Fundamentals of subatomic physics

I GeneralitiesA Introduction Fundamental question: What is an atom?

Very small sphere (R=10-10 m),

Even smaller nucleus (R= 10-14 à 10-15 m)

+ electrons(green pea seen from a distance of 100m!)

B DefinitionsThe nucleus is heavier.(by far! a proton is 2000 times heavier than an electron…).

Sub-atomic physics: physics of infinitively small.

Can be divided in 2: - Particle physics (high energies) - Atomic physics (Nuclear Physics)

The nucleus is made of 2 kinds of particles: proton and neutronMass at rest ~2000 x that of an electron.1 a.m.u. ≈ 1.66 10-27 kg

A nucleus X of atomic mass A and proton number Z: Z protons (A-Z) neutrons

€

ZA X

Isotopes: same Z, ≠ A (i.e. ≠ neutron number)

€

53123I, 53

125I, 53127I, 53

131I

Isobars: same A (mass), ≠ Z

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53131I, 55

131Cs

Isomers: same Z, same A, ≠ energetic state

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4399Tc, 43

99mTc,

Atom

Nucleus

C Fundamental interactions

In order to be able to explain ALL interactions around us:

4 fundamental interactions: Gravity Strong Electromagnetic Weak

Gravitational Interaction Gravitation, tide, astronomy, …Attractive force on any kind of objects (energies) but with a VERY weak intensity: Visible effect on massive objects only

Electromagnetic interactionAt the birth of all electric and magnetic phenomenon.Attractive or repulsive force, On all objects that have an electric charge.

Allows for atomic cohesion.Responsible for chemical and biological interactions.Can create electromagnetic waves (light, radio waves, radars, X rays…).Allow the explanation of almost all day-to-day phenomenon.

From nucleus discovery: How can the nucleus components stay confined in such a small volume?

Not gravitational… Not electromagnetic...

With those 2 interactions only, the protons (positive charge) would strongly repulse each other, to the point of leaving the nucleus…

Hence another interaction : Strong interactionAllows for nucleus cohesion, ties neutrons and protonsAlso responsible for nuclear reactions(star and therefore sun energy source)

Strong interaction

A 4th interaction became necessary to explain ß radioactivity:Disintegrating nuclei emit one electron that is accompanied by another particle, the anti-neutrino (we’ll see that in more details later on).That particle crosses the entire earth without interacting, whereas the electron is absorbed within some mm in matter…The difference was explained by the hypothesis that the anti-neutrino undergo a much weaker interaction than the electron (or photons).

Weak interaction:

Summary

D Elementary particles

a) Standard modelThe standard model is the actual theory that allows to explain all observable phenomenon at the particle scale.The standard model: takes into account ALL KNOWN particles+ 3 interactions that affect the particles (excludes gravity)

The standard model explains all natural phenomenon but gravity (that still resists to theoreticians)

Elementary particles within the standard model:

b) Elementary particles There are 2 kinds of elementary particles:

♦ ”matter” particles,♦ ”radiation” particles that are the vehicle of fundamental interactions.

♦Matter particles

♦12 in total, split in 2 categoriesFundamental “bricks” of atoms:

♦Quarks (6 plus the corresponding anti-particle: anti-quarks)Sensitive to the 3 kinds of interactions.Compose particle called hadrons.Quarks are never free : they stay within hadrons.

♦leptons (6 plus the corresponding anti-particle: anti-leptons)Not sensitive to strong interaction. There are 3 charged leptons (electron, muon et tau) 3 neutral leptons (neutrinos electronic, muon et tau).And again: 3 charged anti-leptons (electron, muon et tau anti-neutrinos) 3 neutral anti-leptons (electron, muon et tau antineutrinos).

Remark: Two hadron types can be distinguished: Baryons made of 3 quarks: nucleons (protons et neutrons).

and therefore anti-baryons (anti-protons and anti-neutrons) are made of 3 anti-quarks. Mesons made of 2 quarks (one quark and one anti-quark). (There is therefore no "anti-meson")

♦ “Radiation” particles:Analogy:

Action-reaction principle, the 2 ships will separate.The interaction is made by exchanging an intermediary object: the balloon.This object is called interaction interaction vector.

Fundamental interactions are explained in the same way. There’s always one (or many) vector particle(s) for each particle interaction.These vector particle are called “radiation” particles. There are 12 of them for the 3 fundamental interactions (excluding gravity):

♦ For strong interaction: 8 gluons♦ For electromagnetic interaction: photon♦ For weak interaction: W+, W- et Z0.

