eee4101f / eee4103f radiation interactions & detection - physics · 2015-10-13 · interaction...

Interaction of Radiation with Matter Charged Particles Electrons Gamma Rays Neutrons

EEE4101F / EEE4103F

Radiation Interactions & Detection1. Interaction of Radiation with Matter

Dr. Steve Peterson

5.14 RW JamesDepartment of PhysicsUniversity of Cape [email protected]

March 02, 2015

EEE4101/3F :: Radiation Interactions & Detection 1 / 70 Dr. Steve Peterson


Interaction of Radiation with Matter

Charged Particles

Electrons

Gamma Rays

Neutrons



References

I J. S. Lilley, Nuclear Physics: Principles and Applications, Wiley2001 - [Ch 5]

I W. R. Leo, Techniques for Nuclear and Particle PhysicsExperiments (2nd ed.), Springer 1994 - [Ch 2]

I G. F. Knoll, Radiation Detection and Measurement (4th ed.),Wiley 2010 - [Ch 2]





Nuclear radiation consists of energetic particles or photons. Theirinteraction with matter lies at the heart of nuclear physics and itsapplication in other areas, like detection, power production andmedical radiation therapy.

Radiation can potentially be dangerous,particularly to living tissue, but theseeffects depend greatly on the intensity,energy and type of the radiation as wellas the nature of the absorbing material.



We will look at how three forms of radiation interact with matter:

I Charged Particles

- Protons, alphas, electrons, positrons

I Photons (electromagnetic radiation)

- Gamma, X-rays

I Neutral Particles

- Neutrons, neutrinos

Each of these interact with matter in different ways, but they all seematter in terms of its basic constituents, i.e., as an aggregate ofelectrons and nuclei.



Interaction of Charged Particles with Matter

Two principal features characterize the passage of charged particlesthrough matter:

1. Continuous loss of energy by the particle

2. Deflection of the particle from its incident direction.

Matter consists of atoms - negatively charged electrons bound tothe positively charged nucleus by an electric field

These effects are primarily the result of two processes:

1. Inelastic collisions with the atomic electrons of the material

2. Elastic scattering from nuclei

Of these processes, the inelastic collisions are primarily responsiblefor the energy loss of charged particles in matter.



Interaction of Charged Particles with Matter

Although all charged particles interact with matter in a similarfashion (primarily by electromagnetic forces), we will split them intotwo groups:

I Heavy Charged Particles (protons, muons, pions, alpha)

I Light Charged Particles (electrons, positrons)

Electrons behave considerably different because of their low massand generally higher speeds



Heavy Charged Particle Energy Loss

A nuclei occupies about 10−15 of the volume of the atom, thus it isroughly 1015 times more probable for the particle to collide with anelectron than with a nucleus. Although the inelastic collisions aremore frequent, they do not transfer much energy.

Using conservation of energy and momentum, a head-on collisionbetween an ion (mass M and energy E) and an electron (mass mand initially at rest) gives a loss of kinetic energy of the ion as:

∆E = E(4m

M) (1)

For example, a 5-MeV α particle would lose about 2.7 keV in ahead-on collision with an electron. This is a maximum value, thetypical energy loss would much smaller.




Although only a small amount of energy is transferred (less than afew keV), the large number of collisions per unit length can producea substantial cumulative energy loss, even in thin layers of material.

I For example, A 10 MeV proton will lose all of its energy in only0.25 mm of copper.

In these collisions, energy is transferred from the particle to theatom by causing ionization (i.e. remove an electron) or excitationin the atom. A typical ionization energy is 10 eV, so most collisionswill transfer enough energy to ionize.

The result is a trail of ionization and excitation of atoms andmolecules along the path of the moving particle.



Charged Particle Range

As the particle travels through the material, it will eventually lose allof its energy, reaching a distance called the range of the particle.




The classical average energy loss per unit pathlength, often calledthe stopping power or simply dE/dx is:

−dEdx

=

[e2

4πε0

]22z2

mv2Ne ln

bmaxbmin

(2)

wheree: charge of electron;z: charge of particle;v: speed of particle;m: mass of electron;Ne: electron density of stopping material;bmax and bmin: short and long cutoff distances (impact parameters)



Bethe-Bloch Equation

A more complete energy loss formula due to Bethe and Bloch,correcting for relativistic effects and atomic structure.

−dEdx

= (ze2

4πε0)2

4πZρNA

Amv2[ln(

2mv2

I)− ln(1− β2)− β2] (3)

v = βc: ion velocityze: ion chargem: electron massNA: Avogadro’s number

A: atomic mass numberZ: atomic numberρ: stopping material densityI: mean ionization energy

The parameter I should in principle be computed by averaging overall atomic ionization and excitation processes, but in practice is anempirical constant, roughly equal to (11eV )Z.



Bethe-Bloch Equation

Characteristics of the Bethe-Bloch equation:

I Dependant on Z/A, z2, and ρ

I Logarithmic dependence on E

I Increases in relativistic region

A non-relativistic version is:

−dEdx

= (ze2

4πε0)2

4πZρNA

Amv2ln(

2mv2

I) (4)

Stopping power is typically quoted in units of energy loss per massper unit area, which is obtained by dividing dE/dx by the density ρ,giving dE/ρdx, called the mass stopping power.



Energy Dependence of Stopping Power

A useful approximation (between 100 keV and 1 GeV) is

dE

dx=

constant

Ek(5)

where k ≈ 0.8.

Beyond E/A ≈ 1 GeV per nucleon (v ≈ 0.96c), −dE/dx passesthrough a point of minimum ionization and then begins to slowlyrise with the growing importance of the relativistic correction terms(containing β2).



Energy Dependence of Stopping Power

The energy dependence of the stopping power is dominated by the1/v2 term.



Projectile Dependence of Stopping Power

For particles in the same material, the Bethe-Bloch formula reducesto:

−dEdx

= z2f(v) (6)

where f(v) is a function of the particle velocity only, so the energyloss in a given material is only dependent on the charge and velocityof the particle.



Bragg curve

As shown above, the rate of energy loss (−dE/dx) increases as theparticle energy decreases, so the number of ions produced per unitlength (ionization density) in the medium will also increase. [α in air]



Calculation of Range

The stopping power equation can be used to calculate thepathlength of an ion with energy E, which in turn can approximatethe range R of the ions in a given material.

R =

∫dx =

0∫E

dE

(dE/dx)(7)

Due to the large mass difference between the heavy charged particleand the electrons, any potential deflection from a glancing collisionis effectively zero, producing a nearly straight-line path. As a result,the range and the pathlength of the ions are essentially identical.



Projectile Dependence of Stopping Power

A corresponding expression for range can be found usingdE = mvdv

R =

∫ 0

E

dE

(dE/dx)∝ m

z2F (v) (8)

for a given energy per nucleon (speed), dE/dx and R vary as z2 andm/z2, respectively. For example, a 40-MeV α particle has four timesthe stopping power and the same range as a 10-MeV proton.

A useful empirical relationship for estimating relative ranges of anion in materials with different mass numbers (A) and densities (ρ) isthe Bragg-Kleeman rule:

R1

R2

≈ ρ2√A1

ρ1√A2

(9)



Range Straggling

Due to the statisticalnature of the stoppingprocess, there will be aspread in the observedrange of monoenergeticparticles; this roughlyGaussian phenomenon iscalled range straggling.

It is not a large effect,produces about a 1%range deviation for a5-MeV α particle.



Interaction of electrons with matter

Like heavy charged particles, electrons lose energy to atomicelectrons via electronic interactions. However, due to their smallmass, for a given energy, their speeds are greater. As a result,−dE/dx is much smaller and electrons are more penetrating thanheavy ions.

I For example, the range of a 1-MeV electron in aluminum isabout 1800 µm compared to about 3 µm for an α particle withthe same energy.



Interaction of electrons with matter

Some other consequences of electron’s smaller mass:

1. An electron will lose a much greater fraction of its energy in asingle collision than does a heavy ion.

2. Electrons will suffer large deflections in collisions with otherelectrons, and therefore follow erratic paths. The range (lineardistance of penetration) will be very different from the electronpathlength.

3. Electrons are more likely to experience sudden changes indirection and speed causing the emission of electromagnetic(bremsstrahlung) radiation.



Electron Energy Loss

Electrons lose energy through two processes, due to collisions withelectrons and due to radiated energy losses.

The total energy loss of electrons and positrons, therefore, iscomposed of two parts:(

dE

dx

)tot

=

(dE

dx

)coll

+

(dE

dx

)rad

(10)



Electron Energy Loss

Solid lines:

I Contribution due tocollisions withelectrons

Dashed lines:

I Contribution due toradiation loss(strongly dependenton Z)

Critical energy: wheretwo terms are equal



Electron Range

The significant probability of large-angle scattering slowly removeselectrons from the incident direction, producing a slow decrease inthe intensity. [in contrast to the α particles]

The extrapolated range Re for electrons is determined by extendingthe linear part of the transmission curve.



