision

l’max= l + Dlmax