Interaction Relative intensity

Interaction range

Particles involved

Vector particles

Strong

1 10-15 m Quarks Gluons

Electromagnetic 10-3 Infinite Charged particles

Photons

Weak 10-12 10-18 m Leptons Quarks

W+, W-, Z0

Gravity 10-40 Infinite all Graviton ?

Summary table:

E Conclusion:

The standard model has never been put in default (until now) …

This is NOT the ultimate theory of physics for a main reason: This model cannot accommodate gravity!In particular, the gravity vector particle (the graviton) has never been observed...There are other limitations, (i.e. particle mass prediction).

Physicist big dream is unifying all interactions (Universal model).

For now, this development still resists to experiments (even though electromagnetic and weak interaction could be merged as electroweak interaction).

This opens nice perspectives, Both from experimental and theoretical point of view...

II Mass-energy relationshipA IntroductionSpectacular example in Nuclear Medicine:At the base of PET imaging: two electrons (charged + and -) annihilate to give birth to 2 gamma photons emitted at 180°…

Conversion of 2 material objects (electrons) into energy (photons)…There is a tight relationship between mass (particle) and energy…

B Concepts of relativistic mechanicsUp to the early XXth century:'mobiles' had a small speed vs. / light speed ( c=3.108 m.s-1).Classical mechanics (Newtonian) remained valid.But since… systems, experimental developments, etc.’Classical' physics was questioned!

Example:In classical physics (mechanics second principle), the ratio between particle acceleration and the force leading to that acceleration is a constant:

€

Fa

= m

But for 'relativistic' particles (whose speed is close to c),When studying On one hand the force applied on the particle, On the other hand the acceleration due to the force,One can see that the ratio varies particle speed…And therefore mass varies with speed:

With:m0: rest massν: particle speedc: speed of light in vacuum

In fact, when one is considering kinetic energy variationsFor a particle of rest mass m0 moving in vacuum,Accelerated in straight line by a constant force:

and:

This means => kinetic energy variation is equivalent to a variation in mass...

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m =m0

1− v2

c2

= γ.m0

€

ΔEc = m −m0( )c2

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ΔEc = Δm.c2

RQ: This is valid only if v ~ c…Example: a plane (100 tons) @ mach 1 (1200 km.h-1)

Kinetic energy:

Ec = 5.5 . 109 JThis corresponds to a mass variation of 0.6 pg...

€

Ec =12mv2

C Einstein general scheme, orders of magnitudeEinstein generalised the previous result:To all total energy variation (ET) of a given system corresponds a mass variation:

And here goes the famous E=mc2…

RQ: Mass m is of course given by equation:

This explains why a photon (rest mass equal to zero) still has an associated energy...

€

ΔET = Δm.c2

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m =m0

1− v2

c2

= γ.m0

Orders of magnitude:Mass and energy are equivalent, hence one can consider equally:

Particle Symbol Mass (kg) Rest mass (MeV) Proton p 1.67261 . 10-27 kg 938.28

Neutron n 1.67492 . 10-27 kg 939.57 Electron e 9.10938 . 10-31 kg 0.511

RQ: 1 eV = 1.6.10-19 J 1 a.m.u. = 1.66043 10-27 kg <=> 931.5 MeV

III Strong interaction - binding energyA Nuclei mass and binding energyA stable nucleus: system bound by strong interaction of A nucleons.It is therefore necessary to provide energy to the nucleus to split it into its components.And mass and energy are equivalent concepts.Hence the bound state as a mass inferior to the sum of the component masses.And, the total energy of a nucleus of charge Z and mass number A(with A=Z+N, where N is the neutron number) can be written:

With: mp: proton mass mn: neutron mass B(A,Z): Nucleus binding energy€

M A,Z( )c2 = Zmpc2 + Nmnc

2 − B A,Z( )

Binding energy: that has to be given to split the nucleus into its nucleons, that is released when the nucleus is created.The nuclear binding energies are enormous!

The nuclear binding energies are on the order of a million times greaterthen the electron binding energies of atoms!

B Nuclear energyFrom experience:When the number of nucleons increases, the binding energy increases.But the interesting parameter is the relative variation of binding energy.

It is possible to define binding energy per nucleon, noted B/A.For A > 16, B/A is independent from A and is around 8 MeV.

C N/Z graphThis graph shows how STABLE nuclei are located in a N/Z representation, and also along the N=Z line.