Example: Stopping Power

Find the approximateenergy loss of 1 MeValpha particles in athickness of 5 µm ofgold.



Interaction of photons with matter

Photons interact with matter quite differently from chargedparticles, primarily due to the lack of electric charge.

The main interactions are:

1. Photoelectric Effect

2. Compton Scattering (including Thomson and RayleighScattering)

3. Pair Production




These reactions explain the two principal qualitative features ofx-rays and γ-rays:

1. X-rays and γ-rays are many times more penetrating in matterthan charged particles.

2. A beam of photons is not degraded in energy as it passesthrough a thickness of matter, only attenuated in intensity.




Why more penetrating?

I Due to the much smaller cross section of the three processesrelative to the inelastic electron collision cross section

Why not energy degraded?

I Due to the fact the three photon processes remove the photonfrom the beam entirely, either by absorption or scattering.

The photons which pass straight through have not suffered anyinteractions at all and retain their original energy.



Photoelectric Effect

The photoelectric effect involves the absorption of a photon by anatomic electron with the subsequent ejection of the electron fromthe atom. The kinetic energy of the outgoing photoelectron is then

T = Eγ −Be (11)

where Be is the binding energy of the electron.

Note: Free electrons cannot absorb a photon and recoil; a heavyatom is necessary to absorb the momentum at little cost in energy.




K-edge: excitation of a tightly-bound electron from the K-shell(only possible above a certain threshold energy value).

The atom which has lost the electron (typically the most tightlybound electrons) may de-excite in one of two ways

1. Emitting Auger electronsI Other, less tightly bound electrons are released from the atom

2. X-ray fluorescenceI Electron from outer shell fill vacancy in inner shell with

emission of a characteristic x-ray



Compton Scattering

Compton scattering is the process by which a photon (Eγ) scattersfrom a nearly free atomic electron, resulting in a less energeticphoton (E ′γ) and a scattered electron carrying the energy (T ) lostby the photon.



Compton Scattering

Using relativistic kinematics, energy conservation gives for thekinetic energy of the electron:

T = Eγ − E ′γ = E −mc2 (12)

where E is the total energy of the recoil electron including its restmass energy mc2.

Momentum conservation requiresthat the momentum be addedvectorially, where pγ = Eγ/c



Compton Scattering

Using the cosine rule, we can write

(pc)2 = E2γ + (E ′γ)

2 − 2EγE′γ cos θ = E2 −m2c4 (13)

using the relation E2 = p2c2 +m2c4

Eliminating E from the equations above, we obtain theCompton-scattering formula for the scattered photon energy:

E ′γ =Eγ

1 + (Eγ/mc2)(1− cos θ)(14)



Compton Scattering

The photon energy varies from the maximum value of Eγ(forθ = 0o) to a minimum value of E ′γ(min) at θ = 180o; E ′γ(min)approaches mc2/2 (≈ 0.25MeV ) when Eγ is large.

The corresponding electron kinetic energy varies from near zero for aglancing collisions to a maximum (always less than Eγ whenθ = 180o.

The probability of Compton scattering is less strongly dependantthan the photoelectroc effect on Eγ and Z.



Pair Production

The process of pair production involves the transformation of aphoton into an electron-positron pair. In order to conservemomentum, this can only occur in the presence of a third body (likewith photoelectric effect), usually a nucleus. Moreover, to create thepair, the photon must have at least an energy of 1.022 MeV.



Pair Production

The total kinetic energy of the electron-positron pair is given by:

T− + T+ = Eγ − 2mc2 (15)

where T− and T+ are the energies of the electron and positron

I Due to the dependence on the presence of a nucleus, there issome Z dependence.

I There is an energy threshold of 2mc2 = 1.022MeV , whichmakes the pair production cross-section unimportant until Eγ isseveral MeV.



Pair Production

Only part of the γ-ray energy is converted into kinetic energy ofcharged particles; the rest goes into the rest masses of the electronand positron.

The positron is an anti-electron, which will slow down, and as itcomes to rest will be attracted to an electron (not the same one),and then annihilate. The annihilation converts the rest masses ofthe electron and positron into two 0.511 MeV γ rays, emitted inopposite directions.




The region of dominance for the three photon interactions



Photon Absorption

The total number of photons is reduced by the number which haveinteracted. The attenuation suffered by a photon beam can beshown, in fact, to be exponential with respect to the thickness, i.e.,

I(x) = Io exp(−µx) (16)

where Io: incident beam intensity; x: thickness of absorber; µ:linear absorption coefficient.



Photon Absorption

The total probability for a photon interaction in matter is the sum ofthe individual photon process cross sections:

σtotal = σphoto + σcompton + σpair (17)

The linear absorption coefficient µ (units: cm−1) is related to crosssection σtotal by:

µ = Nσtotal = σ(Naρ/A) (18)

where N : Nuclear number density, Na: Avogadro’s Number; ρ:density of the material; A: molecular weight or mass number.



Attenuation of gamma rays



Energy deposition of gamma rays



Photon Absorption

The mass attenuation coefficient µm is defined as the linearattenuation coefficient divided by the density of the medium:

µm = µ/ρ (units: cm2/g)

For compounds and mixtures, the total attenuation coefficient maybe calculated using Bragg’s rule,

µ

ρ=∑i

wi(µ

ρ)i (19)



Photon Absorption



Example: Photon Absorption

What thickness of concrete (ρ = 2200 kg m−3) is needed toattenuate a collimated beam of 1-MeV γ rays by a factor of 106?The mass attenuation coefficient of concrete is µm = 0.064 cm2 g−1.



Photon Absorption

We can also calculate the mean distance, λ, traveled by the particlewithout suffering a collision. This is known as the mean free path.Thus,

λ =

∫xP (x)dx∫P (x)dx

=

∫xIo exp(−µx)dx∫Io exp(−µx)dx

=1

µ(20)



Interaction of neutrons with matter

Neutrons interact in matter via nuclear reactions. The type ofnuclear reaction involved depends strongly on:

I The energy of the neutron

I The particular nuclei with which the neutrons collides

The two primary forms of collisions are:

I Scatter - Energy from the neutron is transferred to the recoiling(charged) nucleus

I Absorption - Neutron is captured by the nucleus




Scattering slows the neutron down as energy is lost in successivecollisions. Two possibilities are:

I Elastic scattering from the nuclei, i.e. A(n, n)A. This is theprincipal mechanism of energy loss for neutrons in the MeVregion.

I Inelastic scatter, e.g. A(n, n′)A∗, etc. In this reaction, thenucleus is left in an excited state and may later decay bygamma-ray of some other form of radiative emission. In orderto occur, the neutron must have sufficient energy to excite thenucleus. Below this energy threshold, only elastic scatter mayoccur.




Possible absorption reactions:

I Radiative neutron capture, i.e. n+ (Z,A)→ γ + (Z,A+ 1).In general, the cross-section for neutron capture goesapproximately as 1/v where v is the velocity of the neutron.Absorption is most likely at low energies.

I Other nuclear reactions, such as (n, p), (n, d), (n, α), etc. inwhich the neutron is captured and charged particles are emitted.

I Fission, i.e. (n, f). Most likely at thermal energies.

I High energy hadron shower production - only for very highenergy neutrons (E > 100MeV )




Because of the strong energy dependence of neutron interactions, itis customary to classify neutrons according to their energy:

I High energy neutrons (E > 100MeV )

I Fast neutrons (E ≈ 100keV − 100MeV )

I Epithermal neutrons (E ≈ 0.1eV − 100keV )

I Thermal neutrons (E ≈ 0.025eV )

Thermal neutrons are a unique type of radiation because when it iscaptured, it releases many MeV of energy, making them easilydetectable.




The total probability for a neutron to interact in matter is given bythe sum of the individual cross sections, i.e.

σtot = σelastic + σinelastic + σcapture + ... (21)



Neutron Absorption

When traversing a slab of material, neutrons, like γ rays, can beabsorbed or scattered through large angles, contributing to a loss intransmitted intensity

I = Io exp(−Nσx) (22)

assuming a mono-energetic, well-collimated beam. It can also bewritten as:

I = Io exp(−Σx) = Io exp(−x/λ) (23)

where Σ = Nσ is called the macroscopic total cross section, andλ = 1/Σ is the mean attenuation length.



Neutron Absorption

The attenuation length (λ) can be shown to be equal to the meanfree path (the average distance traveled by a neutron before itinteracts in the medium) and broken into two pieces:

The scattering mean free path λs = 1/Σs

I The average distance between successive scatterings

The absorption mean free path λa = 1/Σa

I The average distance before the neutron is absorbed

Therefore, the total mean free path is given by:

1

λ=

1

λs+

1

λa(24)



Neutron Absorption

The life of a neutron as it travels through material is complex, evenat low energies, depending strongly on the relative magnitudes ofthe scattering and absorption cross-sections.

