Interaction theory Photons. Eirik Malinen

Transcription

1 Interaction theory Photons Eirik Malinen

2

3 Introduction Interaction theory Dosimetry Radiation source Ionizing radiation Atoms Ionizing radiation Matter - Photons - Charged particles - Neutrons Ionizing radiation Effects Molecules Cells Humans H O DNA

4 Objectives To understand primary and secondary effects of ionizing radiation How radiation doses are calculated and measured To understand the principles of radiation protection, their origin and applications

5 Contents FYSKJM4710 Interactions between ionizing radiation and matter Radioactive and non-radioactive sources Calculations and measurement of absorbed doses (dosimetry) Radiation chemistry Biological effects of ionizing radiation Principles of radiation protection

6 Relevant issues X-ray and CT investigations Radiotherapy Positron emission tomography Radiation protection Radiation Biology

7 X-ray contrast: only a matter of differences in density?

8 Cross section 1 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: electron, nucleus... radius r incoming particle radius r 1 (direction into the paper) s is the total area: π (r + r ) 1

9 Cross section N particles move towards an area S with n atoms Atomic cross section, σ Incoming particle S interaction no interaction Probability of interaction: p = nσ/s Number of interacting particles: Np= Nnσ/S

10 Cross section 3 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

11 Cross section 4 Differential cross section with respect to scattering angle dσ number of particles scattered into dω 1 = dω number of particles per unit area dω

12 Photon interactions Photon represented by a plane wave in quantum mechanical calculations A in (r,t) ~ e i( p in r ω t ) in In principle, two different processes: Absorption r A inn ( Ain( r,t ) r,t ) ψ 0 ψ k * θ r A ut ( r,t ) Scattering r A inn ( r,t ) Ain( r,t ) Scattering: coherent (elastic) og incoherent (inelastic) ψ 0 ψ k * θ r A ut ( r,t ) Aout ( r, t)

13 Coherent (Rayleigh) scattering 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: σ R Z hν

14 Incoherent (Compton) scattering 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

15 Compton scattering kinematics Conservation of energy and momentum: hν = h ν ' + T hν h ν ' h ν ' = cosθ + p cos ϕ, sinθ = p sinϕ c c c ( pc ) = T + Tm c e hν hν θ h ν ' =, cotϕ = 1+ tan 1 + (1 cos ) m c hν θ mec e

16 Compton scattering kinematics

17 Compton scattering example 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 50 kev. What is the maximum energy of the backscattered photons? hν hν θ = 180 h ν ' = = hν hν 1 + (1 cos θ ) 1+ m c m c e e 50 h ν = 50 kev h ν ' = = 16 kev

18 Compton scattering cross section 1 Klein and Nishina derived the cross section for Compton scattering, assuming free electron Differential cross section: r 0 d σ ν ' ν ' ν = + sin θ dω ν ν ν ' dω = sinθ dθ dφ r 0 : classical electron radius z θ dω incoming photon along z-axis φ y x

19 Compton scattering cross section Cylinder symmetry results in: 0 d σ ν ' ν ' ν = π r + sin θ sinθ d θ ν ν ν ' ~ probability of finding a scattered photon in the interval [θ, θ+dθ] Total electronic cross section: π ν ' ν ' ν e = 0 + ν ν ν ' 0 σ π r sin θ sinθdθ Atomic cross section: a σ=z e σ

20 Compton scattering cross section 3 Scattered photons are more forwardly directed with increasing photon energy: 10 kev 00 kev MeV

21 Compton scattering cross section 3 Cross section may be modified with respect to energy: dσ dσ dω dσ dθ = = π sinθ d( h ν ') dω d( h ν ') dω d( h ν ') h ν ' = hν hν 1 + (1 cos θ ) m c e dσ π r m c h ν ' hν hν m c = + + d( h ν ') ( h ν ) hν h ν ' h ν ' hν 0 e e 1 1 1

22 Compton scattering cross section 4 Backscatter edge Compton edge

23 Compton scattering cross section 5 Correct atomic cross section: Effect of electron binding energy Klein-Nishina Carbon (Z=6) Copper (Z=9) Lead (Z=8)

24 Compton scattering transferred energy 1 The energy transferred to an electron in a Compton process: T = hν h ν ' The cross section for energy transfer: dσ tr dσ T dσ hν h ν ' = = dω dω hν dω hν Mean energy transferred: T dσ hν h ν ' dσ T dω dω dω hν dω σ tr dσ dω σ σ dω = = = hν

25 Compton scattering transferred energy The fraction of incident energy transferred:

26 Photoelectric effect 1 Photon is absorbered by atom/molecule; the result is an excitation or ionization θ γ e - X X + T a Atom may deexcite and emit characteristic radiation:

27 Photoelectric effect In the kinematics, the binding energy of the ejected electron should be taken into account: T = hν E T hν E b a b Assuming E b =0, the atomic cross section is: 7 / 4 5 mec h e 0 ν e dτ ν = r α Z sin θ 1+ 4 cosθ d Ω h m c e W α: The fine-structure constant Solid angle e Ω gives the direction of the ejected electron

