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Home , Charged particle, Elastic collision, Inelastic collision

Interaction of Heavy Charged with Matter BAEN-625 Advances in Food Engineering Heavy charged particles

y Charged particles other than the and Energy-loss mechanisms y A heavy traversing matter loses energy primarily thru the ionization and excitation of ◦ Except in low velocities, it loses a negligible amount of energy in nuclear y The moving particle exerts electromagnetic forces on atomic and impart energy on them y The transferred energy may be sufficient to knock an electron out of an and thus ionize it y Or it may leave the atom in an excited state Heavy charged particle

y Can transfer only small fraction of its energy in a single electronic y Its deflection in the collision is negligible y Thus it travels an almost straight path thru matter, y It loses energy continuously in small amounts thru collisions with atomic electrons Maximum Energy Transfer in a Single Collision y Assume ◦ the particle is moving rapidly compared to the electron ◦ Energy transferred is large compared with the BE () of the electron in the atom ◦ The electron is free and at rest Maximum Energy Transfer in a Single Collision Maximum Energy Transfer in a Single Collision y Since energy and are conserved 1 1 1 MV 2 MV 2 += mv 2 2 2 1 2 1

1 += mvMVMV 1

y Solving for V1: E=MV2/2 − )( VmM V = Initial KE 1 + mM 1 1 4mME Q MV 2 MV 2 =−= max 2 2 1 + mM )( 2 Incident Particle is an Electron

y Its mass is the same as that of the struck particle, M = m 4mME 4MME Q = = max + mM )( 2 + MM )( 2 4 2 EM Q == E max 4M 2 y Entire energy can be transferred in a single, billiard-ball-type collision Energy Transfer in a Single Collision if Incident Particle is an Electron Maximum Energy Transfer in a Single Collision- Relativistic Expression

y An electron is nonrelativistic as long as T is small compared with the rest energy, mc2 = 0.511MeV 2γ mV 22 Q = max + γ + //21 MmMm 22 γ 1/1 −= β 2 β = / cV Qmax in Collision with Electron

Proton Kinetic Qmax Maximum % Energy, E [MeV] Energy Transfer [MeV] 100Qmax/E 0.1 0.00022 0.22 1 0.0022 0.22 100 0.0219 0.22 100 0.229 0.23 1000 3.33 0.33 10000 136 1.4 100000 1060 10.6 1000000 53800 53.8 10000000 921000 92.1 Elastic or Inelastic

y Equations shown before for Qmax are kinematic in nature y They follow from simultaneous conservation of momentum and KE y The assumption made to calculate energy loss was that the struck electron was not bound y Thus the collision being elastic y Charged-particle energy losses to atomic electrons are, in fact, inelastic Elastic collision y Both momentum and are conserved y This implies that there is no dissipative force acting during the collision and that all of the kinetic energy of the objects before the collision is still in the form of kinetic energy afterward y For macroscopic objects which come into contact in a collision, there is always some dissipation and they are never perfectly elastic y In atomic or nuclear scattering, the collisions are typically elastic because the repulsive Coulomb force keeps the particles out of contact with each other. Examples of Elastic Collision y For a head-on collision with a stationary object of equal mass, the projectile will come to rest and the target will move off with equal velocity, like a head-on shot with the cue ball on a pool table. y This may be generalized to say that for a head-on elastic collision of equal masses, the velocities will always exchange. Examples of Elastic Collision y In a head-on elastic collision where the projectile is much more massive than the target, the velocity of the target particle after the collision will be about twice that of the projectile and the projectile velocity will be essentially unchanged. Examples of Elastic Collision y In a head-on elastic collision between a small projectile and a much more massive target, the projectile will bounce back with essentially the same speed and the massive target will be given a very small velocity.

y Perfectly elastic collisions are those in which no kinetic energy is lost in the collision. y Macroscopic collisions are generally inelastic and do not conserve kinetic energy, though of course the total energy is conserved. y The extreme inelastic collision is one in which the colliding objects stick together after the collision Inelastic Collision Single-Collision Energy-Loss Spectra

y Details about charged-particle penetration are embodied in the spectra of single-collision energy losses to atomic electrons y The collisions by which charged particles transfer energy to matter are inelastic y KE is lost in overcoming the BE of the struck electrons Single-Collision Energy-Loss Spectra

In liquid water y The ordinate gives the probability density 0.06

) 50-eV electrons

W(Q) -1 y W(Q)dQ is the 0.04 5-MeV probability that a W(Q) (eV 150-eV electrons given collision will 0.00 result in an energy 0 50 100 Energy Loss Q (eV) loss between Q and Q + dQ Single-Collision Energy-Loss Spectra

y For fast particles (speed > orbital In liquid water speed) ◦ Similarities in the region from 10- 70eV 0.06 y For slow charged particles 50-eV electrons )

