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- 1. Fundamentals of subatomic physicsI 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 froma distance of 100m!)
- 2. 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 AA nucleus X of atomic mass A and proton number Z: Z protons (A-Z) neutrons Z X €
- 3. Isotopes: same Z, ≠ A (i.e. ≠ neutron number) 123 53 I, 125I, 127I, 131I 53 53 53€
- 4. Isobars: same A (mass), ≠ Z 131 53 I, 131Cs 55€
- 5. Isomers: same Z, same A, ≠ energetic state 99 99m 43Tc, Tc, 43€
- 6. AtomNucleus
- 7. C Fundamental interactionsIn order to be able to explain ALL interactions around us: 4 fundamental interactions: Gravity Strong Electromagnetic Weak
- 8. 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
- 9. 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.
- 10. Strong interactionFrom 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 stronglyrepulse each other, to the point of leaving the nucleus… Hence another interaction : Strong interaction Allows for nucleus cohesion, ties neutrons and protons Also responsible for nuclear reactions (star and therefore sun energy source)
- 11. Weak 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 isabsorbed within some mm in matter…The difference was explained by the hypothesis that the anti-neutrino undergo amuch weaker interaction than the electron (or photons).
- 12. Summary
- 13. D Elementary particles a) Standard modelThe standard model is the actual theory that allows to explain all observablephenomenon 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 resiststo 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.
- 14. ♦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")
- 15. ♦ “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.
- 16. Summary table: Interaction Relative Interaction Particles Vector intensity range involved particles Strong 1 10-15 m Quarks Gluons Electromagnetic 10-3 Infinite Charged Photons particles Weak 10-12 10-18 m Leptons W+, W-, Z0 Quarks Gravity 10-40 Infinite all Graviton ?
- 17. 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 electromagneticand weak interaction could be merged as electroweak interaction).This opens nice perspectives, Both from experimental and theoretical point of view...
- 18. II Mass-energy relationshipA IntroductionSpectacular example in Nuclear Medicine:At the base of PET imaging: two electrons (charged + and -) annihilate to give birthto 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 particleacceleration and the force leading to that acceleration is a constant: F =m a €
- 19. 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: m0m0: rest mass m= = γ .m0ν: particle speed v2 1− 2c: speed of light in vacuum cIn fact, when one is considering kinetic energy variations €For a particle of rest mass m0 moving in vacuum,Accelerated in straight line by a constant force: ΔEc = ( m − m0 )c 2 ΔEc = Δm.c 2 and:This means => kinetic energy variation is equivalent to a variation in mass... € €
- 20. RQ: This is valid only if v ~ c…Example: a plane (100 tons) @ mach 1 (1200 km.h-1) 1Kinetic energy: Ec = mv 2 2Ec = 5.5 . 109 JThis corresponds to a mass variation of 0.6 pg... €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: ΔET = Δm.c 2And here goes the famous E=mc2… m0 m= = γ .m0RQ: Mass m is of course given by equation: v2 1− 2 € cThis explains why a photon (rest mass equal to zero) still has an associated energy... €
- 21. 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
- 22. 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: M ( A, Z )c 2 = Zm p c 2 + Nmn c 2 − B( A, Z )With: mp: proton mass mn: neutron mass B(A,Z): Nucleus binding energy €
- 23. 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!
- 24. 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.
- 25. C N/Z graphThis graph shows how STABLE nucleiare located in a N/Z representation, andalso along the N=Z line.For small Z values,Nuclei that possess the same number ofprotons and neutrons are stable (alongthe N = Z line). NHowever, in order to be STABLE,heavier nuclei need to have a “neutronexcess” (N > Z).It looks as if neutrons where needed to“stabilise” protons in large nuclei... Z
- 26. IV Nuclear instability - radioactivityA DefinitionWithin the incredibly small nucleus, the two strongest forces in natureare pitted against each other. When the balance is broken, the resultant radioactivity yields particles of enormous energy.