For small Z values, Nuclei that possess the same number of protons and neutrons are stable (along the N = Z line).

However, in order to be STABLE, heavier nuclei need to have a “neutron excess” (N > Z).

It looks as if neutrons where needed to “stabilise” protons in large nuclei...

Z

N

IV Nuclear instability - radioactivityA Definition

Within the incredibly small nucleus, the two strongest forces in nature are pitted against each other. When the balance is broken, the resultant radioactivity yields particles of enormous energy.

Transition : Modification of the energetic status of an atom nucleus by disintegration or de-excitation

Disintegration : Spontaneous transition, with a change in atomic number.The resulting nucleus can be: at the fundamental energy state at an excited energy state(the return to the fundamental energy state: one or several γ transitions).

The available energy corresponds to the difference of atomic mass between parent and daughter nuclei at the rest level.

Activity : Ratio between the mean number (dN) of spontaneous nuclear transitions happening from a given energetic state during time interval dt. Activity is noted A and is given by:

International unit:The becquerel (Bq) = 1 disintegration.s-1 The curie (1 Ci = 37 GBq) is also used

€

A = −dNdt

Radioactive decay : the law of radioactive decay applies to ALL radioisotopes :

N0 is the initial number of nuclei,

dN/dt represents the change of the number of nuclei with time.

Activity is proportional to the number of nuclei at the time t

This introduces a constant λ:λ radioactive constant, characteristic of a given radioisotope.

€

dNdt

= −λN

€

dNN

= −λdt

€

dNN

= − λdt0

t∫N 0

N∫

€

lnN[ ]N 0N = −λ t[ ]0

t

€

ln NN0

= −λt

€

N = N0e−λt

Radioactive half-life : Time T it takes for half of the original number of atoms to decay.Also called physical half-life, noted: Tphy or T1/2.

RQ: This concept is characteristic of exponential processes...

€

N0

2= N0e

−λT

€

e−λT =12

€

−λT = ln 12

€

T =ln2λ

=0.693λ

Radioactive decay

Time

B Application: ray spectraRadioactivity: process allowing the evolution from an initial state to a final state of inferior energy.

Example of spontaneous transitions: γ emissionA nucleus, at an excited state Ei,Emits a photon of frequency νTo reach a final state Ef of lesser energy

Hence the relationship:

With Ef represents energy levels < Ei h Planck constant (h = 6,62 × J.s-1).

€

Eγ = hν ≈ Ei −Ef

(a) (b)

(a) (b)

That energy: can be received by photon detectors Gives birth to one or many rays.Example of Fig. (a), 5 different rays of energies:

(a) (b)

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Eγ = Ei

Eγ = Ei −E2Eγ = Ei −E1Eγ = E2 −E1Eγ = E1

Unstable nuclei emit one or ++ particles to reach a state of increased stability…

C Alpha radioactivityα radioactivity: strong interaction.Concerns mainly heavy nuclei (Z>82).Corresponds to the emission of a helium nucleus (alpha particle ):

€

ZAX→ Z−2

A−4Y (*)+24He+ γ( )

(*) Means that Y can also be instable,(γ) Means that alpha emission can be followed by γ emission.Alpha particle (some MeVs) can be emitted at a speed of ~ 20000 km.s-1…Very ionising! (we’ll see that later), but is quickly stopped:some cm in air, some fraction of a mm in water or soft tissues...

213Bi 209Tl

213Po 209Pb

209Bi

α (2.16%)

β (97.84%) β (100%)α (100%)

β (100%)

D beta radioactivityDisintegration ß-:

Characteristic of nuclei with an excess of neutronsCan be explained by the conversion of a neutron into a proton (within the nucleus)€

ZAX→Z+1

AY (*)+−10e− +ν + γ( )

€

01n→1

1p+−10e− +ν

This involves the anti-neutrino, whose mass is considered as null…

Explanation:Conservation of energy gives the following relation :(with only nuclear masses as a first step) :

With :Ecß-: Kinetic energy of the β - particleEcγ: Kinetic energy of the γ ray (eventually)Ecv: Kinetic energy of the anti-neutrino€

mXc2 = mYc

2 +m0c2 +EC

β−+ECν

+ECγ

Replacing nuclear masses by atomic masses gives:

€

MXc2 − Zm0c

2 −EL = MYc2 − (Z +1)m0c

2 +m0c2 +EC

β−+ECν

+ECγ−EL

'

Hypothesis:♦ Binding energy differences are neglected ,♦ γ ray energy is neglected

The relation thus becomes:

€

MXc2 = MYc

2 +ECβ−

+ECν

And the energy involved is:

€

Ed = MXc2 −MYc

2 = ECβ−

+ECν = ECβ( )max

The neutrino (anti-) is therefore a way to explain why a beta spectrum is observed, with all energies between 0 to Eßmax.

