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24.08.2016
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Interaction theory – Photons
Eirik Malinen
24.08.2016
2
Introduction
- Photons
- Charged particles
- Neutrons
Ionizing
radiationAtoms Matter
Interaction
theory
Ionizing
radiation
Dosimetry
Radiation
source
H2O
DNA
Molecules Cells Humans
Ionizing
radiation
Effects
• To understand primary effects of ionizing
radiation
• How radiation doses are calculated and measured
• To appreciate applications of ionizing radiation
Objectives
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• Interactions between ionizing radiation and
matter
• Radioactive and non-radioactive sources
• Calculations and measurement of absorbed doses
(dosimetry)
Contents FYSKJM4710
• X-ray and CT investigations
• Radiotherapy
• Positron emission tomography
• Radiation protection
• Radiation Biology
Relevant issues
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• X-ray contrast: only a matter of differences in
density?
Basic theory of the atom
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Bohr model
Hydrogen
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• Liberation of electron from atom
• Requires energy transfer ~4-25 eV
• A lethal whole-body dose of radiation (5 J/kg)
results in a temperature increase of 0.001 °C
Ionization
Directly ionizing radiation: Fast charged particles,
which deliver their energy to matter directly, through
many small Coulomb-force interactions
Indirectly ionizing radiation: Photons (γ or X-rays)
or neutrons, which transfer their energy to charged
particles in the matter
Ionizing radiation
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• Cross section s: “target area”, effective target
covering a certain area
• Proportional to the interaction strength between
an incoming particle and the target particle
• Consider two discs, one target and one incoming:
• s is the total area:
Cross section 1
incoming particle
radius r1
(direction into the paper)
electron,
nucleus...
radius r2
2 21 2(r r )π +
• N particles move towards an area S with n atoms
• Probability of interaction: p = nσ/S
• Number of interacting particles: Np= Nnσ/S
Cross section 2
Atomic cross
section, σ
S
no interactioninteraction
Incoming particle
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• Separate between electronic and atomic cross
section
• The cross section depends on:
- Type of target (nucleus, electron, ..)
- Type of and energy of incoming particle
(photon, electron…)
• Cross section calculated with quantum mechanics
- here visualized in a classical window
Cross section 3
• Differential cross section with respect to scattering
angle
Cross section 4
ΩΩ=
Ω darea unitper particles ofnumber
d into scatteredparticles ofnumber
d
d 1σ
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• Photon represented by a plane wave
in quantum mechanical calculations
• In principle, two different processes:
– Absorption
– Scattering
• Scattering: coherent (elastic) og incoherent
(inelastic)
Photon interactions
)t,r(Ainn
r
0ψkψ
*
)t,r(Aut
r
θ
)t,r(Ainn
r
0ψkψ
*
)t,r(Aut
r
θ)t,r(Ain
)t,r(Ain
),( trAout
)trpi(in
inine~t)(r,A ω−⋅
• Scattering without loss of energy: hν=hν’
• Photon is absorbed by atom, thereby emitted at a
small deflection angle
• Depends on atomic structure and photon energy
• Atomic cross section:
Coherent (Rayleigh) scattering
2
R
Z
hσ
ν ∝
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• Scattering with loss of energy: hν’<hν• Photon-electron scattering; electron may be
assumed free (i.e. unbound)
• Thomson scattering: low energy limit, hν→0
Incoherent (Compton) scattering
• Conservation of energy and momentum:
→
Compton scattering – kinematics
2 2 2e
h h ' T
h h ' h 'cos p cos , sin p sin
c c c
( pc ) T 2Tm c
ν νν ν νθ ϕ θ ϕ
= +
= + =
= +
2e
2e
h hh ' , cot 1 tan
h m c 21 (1 cos )m c
ν ν θν ϕν θ
= = + + −
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Compton scattering – kinematics
• An X-ray unit is to be installed, with the beam
direction towards the ground. Employees in the
floor above the unit are worried. Maximum X-ray
energy is 250 keV. What is the maximum energy
of the backscattered photons?
