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Interactions with Matter Photons, Electrons and Neutrons

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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” No −µx Nx N(x) = N o e Khan, x Figure 5.1 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. −µx N(x) = N o e −µx I(x) = I o e Attenuation Equation(s) Mass Attenuation Coefficient Linear attenuation coefficient often expressed as the ratio of µ to the density, ρ = mass attenuation coefficient Know these 1 units! 3 2 µ cm 1 = cm = cm cm g g g ρ 3 cm Half Value Layer HVL = Thickness of material that reduces the beam intensity to 50% of initial value. For monoenergetic beam HVL1 = HVL2. I −µ (HVL) 1 = ⇒ e 2 I o Take Ln of both sides 1 Ln2 − µ(HVL) = Ln HVL = Important 2 µ relationship! Half Value Layer Polychromatic Beams After a polychromatic beam traverses the first HVL, it is Khan, Figure 5.3 hardened. low energy photons preferentially absorbed. Beam has higher effective energy after passing through first HVL. More penetrating HVL2>HVL1 Note: Monochromatic beams HVL1=HVL2 Tenth Value Layer - TVL TVL = Thickness of absorber to reduce beam intensity to one tenth of original intensity Ln10 TVL = 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 1 1 N = N N = N o o 2 10 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 TP.E. e- = hν − T P.E. E B e + K L M Photoelectric Effect Cross-sections i.e. probability of an interaction Probability 3 µ α z PE E3 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? 3 µ α z PE E3 Use low Energy Radiation in imaging, so majority of interactions are photoelectric. Radiation is preferentially absorbed in high Z material (bone) achieve good contrast between Bone and soft tissue Photoelectric Effect L-edge K and L edges Note: do not see K and L edge for H 0, occurs 2 at much lower energies A photon with E<B.E.L does not have enough energy to eject L shell electron Low K- edge 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 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 Khan, Figure 5.6 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) 3. A photon with an energy equal 2. L shell e- fills vacancy the difference in the binding excess energy: E=Eb(K) energies is released. – Eb(L) 1. Incident photon ejects K shell electron. e K e + 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) 3. Excess Energy given to 2. L shell e- fills vacancy M shell e-, (auger e-), excess energy: E=Eb(K) which is ejected with – Eb(L) T=Eb(K) – Eb(L) - Eb(m) e 1. Incident photon ejects e k shell electron K e + 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 ' hν hν '= hν 1+ (1− cosφ ) 2 mo c Khan, Figure 5.1 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 hν '= hν 1+ (1− cosφ ) 2 mo c The minimum energy of the scattered photon (max energy of scattered electron) occurs when ϕ=180o (backscattered photon). 1 2 0.511MeV hν '= m = = 0.255MeV 2 0 c 2 Compton Effect The direction of scattered photon depends on the incident photon energy\ Higher Energy is “forward” scattered hν hν '= hν 1+ (1− cosφ ) 2 mo c 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 Electrons per Z • most materials (g/cm3) eff gram1023 (e-/g) except hydrogen have approx. the Fat 0.916 5.92 3.48 same number of electrons per gram. Muscle 1.00 7.42 3.36 Water 1.00 7.42 3.34 N A Z Air 0.001293 7.64 3.01 AW Bone 1.85 13.8 3.0 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)). g g ρx = ×(cm) = 3 2 cm cm For example, in the case of a beam that interacts by Compton effect, the attenuation per g/cm2 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. Electron density = number of electrons per cubic centimeter = density times the number of electrons per gram. N AZ Electron Density (ρ) AW Density Electron per gram Electron density Z (g/cm3) eff 1023 (e-/g) 1023 (e/cm3) Example (ρ ) 5.55 e bone = = Fat 0.916 5.92 3.48 3.19 1.65 (ρe )muscle 3.36 per cm of absorber Muscle 1.00 7.42 3.36 3.36 • attenuation Water 1.00 7.42 3.34 3.34 produced by 1 cm of bone will Air 0.001293 7.64 3.01 0.0039 be equivalent to that produced by 1.65 cm of soft Bone 1.85 13.8 3.0 5.55 tissue. 3. Pair Production Absorption process in which photon disappears and gives rise to an electron/positron pair. e- hν >1.02 MeV e+ e- Nucleus Positron only exists for an occurs in the coulomb instant, combines with force field of the near free e- nucleus Two photons created at annihilation, each with 0.511 MeV and separated by 180o 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 = 2mc2 + KE+ + KE-). Energy Threshold = 1.022MeV Pair Production Kinematics The incident photon must have sufficient Energy 2 + − hν = + + 2mo c T T Average KE of positron/electron hν −1.022MeV T = 2 Average angle of departure of positron/electron 2 Units are in radians mo c θ = to convert to degree multiply by 360o/2p T 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 Khan, Figure 5.11 pair production varies with: Z2 per atom, For a given material, above the threshold energy, the probability of interaction increases as Ln(E).
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