This is a VERY important point to consider in NM and dosimetry!

Disintegration ß+:

€

ZAX→Z−1

AY (*)+10e+ +ν + γ( )

This disintegration is characteristic of nucleus with an excess of protons.It can be considered as the conversion of a proton into a neutron (within the nucleus)

This involves the anti-neutrino, whose mass is considered as null…

Explanation:Conservation of energy gives the following relation :(with only nuclear masses as a first step) :

With :Ecß-: Kinetic energy of the β+ particleEcγ: Kinetic energy of the γ ray (eventually)Ecv: Kinetic energy of the neutrino

€

11p→0

1n+10e+ +ν

€

mXc2 = mYc

2 +m0c2 +EC

β++ECν

+ECγ

Replacing nuclear by atomic masses gives:

€

MXc2 − Zm0c

2 −EL = MYc2 − (Z −1)m0c

2 +m0c2 +EC

β++ECν

+ECγ−EL

'

€

MXc2 = MYc

2 +ECβ+

+ECν+ 2m0c

2

€

Ed = MXc2 −MYc

2 − 2m0c2 = EC

β++ECν

= ECβ+( )

max

Hypothesis:♦ Binding energy differences are neglected ,♦ γ ray energy is neglected

The relation thus becomes:

And the energy involved is:

Remarks :♦ β+ particles (or positrons) are submitted to coulombian repulsion within the nucleus. ♦ β+ decay is only possible if the atomic mass difference between mother and daughter nuclei is more than 2m0c2.

Very famous β + decay:

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918F→ 8

18O+10e+ +ν +γ

E Other means of disintegrationElectron capture“deep layers” electrons (typically K)Have a (small) probability to be found inside the nucleus! The weak interaction is the means to ‘capture’ that electron.The K electron is captured by the nucleus, and a neutrino is emittedHence the equation:

This equation can be explain by the process:

Electron capture is specific to nucleus with a proton excess.Remarks• Electron capture ‘competes’ with β + decay,• The electron vacancy leads to an excited nucleus. This is noted by the ()*.• Electronic capture has a greater probability of occurrence for heavy nuclei (high Z).

€

ZAX+−1

0e(atomic)− → Z−1

AY (*)( )* +ν + γ( )

€

11p+−1

0e(atomic)− →0

1n+ν

As for the preceding disintegrations, it is possible to write:

With:Ecv: neutrino kinetic energyEcγ: γ ray possible energyReplacing nuclear masses by atomic masses gives:

Energetic hypotheses: EL-EL’ is neglected,

And thus:

The disintegration energy is:

€

mXc2 +m0c

2 = mYc2 +ECν

+ECγ

€

MXc2 − Zm0c

2 −EL +m0c2 = MYc

2 − (Z −1)m0c2 +ECν

+ECγ−EL

'

€

MXc2 = MYc

2 +ECγ+ECν

€

Ed = MXc2 −MYc

2 = ECγ+ECν

Important remarks:

Daughter atom has a vacancy in a deep electronic layer.

That will be filled by a more superficial electron (L layer, for example) and will create a characteristic X ray emission.

This is called fluorescence emission X. But then a vacancy will exist for the layer L, etc…This electronic rearrangement will create a lot of X ray emissions.

A Nuclear Medicine example of a radionuclide that decays by electron capture is:

€

53123I + e(atomic)

− → 52123Te( )* +ν +γ

Internal conversion - Auger electronsThis process refers to electrons that can be ejected from the electronic cloud as a consequence of interactions with a photon γ emitted from the nucleus.(The nucleus is therefore originally in an excited state).

Ejected electrons are called internal conversion electrons.These electrons can only be ejected if the γ photon energy (h ν)γis superior to the binding energy of the electron in layer K, L or other (EL)K,L... This is given by the relation:

From a practical point of view, it is impossible to differentiate β - decay electrons from those resulting from internal conversion.They have, however, a characteristic emission energy (ray):€

hν( )γ >> EL( )K ,L ...

€

ECCI= hν( )γ − EL( )K ,L ...