Compton scattering – example
2 2e e
h h180° h '
h 2h1 (1 cos ) 1
m c m c
250h 250 keV h ' 126 keV
2 2501
511
ν νθ ν ν νθ
ν ν
= ⇒ = =+ − +
= ⇒ = =×+
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• Klein and Nishina derived the cross section for
Compton scattering, assuming free electron
• Differential cross section:
Compton scattering – cross section 1
2220rd ' '
sind 2 '
d sin d d
σ ν ν ν θΩ ν ν ν
Ω θ θ φ
= + −
=dΩ
φ
θ
x
y
z
incoming photon along z-axis
r0: classical electron radius
• Cylinder symmetry results in:
• ~ probability of finding a scattered photon in the
interval [θ, θ+dθ]
• Total electronic cross section:
• Atomic cross section: aσ=Zeσ
Compton scattering – cross section 2
22 2
0
d ' 'r sin sin
d '
σ ν ν νπ θ θθ ν ν ν
= + −
22 2
e 0
0
' 'r sin sin d
'
π ν ν νσ π θ θ θν ν ν = + −
∫
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• Scattered photons are more forwardly directed
with increasing photon energy:
Compton scattering – cross section 3
10 keV 200 keV 2 MeV
• Cross section may be modified with respect to
energy:
Compton scattering – cross section 3
2e
22 2 20 e e
2
d d d d d2 sin
d( h ') d d( h ') d d( h ')
hh '
h1 (1 cos )
m c
r m c m cd h ' h h1 1 1
d( h ') ( h ) h h ' h ' h
σ σ Ω σ θπ θν Ω ν Ω ν
νν ν θ
πσ ν ν νν ν ν ν ν ν
= =
=+ −
⇒ = + − + − −
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Compton scattering – cross section 4
Compton edge
Backscatter edge
• Correct atomic cross section:
Compton scattering – cross section 5
Effect of electron
binding energy
Klein-Nishina
Carbon (Z=6)
Copper (Z=29)
Lead (Z=82)
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• The energy transferred to an electron in a
Compton process:
• The cross section for energy transfer:
• Mean energy transferred:
T h h 'ν ν= −
trd d T d h h '
d d h d h
σ σ σ ν νΩ Ω ν Ω ν
−= =
tr
d h h ' dT d d
d h dTd
dd
σ ν ν σΩ Ω σΩ ν Ωσ σ σΩΩ
−
= = =∫ ∫
∫
Compton scattering – transferred energy 1
νh×
• The fraction of incident energy transferred:
Compton scattering – transferred energy 2
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• Photon is absorbered by atom/molecule; the
result is an excitation or ionization
• Atom may deexcite and emit characteristic
radiation:
Photoelectric effect 1
γ
X X+
θ
e-
Ta
• In the kinematics, the binding energy of the
ejected electron should be taken into account:
• Assuming Eb=0, the atomic cross section is:
Photoelectric effect 2
b a bT h E T h Eν ν= − − ≈ −
7 / 222 4 5 2e
e 2e
m cd 2h2 2r Z sin 1 4 cos
d h m c
τ να θ θΩ ν
= +
deW
α: The fine-structure constant
Solid angle eΩ gives the direction of the ejected electron
0
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Photoelectric effect 2
Photoelectric cross section (dσ/dθ)/s
• Energy of characteristic radiation depends on
elektronic structure and transition probabilities
• ”K- and L-shell” vacancies ↔ hνK and hνL
• Isotropic emission
• Fraction of photoelectric interactions:
PK [hν>(Eb)K] and PL [(Eb)L<hν<(Eb)K]
• Probability for emission: YK og YL (flourescence
yield)
• Energy emitted from the atom:
PKYKhνK+(1-PK)PLYLhνL
Characteristic radiation
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• Energy release by ejection of losely bound
electron
• Energy of emitted electron equal to deexcitation
energy
• Low Z: Auger dominates
• High Z: characteristic radiation dominates
Auger effect
• General formula:
• Fraction of energy transferred to photoelectron:
• However: don’t forget Auger electron(s)
• Cross section for energy transfer to
photoelectron:
Photoelectric cross section
n
m
Z , 4<n<5 , 1<m<3
( h )τ
ν∝
bh ET
h h
νν ν
−=
( )K K K K L L Ltr
h P Y h -(1-P )P Y h=
h
ν ν ντ τ
ν−
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• Photon absorption in the nuclear electromagnetic
field where an electron-positron pair is created
• Triplet production: in the electromagnetic field of
an electron
Pair production 1
• Conservation of energy:
• Average kinetic energy after absorption:
• Estimated electron/positron scattering angle:
• Total cross section:
Pair production 2
2eh 2m c T Tν + −= + +
2eh 2m c
T2
ν −=
2em c
