Compton Effect
Theory
Compton effect was first observed by Arthur Compton in 1923 and this discovery led
to his award of the 1927 Nobel Prize in Physics. The discovery is important because
it demonstrates that light cannot be explained purely as a wave phenomenon.
Compton's work convinced the scientific community that light can behave as a stream
of particles (photons) whose energy is proportional to the frequency.
Compton scattering occurs when the incident x-ray photon is deflected from its
original path by an interaction with an electron. The electron is ejected from its
orbital position and the x-ray photon loses energy because of the interaction but
continues to travel through the material along an altered path. Energy and momentum
are conserved in this process. The energy shift depends on the angle of scattering and
not on the nature of the scattering medium. Since the scattered x-ray photon has less
energy, it has a longer wavelength and less penetrating than the incident photon.
The change in wavelength of the scattered photon is given by:
Where: λ
=
λ'
=
h
=
wavelength of incident x-ray photon
wavelength of scattered x-ray photon
Planck's Constant: The fundamental constant equal to the
ratio of the energy E of a quantum of energy to its frequency
ν: E=hν.
mo =
the mass of an electron at rest
c
=
the speed of light
θ
=
The scattering angle of the scattered photon
The scattering of the Compton Effect (also called Compton scattering) is
demonstrated in the picture below is the result of a high-energy photon colliding with
a target, which releases loosely bound electrons from the outer shell of the atom or
molecule. The scattered radiation experiences a wavelength shift that cannot be
explained in terms of classical wave theory, thus lending support to Einstein's photon
theory.
A high-energy photon (generally X-ray or gamma-ray) collides with a target, which
has loosely-bound electrons on its outer shell. The incident photon has the following
energy E, linear momentum p and wavelength λ:
E = hc / λ
p=E/c
The photon gives part of its energy to one of the almost-free electrons, in the form of
kinetic energy then scattered photon has less energy and higher wavelength than
the incident photon λ' > λ.
As expected in a particle collision. We know that total energy and linear momentum
must be conserved.
Energy of the photon before collision
E = hν=hc/λ, after collision E=hc/λ'.
Momentum of the photon before collision
P = h/λ=hν/c and after P=hν'/c.
Energy of electron before collision
Eo=moc2
Energy of electron after collision
E= γmoc2 where γ=1/(1-β)1/2 ,β = v/c.
Momentum of electron
P
Applying conservation of energy
hν + Eo = hν' + E
‫بعد التصادم قبل التصادم‬
hc/λ + mo c2 = hc/λ' +mo c2/(1-(v/c)2)1/2
1
Applying conservation of linear mpmemntum on x-axis
hν/c +0 = hν'/c cosθ + Pcosφ
Pc cos φ=hν-hν' cos θ
2
Applying conservation of linear mpmemntum on y-axis
0 = hν'/c sin θ – Psin φ
Pc sin φ= hν' sin θ
By squaring and summing equations 2 and 3 we get
λ' –λ =h/mo c(1-cosθ).
Where (h/mo c) known as compton wavelength
h/moc = 2.43 pm
3
Setup
Compare the result from diagram with that equation
λ' –λ =h/mo c(1-cosθ).
Where θ=145o and h/moc = 2.43 pm
.