Nombre de β - émis

ECβ−

ECβ −

⎛ ⎝ ⎜

⎞ ⎠ ⎟

max

Raies des électrons de conversion interne

Here again, the electron vacancy will be filled by more superficial electrons. This will create X rays emissions characteristics of the electronic transition.

These fluorescence X rays can leave the atom (i.e. for heavy atoms) or eject superficial electrons (for light atoms) by PE effect (to be seen later).These fluorescence induced electrons are called Auger electrons.And all vacancies created by Auger electrons will in turn be filled, etc...The term “Auger cascade” is used to illustrate this phenomenon.

IC characteristic rays

Electronsemitted

V Conclusion: decay scheme of 99mTcThis illustrate some of the processes seen in the lecture…

ec1 γ1

ec3 γ 3

γ 2

γ 2

γ 3

γ1

ec2

β1−

β2−

β3−

142,681 keV 140,511 keV322,40 keV

89,80 keV

0 keV

293,60 keV

4399mTc

4399Tc4399Tc

4499Ru

6,007 heures

2,14 × 105 ans

Stable

β−

Emission Electronic layer Energy (keV) Absolute yield (%) ec1 M, N 1.73 < E < 2.13 99.11

K 119.467 8.7 L 137.46 1.12 ec2

M, N 139.96 < E < 140.50 0.23 K 121.64 0.61 L 139.64 0.20

IC electrons

ec3 M, N 142.22 < E < 142.64 0.05

ß1- - Max : 204 1.6 × 10-3 ß2- - Max : 293.60 100 ß3- - Max : 436.4 0.001

Decay Energy (keV) Yield (%) 2 (Tc) 140.511 89 3 (Tc) 142.68 0.021 1 (Ru) 89.8 0.7 × 10-5 2 (Ru) 232.7 0.9 × 10-5 3 (Ru) 322.4 0.98 × 10-4

References

- http://hyperphysics.phy-astr.gsu.edu

- http://marwww.in2p3.fr

- http://www.lal.in2p3.fr

- http://www.phys.virginia.edu

- S. Webb et al, The physics of medical imaging, Institute of physics publishing - Bristol and Philadelphia (1993)

- F. Lagoutine, N. Coursol et J. Legrand, Table de radionucléides, CEA/LMRI (1983)

Acknowledgement: T Carlier, NM, Nantes University Hospital

Interactions radiation - matter

Introduction:One can distinguish charged particles interactions with matter X or γ rays interactions with matterPhysics is different, effects are different…

But…X or γ rays interactions with matter can give birth to:

• other electromagnetic radiations…• charged particles (hence the link…)

Furthermore:Nuclear Medicine: Radiopharmaceuticals (radionuclides)

• X or γ rays emissions• Charged particles emissions (ß-, ß+, α)

Consequences on: • Detection• Dosimetry

High energy X and γ rays Interactions with matter

Introduction:In the medical field, electromagnetic radiations: X and γ Radiology, Nuclear Medicine 20 to 500 keV Radiotherapy 1 to 100 MeV

The study of the interactions leads to: Basic rules for radioprotection Physics basis for radiotherapy Physics basis for detectors for medical imaging

Plan of the study: Macroscopic (experimental) aspects µ and HVL Microscopic (atomic) phenomenons PE effect, Compton, pair production Energy ranges Consequences Here it’s the particle side of electromagnetic radiations that’s considered

I Macroscopic effects:

{N0 }N1

Nt

Nd

Détecteur

Na

Np

N0 photons, energy E0=hν, moving in vacuum,Homogeneous medium (a single chemical species),Perfect detector,

N0 = Nt+Na+Ns+Np > Nt+∂Nd = N1

∂ : fraction of scattered photons that are detected

Detector

Ns

A Attenuation coefficient:

Relation between N0 and N1?

N0

0 e x

N(x)N(x+dx)

N(e)

x x+dx

N

N(x) - N(x+dx) = dN = - µ N dx

The number of photons that disappear between x et x+dx is proportional to • The number of photons reaching depth x• The thickness of the layer considered (dx).

µ (proportionality constant) depends upon:• Beam characteristics• Medium crossed

This leads to a first order differential equation:

€

dNN

= −µdx

Ln(N ) = −µx+C

Ln(N )− Ln(N0 ) = Ln( NN0

) = −µx

€

N(x) = N0e−µxTherefore:

N(x) - N(x+dx) = dN = - µ N dx

Consequences:

• If µ is large, the number of interactions increase, The photon number decreases quickly • N(x) is always < N(0) • N(x) is always > 0, one can never stop all photons!