Tθ ≈
2 20r Z Pκ α≈
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Discovery of pair production
e-
e+
Magnetic field
• In the electromagnetic field from an electron, an
electron-positron pair is created
• Energy conservation:
• Average kinetic energy:
• Primary electron is also given energy
• Threshold: 4m0c2
Triplet production
2e 1 2h 2m c T T Tν + − −= + + +
2eh 2m c
T3
ν −=
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• Pair production dominates:
Pair- and triplet production
kp
p/
Z2
or
ktp
/Z
[cm
2]
Pair
• Photon (energy above a few MeV) excites a
nucleus
• Proton or neutron is emitted
• (γ, n) interactions may have consequences for
radiation protection
• Example: Tungsten W (γ, n)
Photonuclear reactions
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Summary, interactions 1
Processes
Kinematics
Atomic
cross section
Mean
energy
transferred
Compton Rayleigh Photoel. eff. Pair/triplet
T=0T = hn - Eb- Ta
= hn - Eb
P: hn = 2m0c2+ 2T
T: hn = 2m0c2+ 3T
Photon-
nuclear
reactions
[ ] )2/tan()cm/h(1cot
'hhT
)cos1)(cm/h(1
h'h
2e
2e
θνϕνν
θννν
+=−=
−+=
−+
∝ θνν
νν
νν
Ωσ 2
2
sin'
''Z
d
d
σσν trhT =
m)hv(
Zn
∝τ)h(PZ
)h(PZ
t
2p
νκ
νκ
∝
∝2
R hv
Z
∝σ
LLLKKKK hYP)P1(hYPhT ννν −−−=
3
cm2hT
2
cm2hT
20
t
20
p
−=
−=
ν
ν
Summary, interactions 2
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• nV atoms per volume = ρ(NA/A)
• Number of atoms:
n=nVV=nVΣdx
• Interaction probability
p=nσ/Σ=nVσdx
• Probability per unit length:
µ=p/dx=nVσ=ρ(NA/A)σµ: linear attenuation coefficient
Attenuation coefficients 1
σdx
Σ
• NA : Avogadro’s constant; 6.022 × 1023 mole-1
• A: number of grams per mole
• NA/A: number of atoms per gram
• NAZ/A: number of electrons per gram
• Number of atoms per volume: r(NA/A)
• Etc.
Attenuation coefficients 2
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• Total mass attenuation coeffecient:
• Coefficient for energy transfer:
• Braggs rule for mixture of atoms:
Attenuation coefficients 3
Rσµ τ σ κρ ρ ρ ρ ρ
= + + +
tr T
h
µ µρ ρ ν
=
=
=
= =
∑∑
ni
i i ni 1mix i
ii 1
mf , f
m
µ µρ ρ
Attenuation coefficients 4
hn [MeV]
Ma
ss a
tte
nu
ati
on
co
eff
icie
nt
[cm
2/
g]
hn [MeV]
Oxygen (Z=8) Uranium (Z=92)
Rayleigh
Compton
Photoelectric
Pair/triplet
Total
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Attenuation coefficients 5
X-ray images at different energies
Conventional X-rays (120 kV) Linear accelerator (5 MV)
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• Beam with N photons impinge absorber with
thickness dx:
• Probability for interaction: µdx
• Number of photons interacting: Nµdx
Attenuation 1
dx
N N-dN
−
= ⇒ =
⇒ =
∫ ∫x
0
dNdN N dx dx
N
N N e µ
µ µ
• Note that µ is the interaction probability per unit
lenght – not the absorption probability
• e-µx corresponds to a narrow beam measurement
geometry:
Attenuation 2
Attenuator, thickness dx
Shield
detector
Scattered photons
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• ’Probability’ for photon not interacting: e-µx
• Normalized probability
• Mean free path:
Attenuation 3
∞− −= = ⇒ =∫
!x x
iv iv iv
0
p Ce , p dx 1 , p eµ µµ
∞ ∞−= = =∫ ∫
xiv
0 0
1x xp dx x e dxµµ
µ
ni ni ni
ni
Attenuation 4
Thickness of absorber (cm)
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• ’Probability’ for photon not interacting: ~e-µx
• Normalized probability
• Mean free path:
• A distance of 3 MFP reduces the beam intensity to
5%
Attenuation 4
∞− −= = ⇒ =∫
!x x
iv iv iv
0
p Ce , p dx 1 , p eµ µµ
∞ ∞−= = =∫ ∫
xiv
0 0
1x xp dx x e dxµµ
µ
ni ni ni
ni
Attenuation - example
• 2 MeV photons
Pb: µ = 0,516 cm-1
H2O: µ = 0,049 cm-1
• 10 times as much water necessary
OH
Pb
Pb
OH
xx
2
2
PbPbO2HO2H
x
x
ee
µµ=⇒
= µ−µ−
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Broad beam attenuation
• Broad-beam geometry: every scattered or
secondary uncharged particle strikes the detector
• e-µx : number of primary photons at a given
depth
• What about the scattered photons?
• Monte Carlo simulations
Scattered photons
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e-µx
Primary and scattered photons, 100 keV
e-µx
Primary and scattered photons, 1 MeV