Compton Scattering

Report
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)
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.
2
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 wavelength
q
3
Experimental data
q = q45 
q=0
Although initially the
incident beam consists of
only a single well-defined
wavelength (l ) the
qq = 90 
q = 135 
scattered x-rays at a given
angle q have intensity
peaks at two wavelength
(l’ in addition), where l
‘>l
4
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
5
Modelling Compton shift as “particleparticle” 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
6
• 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.
7
Initial photon,
E=hc/l,
p=h/l
Initial
electron, at
rest, Eei=mec2,
pei=0
Scattered
photon,
E’=hc/l’,
p’=h/l’
q
1: Conservation of E:
cp + mec2 = cp’ + Ee
f
Scattered
electron, Ee,pe
2: Conservation of momentum:p
8
= p’ + pe (vector sum)
y
x
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 l.mom in ydirection
p’sinq = pesinf
p = p’cosq + pecosf
Conservation of l.mom in x-direction
9
Some algebra…
Mom conservation in y : p’sinq = pesinf
Mom conservation in x : p - p’ cosq = pecosf
(PY)
(PX)
Conservation of total relativistic energy:
cp + mec2 = cp’ + Ee
(RE)
(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 )
10
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)
11
X-ray scattering from an electron (Compton
scattering): classical versus quantum
picture
12
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:
For q = 00  “grazing”
collision => Dl = 0
l’=0.1795 nm
l l’
l
q0
13
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
14

similar documents