• µ is the linear attenuation coefficient (cm-1)° µ increases with the density of the medium° µ decreases with the energy of the incident photons

• µ can be normalised by ρ : µm is the mass attenuation coefficient (cm2g-1)

€

N(x) = N0e−µx

µm =

µρ

B Half value layer (HVL):

µ doesn’t provide an easy way to define how a given material will absorb a given electromagnetic beam…It is useful to define a variable derived from µ that illustrates the attenuation:The HVL is the thickness of material that attenuates the beam by a factor 2:

€

N(HVL) = N0e−µHVL =

N0

2

€

− ln(2) = −µHVL

€

HVL =Ln(2)

µ=0.693

µHVL is expressed as a length (cm).It is independent of N0.1 HVL decreases the photon number by a factor 22 HVL 43 HVL 810 HVL 1000 (1024)

n HVL 2n

II Photoelectric effect:

Most important effect at low energy, easiest to describe…

A Qualitative descriptionPhoton (E0 = hν) interaction with an electron of inner shellsA photon interacts with a K or L shell electron (most likely).If E0 is sufficient (>Eb): - The photon disappears - The electron is ejected out of the atom with Ek and:

E0 = Ek + Eb

E0

Ek

Eb

More likely to happen if E0 is close to Eb.Probability : null if E0 < Eb

high if E0 > Eb low if E0 ≈ Eb

This is a resonance effect: narrow energy range,Around electron binding energies (eV to 100 keV).

After a PE effect, the photon has disappeared This correspond to an absorption, not an attenuation.

Fate of the electron? interactions with medium Useful in radiotherapy!

Fate of the atom? ion, with a vacancy in inner shells... The electronic cloud will reach a stable ionic form by: Fluorescence photon emission Auger electron emission

Fluorescence photon emission

0 Temps

Ei

E'iEa

E

E0Ec

E1ElEp

atome àl'état basal

effetphotoélectrique

émission du photonde fluorescence

The atom is at an initial stable state (Ea <0)PE effect produces an ion (Ei <0), and Ei = Ea + Eb

If an outer shell electron fills the vacancy: The resulting ion has an energy E’i, Ep is the binding energy of the outer shell electron, E’i = Ea + Ep

Eb

Ek

Fluorescencephoton

PhotoelectricEffect

Stableatom

time

0 Temps

Ei

E'iEa

E

E0Ec

E1ElEp

atome àl'état basal

effetphotoélectrique

émission du photonde fluorescence

Ep is smaller than Eb (but the 2 are positive),The resulting ion has an energy inferior to that of the initial ion: E’i < EiIt is more stable, and the transition is spontaneous.The energy difference induces the emission of a photon of energy E1, and E1 = Eb - Ep Eb is close to E0, Eb decreases very quickly from a shell to another (Bohr) => Eb ≈ E0

Eb

Ek

Fluorescencephoton

PhotoelectricEffect

Stableatom

time

But E1 cannot trigger another PE effect in an atom of the same kind!

The direction of emission of the photoelectron is independent of the incidence direction of the photon.

Usually, the fluorescence photon is neglected,And the incident photon is considered as absorbed...Except in the case of scintillation detectors.

The electronic cloud rearrangement can be made by different steps, ++ fluorescence photons The energy is equivalent to the ≠ between energetic levels.

Auger electron emissionIf the energy released by the electronic cloud rearrangement (migration of an electron from the shell of energy Ep to the shell of energy Eb)Is superior to the binding energy (Eq) of an outer shell electron (likely),This electron is expelled from the electronic structure. It’s an Auger electron.

Its kinetic energy is: E’c = (Eb - Ep) - Eq

0 Temps

Ei

E''iEa

E

E0Ec

E'cElEp+Eq

atome àl'état basal

effetphotoélectrique

émission d'électronAuger

Eb

Ek

E’k

PhotoelectricEffect

Stableatom

Auger electronemission

time

The ion is left at an energetic state:

E’’i = Energy (<0) of the atom that experienced PE effect + Electron binding energies (>0) of missing electrons:

E’’i = Ea + Ep + Eq

Most probable event for light elements (C, N, O)Biologic effect equivalent to that of PE electrons

Importance of PE effect in life sciences:

1) Basis of EBRT with photons 2) Basis of radiologic imaging:The X ray flux is modulated by biologic tissues absorption.RQ : If no other effects, detected photons => absorbed photons => µ 3) Detectors in medical imaging: Based on the PE effect (photon flux on a detector)

B Probability of occurrence:In order to estimate the probability of PE effect, Electromagnetic rays characteristics Absorbing medium characteristics=> Probabilistic description of PE effect

Cross section (microscopic) Absorption coefficient (macroscopic)

direction de propagation du photon

A

B

Particle cross section:

Disk of surface σpe Around each particle, In a plane ⊥ photon propagation.This surface : cross sectionIf the trajectory hits the surface: interactionIf the trajectory doesn’t hit the surface : no interactionThe elementary interaction probability is thus given by the target surface.This model can be used for any kind of interaction: For each interaction type, a relevant surface is defined!