28 Photoelectric effect Photoelectric cross section (dσ/dθ)/s

29 Characteristic radiation Energy of characteristic radiation depends on elektronic structure and transition probabilities K- and L-shell vacancies hν K and hν L Isotropic emission Fraction of photoelectric interactions: P K [hν>(e b ) K ] and P L [(E b ) L <hν<(e b ) K ] Probability for emission: Y K og Y L (flourescence yield) Energy emitted from the atom: P K Y K hν K +(1-P K )P L Y L hν L

30 Auger effect 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

31 Photoelectric cross section General formula: τ n Z ( h ν ) m, 4<n<5, 1<m<3 Fraction of energy transferred to photoelectron: T = hν hν E b hν However: don t forget Auger electron(s) Cross section for energy transfer to photoelectron: τ = τ tr ( hν P Y hν -(1-P )P Y hν ) K K K K L L L hν

32 Pair production 1 Photon absorption in the nuclear electromagnetic field where an electron-positron pair is created Triplet production: in the electromagnetic field of an electron

33 Pair production Conservation of energy: + hν = mec + T + T Average kinetic energy after absorption: T = hν mec Estimated electron/positron scattering angle: θ m c Total cross section: e T κ αr0 Z P

34 Discovery of pair production e + e - Magnetic field

35 Triplet production In the electromagnetic field from an electron, an electron-positron pair is created Energy conservation: hν = m c + T + T + T + e 1 Average kinetic energy: T hν mec = 3 Primary electron is also given energy Threshold: 4m 0 c

36 Pair- and triplet production Pair production dominates: k pp /Z or k tp /Z [cm ] Pair

37 Photonuclear reactions 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)

38 Summary, interactions 1 Photonnuclear reactions Processes Compton Rayleigh Photoel. eff. Pair/triplet Kinematics hν hν' = 1 + ( hν / m c )( 1 cosθ ) T = hν hν' cotϕ = e [ 1 + ( hν / m c )] tan( θ / ) e T=0 T = hn - E b - T a P: hn = m 0 c + T = hn - E b T: hn = m 0 c + 3T Atomic cross section d σ ν' ν' ν Z + sin θ dω ν ν ν' Z σ R hv Z n τ ( hv ) m κ Z P( hν ) κ Z P( hν ) t p Mean energy transferred σ T = hν tr σ T = hν P Y hν ( 1 P K K K K )P Y hν L L L T T p t hν m0c = hν m0c = 3

39 Attenuation coefficients 1 n V atoms per volume = ρ(n A /A) Number of atoms: n=n V V=n V Sdx Interaction probability p=nσ/s=n V σdx Probability per unit length: µ=p/dx=n V σ=ρ(n A /A)σ µ: linear attenuation coefficient σ S dx

40 Attenuation coefficients N A : Avogadro s constant; mole -1 A: number of grams per mole N A /A: number of atoms per gram N A Z/A: number of electrons per gram Number of atoms per volume: ρ(n A /A) Etc.

41 Attenuation coefficients 3 Total mass attenuation coeffecient: µ τ σ κ σ = R ρ ρ ρ ρ ρ Coefficient for energy transfer: µ tr µ T = ρ ρ hν Braggs rule for mixture of atoms: µ µ n = f i, fi = n ρ mix i= 1 ρ i m i= 1 i m i

42 Attenuation coefficients 4 Mass attenuation coefficient [cm /g] hn [MeV] Oxygen (Z=8) hn [MeV] Uranium (Z=9) Rayleigh Compton Photoelectric Pair/triplet Total

43 Attenuation 1 Beam with N photons impinge absorber with thickness dx: dx N N-dN Probability for interaction: µdx Number of photons interacting: Nµdx dn dn = Nµ dx = µ dx N = x N N0e µ

44 Attenuation Note that µ is the interaction probability per unit lenght not the absorption probability e -µx corresponds to a narrow beam measurement geometry: Shield Scattered photons detector Attenuator, thickness dx

45 Attenuation 3 Probability for photon not interacting: e -µx Normalized probability! µ x µ x iv ni iv iv µ ni ni 0 p = Ce, p dx = 1, p = e Mean free path: = = µ x iv µ = ni 0 0 x xp dx x e dx 1 µ

46 Attenuation 4 Thickness of absorber (cm)

47 Attenuation - example MeV photons Pb: µ = 0,516 cm -1 H O: µ = 0,049 cm -1 e µ HO x x x H Pb HO O = e µ = µ µ 10 times as much water necessary H Pb Pb O x Pb

48 Scattered photons e -µx : number of primary photons at a given depth What about the scattered photons? Monte Carlo simulations

49 Primary and scattered photons, 100 kev e -µx

50 Primary and scattered photons, 1 MeV e -µx