Interaction theory –PhotonsInteraction theory –Photons Eirik Malinen. 24.08.2016 2...

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


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:


10 keV 200 keV 2 MeV

• Cross section may be modified with respect to

energy:


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 edge

Backscatter edge

• Correct atomic cross section:


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:


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 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.


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• Total mass attenuation coeffecient:

• Coefficient for energy transfer:

• Braggs rule for mixture of atoms:


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


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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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

Interaction theory –PhotonsInteraction theory –Photons Eirik Malinen. 24.08.2016 2...

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