Interactions with Matter Photons, Electrons and Neutrons

Interactions with Matter Photons, Electrons and Neutrons Ionizing Interactions Jason Matney, MS, PhD

Interactions of Ionizing Radiation 1. Photon Interactions Indirectly Ionizing 2. Charge Particle Interactions Directly Ionizing Electrons Protons Alpha Particles 3. Neutron Interactions Indirectly Ionizing

PHOTON INTERACTIONS

Attenuation Attenuation When photons are attenuated they are removed from the beam. This can be due to absorption or scatter Linear Attenuation Coefficient = μ Fraction of incident beam removed per unit path-length Units: 1/cm

Measurement of Linear Attenuation Coefficient - Narrow Beam N o N x N ( x) = N e o µ x Khan, Figure 5.1 x Narrow beam of mono-energetic photons directed toward absorbing slab of thickness x. Small detector of size d placed at distance R >> d behind slab directly in beam line. Only photons that transverse slab without interacting are detected.

N ( x) = N e o µ x I ( x) = e o I µ x Attenuation Equation(s)

Mass Attenuation Coefficient Linear attenuation coefficient often expressed as the ratio of µ to the density, ρ = mass attenuation coefficient g g cm g cm cm cm cm 2 3 3 1 1 = = ρ µ Know these units!

Half Value Layer HVL = Thickness of material that reduces the beam intensity to 50% of initial value. For monoenergetic beam HVL 1 = HVL 2. I I µ o = e ( HVL) 1 2 Take Ln of both sides ( ) µ HVL = Ln 1 2 HVL = Ln2 µ Important relationship!

Half Value Layer Polychromatic Beams After a polychromatic beam traverses the first HVL, it is hardened. low energy photons preferentially absorbed. Beam has higher effective energy after passing through first HVL. More penetrating Khan, Figure 5.3 HVL 2 >HVL 1 Note: Monochromatic beams HVL 1 =HVL 2

Tenth Value Layer - TVL TVL = Thickness of absorber to reduce beam intensity to one tenth of original intensity TVL = Ln10 µ TVL = (3.32)HVL (important relationship for board exams) Most shielding calculations and materials are specified in TVLs

More Important Relationships HVL TVL N = N o 1 2 n N = N o 1 10 n n = number of HVL n = number of TVL

Fundamental Photon Interactions 1. Coherent Scatter 2. Photoelectric Effect 3. Compton Scatter 4. Pair Production

1. Photoelectric Effect Photon interacts with a tightly bound orbital electron (K,L,M) and transfers ALL of its energy to the electron. The electron is ejected from the atom with Kinetic Energy T P.E. e - T = hν P. E. E B e L + K M

Photoelectric Effect Cross-sections i.e. probability of an interaction Probability z 3 µ α PE E 3 Lower Energy P.E more likely P.E interactions are less likely at higher energy Higher Atomic Number: P.E. more likely

Photoelectric effect How is the interaction probability manipulated to achieve good contrast in diagnostic imaging? µ Use low Energy Radiation in imaging, so majority of interactions are photoelectric. α z PE 3 E 3 Radiation is preferentially absorbed in high Z material (bone) achieve good contrast between Bone and soft tissue

L-edge Note: do not see K and L edge for H 2 0, occurs at much lower energies K- edge Photoelectric Effect K and L edges A photon with E<B.E. L does not have enough energy to eject L shell electron Low probability of L shell P.E interaction Dip in curve When E= B.E. L very high probability of L shell P.E interaction Spike in curve Khan, Figure 5.6 As energy increases E> B.E. L probability of L shell P.E decreases Dip in curve When E= B.E. K very high probability of K shell P.E interaction Spike in curve

Photoelectric Effect Results The fast moving photoelectron may participate in 1000s of interactions until it dissipates all of its energy. Other Results Characteristic X-rays Auger Electrons Khan, Figure 5.5

Characteristic X-Ray Production Example (K-shell vacancy) 2. L shell e - fills vacancy excess energy: E=E b(k) E b(l) 3. A photon with an energy equal the difference in the binding energies is released. 1. Incident photon ejects K shell electron. e e + K L M e

Aujer Electrons When an electron displaces inner shell electron an outer shell electron fills the vacancy and rather than giving up the excess energy as characteristic X-Ray, the excess energy is given to a different outer shell electron, which is ejected.