Photon propagation direction

It is possible to show that µpe = σpeP

With P the particle number by volume unit

With: and:

Thus:

One can show that m = 4 and n = 3.

The photoelectric effect is all the more important that :• photon energy decreases,• material density increases• Z is high (heavy elements)

RQ : For energies close to the medium electron binding energies,there can be sharp µ increases (K, L edges).

Relation between cross section and attenuation coefficient:

€

P = ρNZm

A

€

σ pe =εEn

€

µpe =ερNZm

AEn

III Compton effect:

Second important interaction mode for low and medium energy photons.Predominant at medium energies since PE interaction probability decreases sharplyThe incident photon interacts with a free electron of the attenuating medium.Compton Effect: more complex than PE effectPerturbation phenomenon in radioprotection in imaging

Photon incident (E)

Trajectoirede l'électron

percuté

Electronlibre

Photondiffusé

(E')

φ

θ

A Qualitative descriptionThe incident photon (E = hν) hits a free electron.The photon photon energy is imparted: to an electron (kinetic energy) to a secondary (or scattered) photon (θ) same photon? not important...

Incident photon (E)Free

electron

Scatteredphoton

Electrontrajectory

RQ : Free electrons?• Metals• Outer shell electrons, with Eb << E

These electrons can interact by Compton effect

There are relationship between:• Incident photon energy• Scattered photon energy• Deflexion angle

Energy conservation:

E : Incident photon energyE’ : Scattered photon energymc2 : Electron rest energyc : Celerity of light in vacuumUnder the root square : scattered electron energy and momentum after the collision

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E + mc2 = ʹ′ E + p2c2 + m2c4

Conservation of momentum:1) Along the propagation axe of the incident photon:

2) Perpendicular to the propagation axe of the incident photon:

p2 can be calculated through these equations:

Which gives: And if:

We get: This relation binds E and E’ with θ, scattered angle, and E’ ≤ E (normal)

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Ec

=ʹ′ E

ccosθ + pcosφ

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0 =ʹ′ E

csinθ − psinφ

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p2c2 = E − ʹ′ E cosθ( )2 + ʹ′ E sinθ( )2 = E2 + ʹ′ E 2 − 2E ʹ′ E cosθ

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p2c2 = E − ʹ′ E + mc2( )2 −m2c4 = E2 + ʹ′ E 2 − 2E ʹ′ E + 2mc2 E − ʹ′ E ( )

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1− cosθ =mc2

EEʹ′ E −1⎛

⎝ ⎜

⎞ ⎠ ⎟

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α =Emc2

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ʹ′ E E

=ʹ′ ν ν

=λʹ′ λ =

11+α 1− cosθ( )

Importance of Compton effect in life sciences The Compton effect makes everything difficult...

• Explains why one can be irradiated even OUTSIDE the primary beam. (radiation safety consequences for patient and staff)

• Limits radiologic or isotopic imaging quality. blurring effect, parasite...

B Probability of occurrence:Total cross sectionSame principle as for PE effect: one defines σC

Klein et Nishina:

Whereσ0 is the Thompson cross section:r0 = 2.818 10-15 m is the electron radiusAnd fKN is far too complex…

α is the primary photon energy to electron energy at rest ratio.σC variation with E: slower than for PE effect => Compton effect preponderant at medium energies

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σ C =σ 0 fKN (α)

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σ 0 =8πr0

2

3

One finds:

Since Z/A ≈ 0.5 for many chemical species -> µC/ρ almost constant for a given energy.

Differential Cross SectionSince the scattered photon emission is NOT isotropic,It is sometimes useful (imaging) to assess the cross section in a given direction : Differential Cross Section.

Example : Proportion of primaries vs. scattered photons in an image? depends upon patient and detector geometry.

The differential cross section give the probability of photon interaction by CEand the probability that the scattered photon is emitted with an angle θ.The calculation (Klein et Nishina) gives a complex result, The representation can be cartesian or polar:The Compton effect is anisotropic and anisotropy increases with energy.