◦ The energy-loss spectra differ -1 from one another ◦ The time of interaction is longer 0.04 1-MeV protons

than for fast particles W(Q) (eV ◦ The BE is more important 150-eV electrons

◦ Energy losses are closer to Qmax ◦ Slow particle excites atoms instead of ionizing them 0.00 0 50 100 y A minimum energy Qmin >0 is Energy Loss Q (eV) required for excitation or ionization of an atom Stopping Power

y The average linear rate of energy loss of a heavy particle in a medium [MeV/cm] y Also referred as linear energy transfer (LET) of the particle Stopping Powers y Can be calculated from energy-loss spectra y For a given type of charged particle at a given energy, the SP is given by ◦ The probability μ per unit distance of travel that an electronic collision occurs

◦ The average energy loss per collision, Qmax

Qmax Qavg = )( dQQQW ∫Q min

dE Qmax Qavg ==− μμ )( dQQQW dx ∫Qmin

[1/cm] [MeV] [MeV/cm] Stopping Power-Semi Classical Calculation

V Y ze Coulomb force

zek 2 r F = 0 r 2 b Fy θ X Fx m -e Representation of the sudden Collision of a heavy charged Particle with an electron, Located at the origin XY Stopping Power-Semi Classical Calculation

y The total momentum imparted to the electron is the collision is: ∞ ∞ ∞ 2 cosθ y == cosθ = 0 zekdtFdtFp dt ∞− ∫∫ ∞− ∫ ∞− r 2 t = 0 thetime( heavy particle cross axis)-Y θ = b/rcos ∞ cosθ ∞ b ∞ dt dt = 2 = 2bdt ∞− r 2 0 r 3 ∫∫∫ 0 + tVb )( 3/2222 ∞ ⎡ t ⎤ 2 = 2b⎢ 2/12222 ⎥ = ⎣ + tVbb )( ⎦0 Vb 2 zek 2 p = 0 Vb p2 2 ezk 42 Q == 0 2m V bm 22 Stopping Power-Semi Classical Calculation y In traversing a distance dx in a medium having a uniform density of n electrons per unit volume y The heavy particle encounters 2πnb db dx electrons at impact parameters between b and b + db y The energy lost to these electrons per unit distance traveled is 2πnQb db y The total linear rate energy loss is: 422 Q b dE max 4π 0 nezk max db =− 2π dbQbn = 2 = dx ∫Qmin mV ∫bmin b 422 dE 4π 0 nezk bmax =− 2 ln dx mV bmin Relativistic Stopping Power (Bethe’s Equation)

y The linear rate of energy loss to atomic electrons along the path of a heavy charged particle in a medium is the basic physical quantity that determines the dose that the particle delivers in the medium

422 22 dE 4π 0 nezk ⎡ 2mc β 2 ⎤ =− 22 ⎢ln 2 − β ⎥ dx mc β ⎣ I − β )1( ⎦

−229 k0 ×= 1099.8 CNm z = atomic theofnumber heavy particle; e = magnitude of electron charge; n = ofnumber electrons per volumeunit in the medium m = electron mass;rest c = speed of light; β V/c == speed theof particle torelative c I = mean excitation theofenergy medium Stopping Power y Depends only on the charge ze and velocity β of the heavy particle y The relevant properties of the medium are its mean excitation energy I and the electronic density n y m is the mass of the target atomic electrons y Units: MeV/cm, mass stopping power-[MeV cm2/g] Stopping Power, general

y For any heavy charged particle in any medium

dE ×1009.5 − 231 nz =− β − IF ],ln)([ MeV/cm dx β 2 eV ×101.02 β 26 β = ln)F( − β 2 1− β 2 Mass Stopping Power y Useful quantity because it express the rate of energy loss of the charged particle per g/cm2 of the medium traversed y In –dE/dx depends on pressure, but –dE/ρdx does not y MSP does not differ greatly for materials with similar atomic composition (primarily light elements) 2 y For 10 MeV protons the MSP of H2O is 45.9 MeV cm /g and for 2 2 C14O10 44.2 cm /g, however for Pb(Z=82) the MSP is 17.5 cm /g y Heavy elements are less efficient on a g/cm2 basis for slowing down heavy charged particles (many of their electrons are too tightly bound in the inner shells to participate effectively in the absorption of energy) Mean Excitation Energies

y Can be calculated from SP equation y It is the material parameter describing the ability of the target system to absorb energy y Empirical expressions: ⎧ eV ,0.19 Z = 1 ⎪ I ≅ ⎨ + eV.. ,711211 ZZ ≤≤ 132 ⎪ ⎩ + 71.88.52 eV, ZZ > 13 Mean Excitation Energies y When a material is a compound or mixture, the SP can be calculated by simply adding the separate contributions from the individual components 3 y If there is Ni atoms/cm of an element with atomic number Zi and mean excitation Ii:

ln = ∑ ln IZNIn iii i

Total # of electrons/cm3 in the material

(n=ΣNiZi) Example

y Calculate the mean excitation energy of H2O

H == 0.19),1( eVIZH

IZO O =×+== 10587.112.11),8( eV ZN ×12 ×81 ln I = ii ln I = 0.19ln + = 312.4105ln n i 10 10 = 6.74 eVI Table for Computation of SP’s

y Use of table to facilitate the computation of SP for heavy charged particles

dE ×1009.5 − 231 nz =− β − IF ],ln)([ MeV/cm dx β 2 eV ×101.02 β 26 β = ln)F( − β 2 1− β 2 Table for Computation of SP’s

Proton KE β2 F(β) [MeV] 0.01 0.000021 2.179 0.06 0.000128 4.873 0.1 0.000213 5.161 0.4 0.000852 6.771 1 0.002129 7.685 6 0.01267 9.753 10 0.02099 9.972 60 0.1166 11.96 100 0.1834 12.16 SP of Water for Protons

y Protons, Z = 1 y MW of water = 18.0g/mol y Number of electrons/, n = 10 ele/mol y 1 m3 of water = 106 g y Density of electrons is

106 g n 1002.6 23 ×=×××= 1034.310 m−329 0.18 g SP of Water for Protons

y Ln Iev = 4.3212 y SP of water for a proton of speed β is: dE 170.0 =− []F β − 31.4)( MeVcm−1 dx β 2 69.7)(;00213.0 - Table - MeV 1at 1at MeV - Table - β 2 = F β = 69.7)(;00213.0 dE 170.0 =− []=− 27031.469.7 MeVcm−1 dx 00213.0 SP of Water for Any Particle

y Previous table can be used for any other particle –get F(β) and β2 by using the following relationship:

T M partic = partic Tproton M proton Example y What is the SP for an 10-MeV in water T 4 alpha == 4 Tproton 1

Tproton = MeV = 5.24/10 MeV 5.2 Table from and )F( find )F( and ββ2from Table 5.2 dE/dx- calculate dE/dx- Range y Of a charged particle is the distance it travels before coming to rest y The reciprocal of the stopping power gives the distance traveled per unit energy loss y So, R(T) of a particle of kinetic energy T is:

−1 T ⎛ dE ⎞ TR )( ∫⎜−= ⎟ dE 0 ⎝ dx ⎠ 1 T dE TR )( = 2 ∫ z 0 G β )( Range y For a heavy particle:

M R β = Rp β )()( z 2

The proton range Particle’s velocity Example y Use Table 5.3 to find the range of an 80-MeV 3+ 2He in soft tissue M R β = Rp β )()( z 2 2 Mz == 3,4 3 )( = RR ββ)( 4 p Example

3 )( = RR ββ)( 4 p 80 Proton Energy == 7.26 MeV 3 interpolation Tablein 5.3 −2 Rp = 705.0 cm 3 R β )( == 529.0705.0 gcm−2 4

cm /529.0 ρ tissuesoft Ranges in cm of protons, alpha, and electrons in air at STP

105

104 y For alpha particle at electrons o 103 15 C and 1 atm: R = E,56.0 E < 4 2 protons 10 ER −= ,62.224.1 E << 84

Range in air (cm) Range Alpha

1 particles 10 y R in cm and E in MeV

100

10-1 10-2 10-1 100 101 102 103 Energy (MeV) Slowing-Down Time

y We can use the SP formula to calculate the rate at which a heavy charged particle slows down y The time rate of energy loss, -dE/dt can be expressed in terms of the SP by:

dE ⎛ dE ⎞⎛ dt ⎞ ⎛ dE ⎞ −=− ⎜ ⎟⎜ ⎟ V ⎜−= ⎟ dt ⎝ dx ⎠⎝ dx ⎠ ⎝ dx ⎠ for proton wit = 5.0h MeVT in water dE ×=− 1019.4 11 MeVs−1 dt Slowing-Down Time y A rough estimation can be made of the time it takes a heavy charged particle to stop in matter, if one assumes that the slow-down rate is constant T T τ ≈ = − dtdE − dtdEV )/(/ for proton wit = 5.0h MeVT in water τ ≈ MeV × 11 MeVs−1)1019.4/()5.0( τ ≈ ×102.1 −12 s Limitations of Beth’s Equation y It is valid at high energies as long as γm/M<<1 holds (e.g. up to ~ 106MeV for protons y At higher energies it needs to consider ◦ Forces on the atomic electrons due to the particle’s spin and magnetic moment y It is based on the assumption that the particle moves much faster than atomic electrons ◦ At low energies it fails because the term 2mc2β2/I becomes negative (given a negative value for stopping power) y many more (see text)