Aujer Electron Production Example (K-shell vacancy) 2. L shell e - fills vacancy excess energy: E=E b(k) E b(l) e 3. Excess Energy given to M shell e-, (auger e-), which is ejected with T=E b(k) E b(l) - E b(m) 1. Incident photon ejects k shell electron e e + K L M e

1b. Coherent Scatter (Low Energy) Coherent scatter occurs when the interacting photon does not have enough energy to liberate the electron. Energy photon < binding energy of electron Photon energy is re-emitted by excited electron. The only change is a change of direction (scatter) of the photon, hence 'unmodified' scatter. Coherent scattering is not a major interaction process encountered in at the energies normally used in radiotherapy

2. The Compton Effect (E>100 KeV) A photon with energy, E=hv, incident on unbound stationary FREE electron (for purposes of easier calculation). The electron is scattered at an angle θ with energy T and the scattered photon with E=hν departs at angle f with energy, hv. T = hν hν ' Khan, Figure 5.1 hν ' = 1+ hν hν moc ( 1 cosφ ) 2

Compton Effect The incident photon can never transfer ALL of it s energy to the electron, but it can transfer most of its energy. hν ' = 1+ hν hν moc ( 1 cosφ ) 2 The minimum energy of the scattered photon (max energy of scattered electron) occurs when ϕ=180 o (backscattered photon). 1 MeV h m c 2 0.511 ν ' = 0 = = 0. 255MeV 2 2

Compton Effect The direction of scattered photon depends on the incident photon energy\ Higher Energy is forward scattered hν ' = 1+ hν hν moc ( 1 cosφ ) 2

Compton Probability of an Interaction Compton effect is independent of Z Compton effect does depend on e- density Let s consider these statements in more detail..

Compton Probability of an Interaction Because the Compton interaction involves essentially free electrons in the absorbing material, it is independent of atomic number Z. It follows that the Compton mass attenuation coefficient (σ/r) is independent of Z and depends only on the number of electrons per gram. Although the number of electrons per gram of elements decreases slowly but systemically with atomic number, most materials except hydrogen can be considered as having approximately the same number of electrons per gram.

Electrons per Gram Density (g/cm3) Z eff Electrons per gram10 23 (e - /g) Fat 0.916 5.92 3.48 Muscle 1.00 7.42 3.36 Water 1.00 7.42 3.34 Air 0.001293 7.64 3.01 Bone 1.85 13.8 3.0 most materials except hydrogen have approx. the same number of electrons per gram. N A A W Z

Compton Scatter Interactions If the energy of the beam is in the region where the Compton effect is the most common mode of interaction (i.e. megavoltage therapy beams) get same attenuation in any material of equal density thickness (density (ρ) times thicknes (x)). ρ = g cm ( cm) x = 3 2 For example, in the case of a beam that interacts by Compton effect, the attenuation per g/cm 2 for bone is nearly the same as that for soft tissue. However, 1 cm of bone will attenuate more than 1 cm of soft tissue, because bone has a higher electron density. cm Electron density = number of electrons per cubic centimeter = density times the number of electrons per gram. g

Electron Density N A A W Z ( ρ) Density (g/cm3) Z eff Electron per gram 10 23 (e - /g) Electron density 10 23 (e/cm 3 ) Fat 0.916 5.92 3.48 3.19 Muscle 1.00 7.42 3.36 3.36 Water 1.00 7.42 3.34 3.34 Air 0.001293 7.64 3.01 0.0039 Bone 1.85 13.8 3.0 5.55 ( ρe ) ( ρ ) e Example bone muscle = 5.55 3.36 = 1.65 per cm of absorber attenuation produced by 1 cm of bone will be equivalent to that produced by 1.65 cm of soft tissue.

3. Pair Production Absorption process in which photon disappears and gives rise to an electron/positron pair. h ν >1.02 MeV e- Nucleus occurs in the coulomb force field of the near nucleus e + e - Positron only exists for an instant, combines with free e- Two photons created at annihilation, each with 0.511 MeV and separated by 180 o

Pair Production: Threshold Energy The incident photon must have sufficient energy to create a positron and an and electron (need rest mass of each, 0.511 MeV), any extra energy is kinetic energy for the positron and electron (hν = 2mc 2 + KE + + KE - ). Energy Threshold = 1.022MeV

Pair Production Kinematics The incident photon must have sufficient Energy hν = 2moc 2 + T + + T Average KE of positron/electron T h = ν 1.022MeV 2 Average angle of departure of positron/electron = θ moc 2 T Units are in radians to convert to degree multiply by 360 o /2p