If σa >> σs, then before any scattering occurs, neutrons areremoved (attenuated) according to equation (22), where σtot ≈ σa.

If σs >> σa, then many scatterings occur with successive energylosses, slowing the neutron to low, possibly thermal energies beforeit is captured. This slowing-down process is called moderation, acritical process in thermal fission reactors.



Neutron moderation

Consider a non-relativistic, elastic collision between a neutron (massm), with energy E0 and speed v0, incident on a target nucleus(mass M) initially at rest.

In the lab frame, the scattered neutron as energy E1 and speed v1.



Neutron moderation

In the c-m frame, the centre of mass is at rest and, therefore, thespeed of the target nucleus w∗ is equal to Vcm, wheremvo = (m+M)Vcm.

Also, since kinetic energy is conserved in an elastic collision, theinitial and final speeds of m and M in the c-m frame (v∗ and w∗)and unchanged by the collision.



Neutron moderation

Using law of cosines,

v21 = (v∗)2 + V 2cm + 2v∗Vcm cos θcm (25)

the lab energy of the scattered neutron is

E1 = E0

(M2 +m2 + 2Mm cos θcm

(M +m)2

)(26)

Using m = 1 and M = A gives

E1 = E0

(A2 + 1 + 2A cos θcm

(A+ 1)2

)(27)



Neutron moderation

The energy of the scattered neutrons is limited to the range

αE0 < E1 < E0, where α =

(A− 1

A+ 1

)2

(28)

In the particular case of scattering protons, A = 1, then the range is0 < E1 < E0.

Intuitively, the lighter the nucleus, the more recoil energy it absorbsfrom the neutron. This implies the the slowing down of neutrons ismost efficient when protons or light nuclei are used, such as water orparaffin (CH2).



Neutron moderation

To determine the effect of scattering on the average neutron energy,we can calculate the probability distribution P (E1)dE1 of thescattered neutron having energy between E1 and E1 + dE1. Fromequation (27), there is a direct correlation between E1 and θcm.Thus,

−P (θcm)dθcm = P (E1)dE1 (29)

where the negative sign allows for the fact that the energy decreasesas the angle increases.

For neutron energy below 15 MeV, the scatter angular distribution isisotropic in the c-m system, giving

P (θcm)dθcm =2π sin θcm

4πdθcm =

1

2sin θcmdθcm (30)



Neutron moderation

From equation (27), we get

dE1

dθcm= −2AE0 sin θ

(A+ 1)2(31)

and

P (E1)dE1 =(A+ 1)2

4AE0

dE1 =dE1

(1− α)E0

(32)

After one scatter, the energy distribution of an originallymono-energetic neutron is constant over the energy range αE0 toE0.



Neutron moderation

The energy distribution of singly scattered neutrons

P (E1)dE1 =(A+ 1)2

4AE0

dE1 (33)



Neutron moderation

After the first scatter, the average neutron energy isE1 = 1

2(1 + α)E0 and the average energy loss per collision is

12(1− α)E0 (independent of initial energy). After a second collision,

each of the neutrons in the distribution loses on average the samefraction of its energy again, and so on. After n collisions, theaverage energy becomes

En = E0 ×(E1

E0

)n(34)

Unfortunately, this misrepresents the actual neutron energydistribution.



Neutron moderation

Energy distribution of neutrons after several elastic scatters



Neutron moderation

Instead of using average energy loss, the distribution is betterrepresented by the logarithmic energy decrement ξ.

ξ = lnE0 − lnE1 =

[E0

E1

]average

(35)

and after n collisions, the average value of lnE0 is given by

lnEn = lnE0 − nξ (36)



Neutron moderation

Using equation (27) and integrating over the solid angle, we get

ξ = 1 +(A− 1)2

2Aln

(A− 1

A+ 1

)(37)

For a neutron to slow down from energy E0 to an energy E ′, thetotal number of collisions n required would be

n =1

ξlnE0

E ′(38)



Neutron moderation

Lilley Table 5.1 Scattering properties of several nuclei

Nucleus α ξ n (to thermalize)1H 0 1.00 182H 0.11 0.725 254He 0.360 0.425 4312C 0.716 0.158 115238U 0.983 0.0084 2200


eee4101f / eee4103f radiation interactions & detection - physics · 2015-10-13 · interaction...

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