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µC =σ CρNZA

Relation between cross section and attenuation coefficient:

IV Pair production:

Electron rest energy: 511 keV.An incident photon with energy above 2 x 511=1022 keV can induceone electron (ß-) and one positron (ß+).This phenomenon is called materialisation, pair creation.

Conservation of:• Energy• Momentum• Electric charge

If E > 1022 keV, residual energy -> EkIf Ek low, almost immediate recombinationIf Ek high, the 2 particles can move apart, and the pair production probability increases.

From a practical point of view: • negligible if E < 5 MeV• predominant if E > 100 MeV

Impact restricted to High Energy EBRT

Summary of X or gamma interactions

Charged particles matter interactions

Introduction:Charged particle arriving in the neighbourhood of an atom:

• interactions with the nucleus• interactions electrons

In all cases, the particle looses kinetic energy...(Neutrons are not charged particles and won’t be considered here).

I Interaction with an electron of the target atomParticle-electron interaction: Energy ∆Q transferred to the electron.

According to the ratio of ∆Q and W (electron binding energy):•∆Q > W: target electron is expelled from its orbit.With a kinetic energy (∆Q-W).The target atom is ionised.Expelled electron (called secondary electron) is capable to create ionisationson its track (if its kinetic energy is high enough).

•∆Q < W: Target electron can be taken to a higher energetic state (if ∆Q is sufficient). The atom is excited. The energy imparted is released(as thermal or low energy electromagnetic energy).

•∆Q << W: ∆Q transformed in thermal energy (translation, rotation, vibration).

Phenomenon called collision (improper term).Radiobiologic consequences VERY importantThe incident particle energy loss is (on average) a characteristic of target atoms.

In the case of water: ionisation if ∆Q > 16 eVBut: for a ionisation, there are (on average) 3 excitations and ++ thermal effects => Mean energy per ionisation: 32 eV (34 eV in air).

The LET (Linear Energy Transfer):Quantity of energy transferred to target medium by the incident particle,per unit length of incident particle trajectory.LET is expressed in keV/µm.

The Ionisation Density (ID) is the number of ion pairs created,per unit length of incident particle trajectory.It is expressed in ion pairs/µm.

If W is the mean energy per ionisation, then:

LET = ID x W

RQ: Charged particle beams can be completely stopped...

II Interaction with the nucleus of the target atomIncident particle : according to the charge, attraction or repulsion from the nucleus.Inflexion of the incident particle trajectory.Energy loss is emitted under the form of X rays (Bremsstrahlung).Photon energy < Ek of incident particle (never equal).

Slowing spectrum, continuous, from 0 to Ec.Main medical application: X ray production.

Non-frequent event:Frontal collision between a very energetic particle (α or proton) and the target nucleus:Nuclear reaction (used for radioisotope production).

III Electrons and positronsWe’ll call both e+ and e- electrons…If kinetic energy < 100 MeV, energy losses mainly by collision.(Bremsstrahlung can usually be considered as negligible).

In waterElectron LET is low and decreases with electron kinetic energy.For electron kinetic energies > 1 MeV: LET ~ 0.25 keV/µm, ID ~ 8 ion pairs/µm

Trajectories are sinuous (zigzag).Changes in direction: high energy transfer (collision or bremsstrahlung).Trajectory ends when the electron lost all its Ek.

Path length in water: L (cm) = E (MeV) / 2 (for energies above 1 MeV)

Mean depth of penetration:Always inferior to total particle track length (zigzags).2 electrons that have the same energy may have a different depth penetration…The maximum range can be obtained through tables…

• in air: some m• in water (soft tissues): 1 to 2 cm (and less for most MN ß emitters).

In the case of positrons:When the particle lost all kinetic energy Ek,it interacts with an electron of the medium and annihilates

IV Protons and alpha particlesHeavy particles when compared to electrons:Energy losses have little impact on particle trajectory…For equivalent energies, these particles are MUCH slower than electrons,The LET is much higher:

In water, LET ~150 keV/µm, ID ~ 4500 ion pairs/µm

Straight trajectories

Energy loss by bremsstrahlung is always negligible...

Mean penetration range: ~ particle track length (straight)Varies little for 2 identical particles (with identical initial energy).Order of magnitude: some cm in air,Order of magnitude: some 10 µm in water and soft tissues.-> no risk of external irradiation...

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L(cm) ≈ E(keV )LET (keV /µm)

Summary: