Photoelectric effect  
  Moseley's law  
  Bohr's atom  
  ask us  

Projectile's charge Z Projectile's E (MeV)


The nucleus


Radiation Measurements

Black body radiation

Statistical Mechanics

Radiation and scattering

Related topics

Related links

© The scientific sentence. 2010

The 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, known as incoherent 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 as:

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
Pe12 = (hv0/c)2 + (hv1/c)2 -2 (hv0/c)( hv1/c) cos (b) (Eq.1)

The conservation of energy is written as:

E0+ Ee0=E1+Ee1
We know that :
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.

4. Differential cross section for the Compton scattering

For an incident photon (x-rays, γ-rays) of energy Eγ, the differential cross section is given by the Klein-Nishina formula:

dσ/dΩ = re2/2 {P - P2sin2θ + P 3}
Where: P = P(Eγ,θ) = 1/[1 + (Eγ/mec2(1 - cosθ)]
re is the classical electron radius, me is the electron rest mass, θ is the scattering angle, and dΩ = 2πsinθ dθ.

  Web ScientificSentence

© Scientificsentence 2010. All rights reserved.