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12/31/2011 1 It ti f Ch dP ti l Interactions of Charged Particles With Matter Chapter # 4 Mode of Interaction depends: – Type of Radiation – Energy of Radiation T f M il Type of Material
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Page 1: Interactions of Charged Particles With Matter.ppt › PHYS › akmekki › Phys212-Chapter14-K… · Interaction Processes for Photons with Matter • Photoelectric Effect An incident

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I t ti f Ch d P ti lInteractions of Charged Particles With Matter

Chapter # 4

Mode of Interaction depends:

– Type of Radiation

– Energy of Radiation

T f M i l– Type of Material

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General Properties of Radiation

Name Description Charge Rest Mass Range Ionizing g gPower

Nucleus of He Atom

+2e 6.61027 kg Very Short

(<0.1 mm in tissue)Very High

Electron/

Positron-1e/

+1e

9.11031 kg > (Few mm in tissue)

<

Electro 0 0 >> << Electro-magnetic

0 0 >> (Whole Body)

<<

n Nucleon 0 1.671027 kg >> (Several cm in tissue)

<<

Benefits of Radiation Interaction Studies

Nuclear Radiation is invisible to the eye has noNuclear Radiation is invisible to the eye, has no smell or taste. Only through its effects on matter one can

• Identify the type, intensity and energy of the radiation

• Understand radiation hazards to biological tissue• Design suitable Radiation Shielding• Build appropriate Radiation detectors

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Radiation Groups with similar Mode of Interaction:

• Charged Particles– Heavy Particles: p, d, , Heavy Ions

– Light Particles: e, e

• Gamma Rays and X-Raysy y

• Neutrons

Interaction of Heavy Charged Particles with Matter

• Excitation of Atoms of matter

• Ionization of Atoms of Matter

• Bremsstrahlung Interaction with Coulomb Field of Atomic Nuclei of Matter

• Nuclear Reaction with Nuclei of Matter

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Some Important Terms in Charged ParticleInteractions:

• Stopping Power or Specific Energy Loss (S)Stopping Power or Specific Energy Loss (S)Rate of energy Loss per unit length in the medium

S = dE/dX

• Linear Energy Transfer (LET)Mean energy transferred to the medium per unit length

• Specific Ionization (Is)N b f i i d d i h di i l h GNumber of ion pairs produced in the medium per unit length. Greaterfor heavy particles than electrons

• W Value (Eav)Average energy spent to create one ion pair. In air W isabout 35 eV per ion pair

Some Important Terms in Charged ParticleInteractions

• Range (R)Range (R)Distance traveled by a particle before coming to a full stop

Range of a particles in Air: Rair = 0.325 E3/2

(R is in cm and E is in MeV)

Range in other materials: RM = 3.2 104 Rair M1/2 /(RM is range is range in material of density and atomic weight M)

• Relative Stopping Power:Ratio of the range of a particle in air to its range in some other material

Relative stopping power = Rair/RM = 3100 /M1/2

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Interaction of Beta Particles with Matter

• Excitation and IonizationAppreciable at low energy and in light (low Z) materialspp gy g ( )

• Bremsstrahlung RadiationDominant at high energy (> 10 MeV)

Greater in heavy (high Z) materials

• Range of Beta particlesR = 0.412E1.265 – 0.0954lnE for E < 2.5 MeVR = 0.412E 0.106 for E > 2.5 MeV

R(range) in cm, (density) in g/cm3, E(energy) in MeV,

R(equivalent thickness) in g/cm2

Energy Straggling

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Range Straggling

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Example 3

Tritium emits beta- particles with a maximum energy of 18.6 keV.What is the maximum range of these beta rays in air and in water?

SolutionR = 0.412 0 .0186 1.265-0.0954 ln(0.0186) = 0.000586 g/cm2

- At standard ambient temperature and pressure (25 °C and 100 kPa) dryair has a density of ρSATP = 1.168 kg/m3 = 0.001168 g/cm3

Rair = 0.5 cm

- Water = 1 g/cm3

RWater = 0.000586 cm = 5.86 m

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Interactions of Neutrons with Matter

Chapter # 8

General Properties of Neutrons

S b lSymbol = n

Nature = Nucleon

Charge = 0

Rest Mass = 1.6 10–27 kg

Range = Much larger than or particles ( l f i )(can penetrate several cm of tissue)

Ionizing Power = Much less than or particles

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Sources of Neutrons

A Radioisotope SourcesA. Radioisotope Sources

1. (,n) Sources: 241Am-Be, 210Po-Be

2. (,n) Sources: 124Sb-Be, 24Na-Be

3. Spontaneous Fission: 252Cf

Sources of Neutrons

B. Accelerator Sources1. Electron Accelerators: High Energy Bremsstrahlung radiation g gy gfrom these accelerators is used in nuclear reactions to produce neutrons

9Be + = 8Be + n

2. Low Energy Positive Ion Accelerators (~ 300 keV):Deuterons and Tritons from these accelerators are used innuclear reactions to produce neutrons

a. D-T reactions: 2H+3H =4He+n (En = 14 MeV)

b. D-D Reactions: 2H+2H =3He + n (En = 3 MeV)

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Sources of Neutrons

B. Accelerator Sources3. High Energy Positive Ion Accelerators (~ 1-10 MeV): g gy ( )Particles with MeV energies from these accelerators are used in nuclear reactions to produce neutrons

7Li + p = 7Be + n

4. Fission Reactor Sources

Reactors are good sources of high flux neutrons arising from theg g gfission process

