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4/2003 Rev 2 I.3.4 – slide 1 of 24
Session I.3.4
Part I Review of Fundamentals
Module 3 Interaction of Radiation with Matter
Session 4 Photon Interactions
IAEA Post Graduate Educational CourseRadiation Protection and Safety of Radiation Sources
4/2003 Rev 2 I.3.4 – slide 2 of 24
In this session we will discuss photon interactions including:
Photoelectric effect Compton scattering Pair production
Overview
4/2003 Rev 2 I.3.4 – slide 3 of 24
Photoelectric Effect
4/2003 Rev 2 I.3.4 – slide 4 of 24
Photoelectric Effect
-Example: Eincident photon = 80 keV
Ebinding energy = 20 keVEphotoelectron = 60 keV
Ephotoelectron = Eincident photon – Ebinding energy
4/2003 Rev 2 I.3.4 – slide 5 of 24
Photoelectric Effect
The photoelectric effect is predominant for:
Low energy photons High atomic number “Z” materials
Probability is proportional to: Z4
E3
4/2003 Rev 2 I.3.4 – slide 6 of 24
Photoelectric Effect
4/2003 Rev 2 I.3.4 – slide 7 of 24
Compton Scattering
incidentphoton (Eip)
scatteredphoton (Esp)
scatteredelectron (Ese)
loosely boundelectron (Eie)
4/2003 Rev 2 I.3.4 – slide 8 of 24
Compton Scattering
Eie = moc2 Ese = mc2
Eip =hcip
E*sp =hcsp
Conservation of Energy: hcip
+ moc2 = hcsp
+ mc2
4/2003 Rev 2 I.3.4 – slide 9 of 24
Compton Scattering
Pse = mv
Pip = hip
P*sp = hsp
Conservationof Momentum:
hip
= hsp
cos + mv cos
0 = hsp
sin + mv sin
horizontal
vertical
Pie = 0
4/2003 Rev 2 I.3.4 – slide 10 of 24
Compton Scattering
(1 - cos) = sp - ip =h
moc
Solving both the energy and momentum equations yields:
The energy transferred to the scattered electron is:
Ese = Eip – Esp = -hcip
hcsp
4/2003 Rev 2 I.3.4 – slide 11 of 24
Compton Scattering
Substituting E =hc
into the momentum equations gives the energy of the scattered photon:
Esp =1 + (1 - cos)
Eip
moc2
Eip
moc2 = rest mass energy of the electron = 0.511 MeV
4/2003 Rev 2 I.3.4 – slide 12 of 24
Compton Scattering
Esp =1 + (1 - cos)
Eip
moc2
Eip For simplicity let
Eip
moc2= f
Esp =1 + f (1 - cos)
Eip
4/2003 Rev 2 I.3.4 – slide 13 of 24
Compton Scattering
When = 90º, Esp = the energy of the scattered photon is reduced
When = 180º, Esp = the energy of the scattered photon is minimum
When = 0º, Esp = Eip there is no interaction
Eip
(1+f)
Eip
(1+2f)
4/2003 Rev 2 I.3.4 – slide 14 of 24
Compton Scattering
Example 1: A low energy photon (10 keV) scattered by 90º and 180º
f = 10 keV/511 keV = 0.02; cos 90º = 0; cos 180º = -1
Esp =1 + 0.02 (1 - 0)
10 keV= 10/1.02 = 9.8 keV
Esp =1 + 0.02 (1 - -1)
10 keV= 10/1.04 = 9.6 keV
the scattered electron
receives only 2-4% of the
incident energy
4/2003 Rev 2 I.3.4 – slide 15 of 24
Compton Scattering
Example 2: A high energy photon (1000 keV) scattered by 90º and 180º
f = 1000 keV/511 keV = 1.96; cos 90º = 0; cos 180º = -1
Esp =1 + 1.96 (1 - 0)
1000 keV= 1000/2.96 = 340 keV
Esp =1 + 1.96 (1 - -1)
1000 keV= 1000/4.92 = 200 keV
the scattered electron receives
about 66-80% of the incident
energy
4/2003 Rev 2 I.3.4 – slide 16 of 24
Compton Scattering
Example 3: What is the maximum energy of a photon scattered through 90º and 180º for a very high energy incident photon
If f = Eip/511 keV >> 1 then (1 + f) f
1 + f
Eip
f
EipEsp = = 511 keV= =
Eip
Eip
511
(90º)
(180º)1 + 2f
Eip
2f
EipEsp = = = = = 255 keV
511
2
Eip
2Eip
511
4/2003 Rev 2 I.3.4 – slide 17 of 24
Compton Scattering
E =hc
h = 6.62 x 10-34 J-secc = 3 x 108 m/sec
E =(6.62 x 10-34 J-sec)(3 x 108 m/sec)
(1.6 x 10-19 J/eV)(103 eV/kev)
E = 1.24 x 10-9 keV-m (m)
The energy of a photon relative to its wavelength:
4/2003 Rev 2 I.3.4 – slide 18 of 24
Pair Production
Photon converted into two particles
(energy into mass)
electron (-)positron (+)
4/2003 Rev 2 I.3.4 – slide 19 of 24
Pair Production
The rest mass energy of a positive or negative electron is 0.511 MeV
To create these two particles requires a minimum energy of 2 x 0.511 MeV = 1.02 MeV
4/2003 Rev 2 I.3.4 – slide 20 of 24
Pair Production
A positron cannot exist at rest. It combines with an electron. The two particles annihilate each other converting mass back into energy.
Since the rest mass energy of each particle is 0.511 MeV, the two photons created must each possess an energy of 0.511 MeV. Two photons must be created, traveling in opposite directions, to satisfy the Law of Conservation of Momentum.
4/2003 Rev 2 I.3.4 – slide 21 of 24
Photon Energy (MeV)
PairProduction
Compton
Photoelectric
Combined
Pro
bab
ility
WATER
Photon Interactions
4/2003 Rev 2 I.3.4 – slide 22 of 24
Photon Energy (MeV)Pair
Production
Compton
Photoelectric
Combined
Pro
bab
ility
LEAD
K shellBindingEnergy
Photon Interactions
4/2003 Rev 2 I.3.4 – slide 23 of 24
Summary
We have discussed Photon Interactions including:
Photoelectric Effect low energy photons photon electron
Compton Scattering medium energy photons photon electron + new photon
Pair Production photon ( 1.02 MeV) photon e- + e+ 2 photons (0.511 MeV each)
4/2003 Rev 2 I.3.4 – slide 24 of 24
Where to Get More Information
Cember, H., Johnson, T. E., Introduction to Health Physics, 4th Edition, McGraw-Hill, New York (2008)
Martin, A., Harbison, S. A., Beach, K., Cole, P., An Introduction to Radiation Protection, 6th Edition, Hodder Arnold, London (2012)
Jelley, N. A., Fundamentals of Nuclear Physics, Cambridge University Press, Cambridge (1990)
Firestone, R.B., Baglin, C.M., Frank-Chu, S.Y., Eds., Table of Isotopes (8th Edition, 1999 update), Wiley, New York (1999)