Pair Production X-Sections Probability of an interaction Because the pair production results from an interaction with the electromagnetic field of the nucleus, the probability of this process increases rapidly with atomic number. The attenuation coefficient for pair production varies with: Z 2 per atom, For a given material, above the threshold energy, the probability of interaction increases as Ln(E). Khan, Figure 5.11

Probability of Interaction Three photon interactions dominate at the energies we use in radiotherapy Energy Increases Z Increases Photoelectric Effect ( 1 EE 3 ) (Z 3 ) Compton Scatter ( 1 EE) No change Pair Production (E > 1.02 MeV) Z

Let s Test our knowledge

Photon Interactions Which of the following is FALSE? A photon can undergo a followed by a interaction. 1. Compton, Pair Production 2. Compton, another Compton 3. Compton, photoelectric 4. Photoelectric, Compton No photon remains after photoelectric interaction! KEY: So long as incident photon has sufficient energy following first interaction, can undergo another interaction

Photoelectric Effect Which of the following statements regarding Photoelectric Interactions is FALSE? 1. They are mainly responsible for differential attenuation in radiographs 2. The incident photon is absorbed 3. Bound electrons are involved 4. The probability increases rapidly with increasing energy

Photoelectric Effect Two materials are irradiated by the same energy photons. Material A has an atomic number of 14 and B has an atomic number of 7. The photoelectric interaction cross-section (probability) of A is times that of B? µ α Z E 3 PE 3 A B = ( 14) ( 7) 3 3, A = 8B Using above equation, set up a ratio and solve (energy cancels out).

Photoelectric Effect A photon detected following a photoelectric interaction is most likely to be: 1. The scattered incident photon 2. A gamma ray 3. An annihilation photon 4. Crenkov Radiation 5. A characteristic X-Ray

Trivia: Cherenkov Radiation Cherenkov radiation is EM radiationemitted when a charged particle (such as an electron) passes through an insulator at a constant speed greater than the speed of light in that medium. The characteristic blue glow of nuclear reactors is due to Cherenkov radiation. It is named after Russian scientist Pavel Cherenkov, the 1958 Nobel Prize winner who was the first to characterize it rigorously. http://en.wikipedia.org/wiki/cherenkov_radiation

Photoelectric, Compton Scatter & Pair Production What is the overall effect of Photon Beam?

Total Attenuation Coefficient Mass attenuation coefficient for photons of a given, relevant energy in a given material composed of the individual contributions from the physical processes that can remove photons from the narrow beam. µ = ρ µ ρ PE µ + ρ CS µ + ρ PP

Photon Interactions in Water as a function of energy Some important Energies: 26 kev, 150 kev, and 24 MeV Nomenclature: T = Photoelectric, s = Compton, P = Pair Production PE most likely at Low E 1:1 PE:Compton All Compton 1:1 Compton:PP PE most likely at very High E

Photon Interactions (m/r) Two materials: H 2 0 and Pb Photoelectric effect Higher for Pb than H 2 0. PE Pair Production Higher for Pb than H 2 0. Compton effect Similar for both materials. Compton Pair Production Khan, figure 5.12

Total Attenuation Coefficient Rule of Thumb: Compton scattering interaction dominates from ~25 kev to ~25 MeV in water

ELECTRON INTERACTIONS

Electron Interactions As electrons travel through a medium they interact with atoms through a variety of processes due to COULOMB interactions. Directly ionizing Through these collisions the electron may Lose kinetic energy (collision and radiative loss) Change direction (scatter)

Electron Interactions Two Types of Electron Interactions Collisional Interactions: Incident electron interacts with atomic electrons in the absorbing medium Radiative Interactions: Incident electron interacts with atomic nuclei in the absorbing medium. e- e-

Electron Interactions e - e - passes atom at some distance, b b a=radius of atom a Distance of electron from atom in relation to size of the atom will determine the type of e- interaction Three Possibilities 1. b>>a 2. b=a 3. b<<a Collision Interactions Radiative Interactions

Collision Interactions Two types of collision interactions: 1. Hard collisions 2. Soft Collisions

Collision Interactions Soft Collision (b>>a) Soft Collision: when an e - (primary e - ) passes an atom at a considerable distance, the e - s coulomb force affects the atom as a whole. Result excite and sometimes ionize valence electrons. Energy of transition Transfer a small amount of energy (few ev) to atom of abs medium. Probability Very Likely, Most numerous type of collision interaction. Net Effect accounts for about ½ of total energy transferred to absorbing medium from collision interactions.