235U + n = Fission Fragments + (2-3) neutrons

Average energy of fission neutrons En = 2 MeV

Classification of Neutrons

1 Thermal Neutrons: E = 0 025 eV1. Thermal Neutrons: E 0.025 eV

2. Slow Neutrons: E < 0.5 eV

3. Intermediate Energy Neutrons: 0.5 eV < E < 10 keV

4. Fast Neutrons: E > 10 keV

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Neutron Interaction Cross Section

Defined as the probability of interaction with a materialDefined as the probability of interaction with a material

• Macroscopic Cross Section: Probability of Interaction per unit length in a material:

Unit: cm1

• Microscopic Cross Section: Probability of interaction per nucleus of material

Unit: Barn1 Barn = 1024 cm2

Relation Between Macroscopic and Microscopic Cross Sections

= N N is the number of atoms per cm3 of the material

N = NA/M

= Density of the material (g/cm3) y (g )

M = Molar Mass of the material (g/mole)

NA = Avogadro’s Number = 6.02 1023 (atoms/mole)

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Neutron Interaction Processes

A. Elastic Scattering from Nuclei of Material: (Billard Ball)

Most probable Interaction Process

-Elastic Scattering:

KEn + KEN (before) = KEn + KEN (after)

-Amount of energy lost by a neutron in each elastic collision

E [(1 )/2]EE = [(1-)/2]Eo

Eo = Initial Neutron energy

= [(A-1)/(A+1)]2

A = Atomic mass of the target

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Neutron Interaction Processes

B. Inelastic Scattering

KEn + KEN (before) = KEn + KEN (after)

C. Radiative Capture Process

Neutrons are absorbed giving radiation:1H + n = 2H + (E = 2.52. MeV)

D Charged Particle ReactionsD. Charged Particle Reactions10B + n = 7Li +

E. Fission Reactions235U + n = Fission Fragments + (2-3) n

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Mean Free Path

Average distance that a neutron travelsAverage distance that a neutron travels between collisions.

It is related to the linear absorption coefficient by:

=1/ /

Relaxation Length Rl

Thickness of Material necessary to attenuateThickness of Material necessary to attenuate a neutron beam by 1/e (37%); e =2.72 is the base of natural logarithm.

I(X=Rl) /Io = eRl =1/e = e1

Rl = 1

Rl (cm) = 1/

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Relative Hazards of Charged Particles

• Hazard from external a sources is minimal since a particleshave short ranges. ’s are easily stopped by the skin.Protective eye cover requiredProtective eye cover required.

• Hazard from internal a sources can be significant in somesensitive organs from direct ionization.

• Because of their greater range, external beta particles canpenetrate the skin and deposit their energy in sensitivetissues.

ll b i ifi d b• Internally, beta sources can cause significant damage byionization.

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Relative Hazards Gamma Rays

• -Rays and X-rays from external sources aresignificant hazards because of their long range.

• Because photons penetrate large thicknesses, thedamage to tissue will extend throughout the body.

Relative Hazards of Neutrons

• Neutrons are uncharged, and so they travel long distances,and can be an external hazard.

D iti ti d t t l t• Deeper, sensitive tissues are exposed to external neutronfield

• Damage to tissues is dependent on neutron energy. 50% ofneutron energy is lost in tissues in a single collision withhydrogen nuclei.

• Absorption and inelastic collisions with tissue materialsi i di i hi h d h blgives rise to radiation which compounds the problem

• Neutrons can also produce recoil charged particles in tissuewhich can cause ionization damage

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I i f Ph i hInteraction of Photons with Matter

Chapter # 6

General Properties of Photons

• Category = Gamma rays () and X-raysCategory Gamma rays () and X rays

• Nature = Electromagnetic Radiation

• Charge = 0

• Rest Mass = 0

• Range = Much larger than and particles (can penetrate whole body)

• Ionizing Power = Much less than and particles

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Interaction Processes for Photons with Matter

• Photoelectric EffectAn incident photon spends all its energy to eject an electron p p gy j(photoelectron) from the atom of the material

• Compton ScatteringIncident Photon loses part of its energy in ejecting an electron from the atom, and is itself scattered with a lower energy

• Pair Production ProcessA h i l f h i l di i iA photon passing near a nucleus of the material disappears, giving up all its energy to create an electron-positron pair. A positron is an electron with a positive charge

3

4

E

ZE

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E

Z

)cos1)(511.0/(1

MeVE

EE

EnZ 2

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Gamma Ray

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Total Absorption Coefficient for a Material

= + + total = pe + c + pp

where

pe = photoelectric process

c = Compton Scattering processC p S g p

pp = pair production process

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Attenuation Coefficient for a Mixed Material

= [W + W + W + ]/100 = [W11 + W2 2 + W3 3 + …]/100

where

1, 2, 3 = attenuation coefficient for individual components in the mixture

W W W percents b eight of the constit entsW1, W2, W3 = percents by weight of the constituents

Mean Free Path

Average distance a photon travels between collisions withAverage distance a photon travels between collisions withthe atoms in the absorber. It is related to the linearabsorption coefficient.

=1/

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Relaxation LengthThickness of material necessary to attenuate the photon beam by 1/e(or by 37%)

e is the base of the natural logarithm = 2.72

I = Io ex

I/Io= 1/e = e Rl

Take ln of both sides 1 = Rl

Therefore Rl = 1/

Half Value Layer (HVL)

Thickness of material needed to reduce initial radiation c ess o ate a eeded to educe t a ad at ointensity by half

Putting I(X = HVL) =Io/2 in the attenuation equation I = Io eX

½ = eHVL

Taking ln of both sides0.693 = HVL

HVL = 0.693/

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