Dosimey of indirectly ionizing radiation FYS-KJM 4710 Audun Sanderud
Indirectly ionizing radiation Indirectly ionizing radiation has relative few interactions with matter, but loses a relative large amount of the energy in each interaction Examples: photons, neuons Secondary charged particles (often elecons) will then depose the energy in a relative small distance What is the energy ansferred from interactions between photons and matter in a given volume? Energy accounting important?
Energy ansferred, ε 1) A photon field of total energy R in,u entering a volume and R out,u-rl is the amount of photon energy of interest which leaving the volume: R in,u V R out,u-rl Energy ansferred: ε = R in,u R out,u-rl + Q ε is the total energy ansferred from the photons to the charged particles and the sum of all kinetic energy ansferred to charged particles
Energy ansferred, ε 2) u-rl: uncharged minus radiative loses; if the secondary elecons have lost energy by bremssahlung shall these photons not be included in the Rout,u-rl ε is a stochastic value Q: energy derived from rest mass in V (m E positive, E m negative) Ex: pair-production: Q=-2m e c 2 annihilation: Q=+2m e c 2
KERMA 1) Kinetic Energy Release per Mass: dε K unit [J/kg] dm K is the energy ansferred to charged particles per mass unit in a point of interest Consider a monoenergetic photon field (energy hν) which pass through a thin layer in a object Ψ S: cross-area of photon field Δx
KERMA 2) Probability per length unit of a photon interaction with a given fraction ansferred to elecons is µ Transferred energy to elecons: ε =N(hν)µ Δx Energy fluence of monoenergetic photons: ( ) Ψ= hν Φ= N( hν) S Kerma then: ε ( ) ( ) N hν μ Δx N hν μ Δx K= = = =Ψ m ρv ρsδx ρ μ
KERMA 3) Kerma is the dependent of the energy fluence and massenergy ansfer coefficient For a disibution (a specum) of photons are given: hν max μ ( ) K= Ψ d hν ρ 0 Remember that µ /ρ dependent of atom number and photon energy
Kerma components Kerma includes all kinetic energy given to elecons and the energy will be lost by: Collision Radiation losses Kerma can be divided in components: K = K c + K r K c : collision kerma; will give a value of the energy losses per mass unit from photons which results in collision losses of elecons!
n ε Net energy ansferred, defined from a volume: R in,u V R out,u n ε n ε = R in,u R out,u + ΣQ R out,u is all energy emitted out of the volume as photons (also the bremssahlung) n ε is then the total kinetic energy past to elecons which is not lost in bremssahlung
Defined by: n ε K= d c dm Collision kerma 1) Account from the bremssahlung by inoducing g; the fraction of kinetic energy resulting in bremssahlung: μ K=K1-g=Ψ c ( ) ( 1-g) ρ Defines: μ en ρ μ ( ) = 1-g ρ µ en /ρ: mass energy-absorption coefficient
K c is then given as: μen K=Ψ c ρ Collision kerma 2) Generally: K c < K Special case: Low energetic photons give origin to low energetic secondary elecons in a substance of low Z. Bremssahlung is then nelectable, g=0 and K c = K
Absorbed energy and dose, ε and D Consider all ansport of energy (both charged and uncharged particles) ough a volume: R in,u + R in,c V R out,u + R out,c ε = R in,u + R in,c R out,u R out,c + Q Absorbed dose is defined (and only defined) as: D = dε dm [ ] [ ] unit: Gy = J/kg
Dose The dose is all energy which stays in the volume per mass unit Can not be related to photon coefficients as K and K c But: in some special cases can absorbed dose be approximated with K c
ε, ε, dsa ε: example Two photons interact in volume V ( Q = 0): n hν hν T 1 hν 2 hν T 3 T 2 T 1
Score form photon 1: n ε, ε Dose, ε: example ( ) ( ) ( ) ε = R R = hν hν T = T in,u out,u-rl 1 1 ε = R R = hν hν T hν = T hν n in,u out,u 1 2 1 2 ε = R R +R R in,u out,u in,c out,c = hν + 0 hν T hν T = T hν T 1 2 1 1 2 1 Score form photon 2: ε = hν 0= hν ε = hν 0= hν n ε = hν 0 T T = hν T T 2 3 2 3
Charged Particle Equilibrium (CPE) 1) Photons entering a volume V, which includes a smaller volume v: V Photons V v Charged particle equilibrium: The number of charged particles of given type and energy which entering v is equal the number of particles of same type and energy which leaves the volume v Certain conditions must be present to have CPE the substance most be homogenous and the photon field attenuation most be nelectable
CPE: v CPE 2) When CPE is present: R in,u = R out,u ε = R R +R R = R R = ε n in,u out,u in,c out,c in,u out,u And the dose equals collision kerma: ε ε µ D= = = K c = Ψ m m ρ CPE n en
Dose at CPE The dose at CPE is Ψµ en /ρ, and is then proportional with the interaction probability of the photons Two substances A and B is placed in the same point in a radiation field, will then get doses related to each other by: µ en µ en Ψ ρ ρ A = D B µ en µ en Ψ ρ ρ D A = B A B
CPE, example Two volumes of water and air is positioned in the same point in a radiation field (1 MeV photons) with CPE present. If the dose two the air is 1 Gy what is the dose to water? Use the tabulated values of µ en /ρ (Attix) a get: (µ en /ρ) water = 0.0309 (µ en /ρ) air = 0.0278 D water =D air (µ en /ρ) water /(µ en /ρ) air = 1.11Gy
CPE, problem 1) If the photon energy increase, will the peneation ability of the secondary elecons increase more than that of the photons Attenuation of the photon field ough the range of the secondary elecons: Photon energy (MeV) 0.1 1 10 30 Photon attenuation (%) in water ough the range of secondary elecon 0 1 7 15
CPE, problem 2) γ e - Then is R in,c > R out,c ε = R R + R R > ε D > K n in,u out,u in,c out,c c upseam e - entering have higher energy then e - emitted and leaving; due to the photon attenuation The case of high photon energies
TCPE Transient Charged Particles Equilibrium: elecons from upseam conibute to the dose and the photon conibution (R in,u R out,u ) is the collision kerma Dose proportional with K c : TCPE D = K f 0 TCPE c 1+ f ( ) TCPE