Collision Interactions Hard Collision (b=a) Hard Collision (knock-on) b=a, incident e - (primary e - ) interacts with atomic e -. Result d-ray = atomic electron ejected with considerable kinetic energy (deposits its energy along separate track from 1 o e - ). Energy of transition considerable amount of energy to atom of abs medium. Probability Less Likely. Net effect accounts for about ½ of total energy transferred to absorbing medium from coulomb interactions.

Radiative Interaction b<<a e - passes in close proximity to nucleus and interacts with coulomb force of nucleus. Two types of radiative interactions possible: 1. Elastic Interaction No Energy change 2. Inelastic Bremsstrahlung photons produced

Elastic Radiative Interaction e- is scattered with NO change in energy No energy Transferred to absorbing Medium. Probability increases with Z 2, (Z = atomic number of medium). Thin High Z foil can be used to elastically scatter electrons (e.g. scattering foil). Note: foil must be thin or too much Bremsstrahlung production. Accounts for 97 98% of radiative interactions

Inelastic Radiative Interaction e- interacts with coulomb field of nucleus: Bremsstrahlung Radiation Energy carried away via photon emission Accounts for 2-3% of radiative interactions (for electrons).

NEUTRON INTERACTIONS

Why do we care about neutrons? Neutrons in Radiation Therapy Neutron Therapy (very few centers exist) Contamination Neutrons in X-Ray Therapy Contamination Neutrons in Proton Therapy Neutron Shielding

Neutron Interactions with Tissue Neutrons are indirectly ionizing Neutrons interact by setting charged particles in motion i.e. give rise to densely ionizing (high LET) particles: recoil protons, a-particles, and heavier nuclear fragments. These particles then deposit dose in tissue. The type of interaction and the amount of dose deposited in the body is strongly dependent on neutron energy.

Exponential Attenuation similar concept to photon attenuation Neutrons are removed exponentially from a collimated neutron beam by absorbing material. I o I I = I o e Σ total where N = number of absorber atoms per cm 3 (atomic density) s = the microscopic cross section for the absorber, cm 2 t = the absorber thickness, cm t s total N

Classification of Neutrons by Energy Category Fast Intermediate Epithermal Thermal Energy Range > 500 kev 10 kev 500 kev 0.5 ev 10 kev < 0.5 ev The classification of neutrons by energy is somewhat dependent on the reference text. Some sources may include an epithermal category while others only include fast and slow (thermal).

Fast Neutron Interactions in Tissue Higher energy neutrons result in the release of charged particles, spallation products, from nuclear disintegrations, [(n, D), (n,t), (n,alpha), etc]. These charged particles then deliver dose to tissue. Examples of (n,alpha) Oxygen Carbon

Fast Neutron Interactions with Oxygen and Carbon Recoil alpha-particles A neutron interacts with a Carbon atom (6 protons and 6 neutrons), resulting in three a-particles. (Hall, Fig 1.10) A neutron interacts with an Oxygen atom (8 protons and 8 neutrons), resulting in four a-particles. (Hall, Fig 1.10)

Intermediate Neutron Interactions in Tissue Interact primarily with Hydrogen in Soft Tissue For intermediate energy neutrons, the interaction between neutrons and hydrogen nuclei is the dominant process of energy transfer in soft tissues. 3 Reasons Hall, Figure 1.9 1.Hydrogen is the most abundant atom in tissue. 2.A proton and a neutron have similar mass, 938 MeV/cm 2 versus 940 MeV/cm 2. 3.Hydrogen has a large collision cross-section for neutrons.

Thermal Neutron Interactions in Tissue The major component of dose from thermal neutrons is a consequence of the 14 N(n,p) into 14 C + 0.62 MeV photon Dominant energy transfer mechanism in thermal and epithermal region in body Another thermal neutron interaction of some consequence is the 1 H(n,γ) into 2 H + 2.2 MeV photon 2.2 MeV to gamma (non-local absorption) Small amount of energy to deuterium recoil (local absorption) *This 2.2 MeV photon interaction is important in shielding high energy accelerators*

References The Physics of Radiation Therapy 3 rd edition, Faiz Khan, Ch 5 Herman Cember. Introduction to Health Physics 3 rd Ed. (1996) Eric J. Hall. Radiobiology for the Radiologist 5 th Ed. (2000) Frank H. Attix. Introduction to Radiological Physics and Radiation Dosimetry. (1986)