The effects
in PHYSICS

 Lorentz Transformations Cerenkov effect Doppler effect Auger effect Photoelectric effect Hall effect Compton effect Pair production effect X rays Sagnac Effect Mossbauer effect Raman effect Zeeman effect Lasers

## Compton Effect

```1. Abstract
The photoelectric effect used the fact that
energy is conserved with a collision between a
photon and an electron at rest in a metal.
the involved energy of the incident photon is on the same
order of magnitude as the binding energy of an
electron to a nucleus , that is  few eV.
However, if the energy of the photon is large
compared to the binding energy of the electron, that is
several KeV, therefore, both conservation of momentum
and energy could be considered.
Compton used this fact in an experiment of
scattered x-ray radiation off of a
graphite block to measure the inrease of the
wavelength of the x-rays.
2.Introduction
The Compton effect was observed by Arthur Compton in 1923.
This Compton Scattering is about an interaction between an
incident gamma photon and an electron at rest.
The incident particle-wave loses enough energy to an
orbital electron to cause its ejection, that is to ionize the
related atom.
After the collision, the incident photon becomes lower in energy,
then infrequency, and then  large in wavelength with an emission
direction different from that of before the collision.
Compton scattering is considered to be the principal absorption
mechanism for gamma rays in the intermediate energy range
100 keV to 10 MeV.

- P0 Incident photon's momentum ,
- Pe0 Stationary electron's momentum,
- P1 Scattered photon's momentum,
- Pe1 Recoil electron's momentum,
-  a , b  Scattering and recoil
angles respectively.
3. Compton wavelength
The conservation of momentum is written by:
P0 + 0 = P1 + Pe1
Solving for Pe1, we have:
Pe12 =( P0- P1)2=P02 + P12-2 P0P1
= P02 + P12 -2 P0P1cos (b)

We know that:
P0 = hv0/c
P1 = hv1/c
Then:
Pe12 = (hv0/c)2 + (hv1/c)2
- 2(hv0/c)(hv1/c) cos(b) (Eq.1)

The conservation of enegy is written by:
E0+ Ee0=E1+Ee1
We know that :
E0=hv0and
Ee0=mec2and

E1=hv1 and
Ee1=sqrt (pe12c2+me2c4)

Solving for pe12, we have:
pe12c2= Ee12 - me2c4
Using :
Ee1 = E0+ Ee0-E1
we get:

pe12c2 = (hv0+mec2-hv1)2 - me2c4   (Eq.2)

Equating (Eq.1) and (Eq.2), we have:

(hv0)2 + (hv1)2 -2 (hv0)( hv1) cos (b) = (hv0)2+  (hv1)2 - 2hv0hv1 +
2  (hv0 -hv1) m ec2
(hv0)( v1) cos (b) = hv0 v1 - (v0 -v1) m ec2

hv0 v1(1-cos (b)) = (v0 -v1) m ec2
hc/w0 c/w1(1-cos (b)) = (c/w0 -c/w1) m ec2
hc (1-cos (b) = (w1 - w0) m ec2
Where w =c/v is the incident photon's wavelength.
w1 - w0 =h (1-cos (b))/ m ec
That is :

w1 - w0 = Wc(1-cos b)

Where WC= h/ m ec
is known as the Compton wavelength.
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