- 27. Transition : Modification of the energetic status of an atom nucleusby disintegration or de-excitationDisintegration : 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 parentand daughter nuclei at the rest level.Activity : Ratio between the mean number (dN) of spontaneous nuclear transitionshappening from a given energetic state during time interval dt. Activity is noted Aand is given by:International unit: dNThe becquerel (Bq) = 1 disintegration.s-1 A=−The curie (1 Ci = 37 GBq) is also used dt €
- 28. Radioactive decay : the law of radioactive decay applies to ALL radioisotopes :N0 is the initial number of nuclei, dN = −λN dtdN/dt represents the change of the number of nuclei with time. dN = −λdtActivity is proportional to the number of nuclei at the€ t time NThis introduces a constant λ: N dN t €λ radioactive constant, characteristic of a given radioisotope. ∫ N0 N = − ∫ λdt 0 N t [ln N ] N = −λ [t ]0 0 € N ln = −λt N0 € N = N 0 e− λt € €
- 29. 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 Radioactive decay = N 0 e− λT 2 1 e− λT = 2€ 1 −λT = ln 2€ ln 2 0.693 T= = Time λ λ€€
- 30. B Application: ray spectraRadioactivity: process allowing the evolution from an initial state to a final stateof 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 energyHence the relationship: Eγ = hν ≈ Ei − E fWith Ef represents energy levels < Eih Planck constant (h = 6,62 × J.s-1). € (a) (b)
- 31. (a) (b) (a) (b)That energy: can be received by photon detectorsGives birth to one or many rays. Eγ = EiExample of Fig. (a), 5 different rays of energies: Eγ = Ei − E2 Eγ = Ei − E1Unstable nuclei emit one or ++ particles Eγ = E2 − E1to reach a state of increased stability… Eγ = E1 €
- 32. C Alpha radioactivity α radioactivity: strong interaction. Concerns mainly heavy nuclei (Z>82). Corresponds to the emission of a helium nucleus (alpha particle ): A Z X→ A−4 Y (*) +2 He + (γ ) Z −2 4(*) 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 α (2.16%) 209Tl β (97.84%) β (100%) 213Po α (100%) 209Pb β (100%) 209Bi
- 33. D beta radioactivity Disintegration ß-: A Z X→Z +1Y (*) +−1 e− + ν + (γ ) A 0 Characteristic of nuclei with an excess of neutrons€ Can be explained by the conversion of a neutron into a proton (within the nucleus) 1 1 0 0 n→1 p+−1 e− + ν €
- 34. 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) : 2 2 2 m c =m c +m c +E X Y 0 C − + EC ν + EC γ β With : Ecß-: Kinetic energy of the β - particle Ecγ: Kinetic energy of the γ ray (eventually) Ecv: Kinetic energy of the anti-neutrino€
- 35. Replacing nuclear masses by atomic masses gives: M X c 2 − Zm0 c 2 − E L = M Y c 2 − (Z +1)m0 c 2 + m0 c 2 + EC − + ECν + ECγ − E L β Hypothesis: ♦ Binding energy differences are neglected , ♦ γ ray energy is neglected€ The relation thus becomes: 2 2 M X c = M Y c + EC − + EC ν β And the energy involved is: 2 β ( ) E d = M X c − M Y c 2 = EC − + EC ν = EC β 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!
- 36. Disintegration ß+: A Z X→Z −1Y (*) +1 e+ + ν + (γ ) A 0This 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) 1 1 p→01 n+1 e+ + ν 0This 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) : m X c 2 = mY c 2 + m0 c 2 + EC + + ECν + ECγ βWith :Ecß-: Kinetic energy of the β+ particleEcγ: Kinetic energy of the γ ray (eventually)€ v: Kinetic energy of the neutrinoEc
- 37. Replacing nuclear by atomic masses gives: M X c 2 − Zm0 c 2 − E L = M Y c 2 − (Z −1)m0 c 2 + m0 c 2 + EC + + ECν + ECγ − E L β Hypothesis: ♦ Binding energy differences are neglected , ♦ γ ray energy is neglected€ The relation thus becomes: M X c = M Y c + EC + + ECν + 2m0 c 2 2 2 β And the energy involved is: € ( ) Ed = M X c 2 − M Y c 2 − 2m0 c 2 = EC + + ECν = EC + β β max €
- 38. 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. 18Very famous β + decay: 9 F→18 O+1 e+ + ν + γ 8 0 €
- 39. 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: * A Z X+−1 e(atomic) → ( Z −1Y (*) ) + ν + (γ ) 0 − AThis equation can be explain by the process: 1 0 − 1 p+−1 e(atomic) →0 n + ν € 1Electron 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).
- 40. As for the preceding disintegrations, it is possible to write: 2 2 2 With: m X c + m0 c = mY c + ECν + EC γ Ecv: neutrino kinetic energy Ecγ: γ ray possible energy Replacing nuclear masses by atomic masses gives: € M X c 2 − Zm0 c 2 − E L + m0 c 2 = M Y c 2 − (Z −1)m0 c 2 + ECν + ECγ − E L Energetic hypotheses: EL-EL’ is neglected, And thus: M X c 2 = M Y c 2 + EC γ + EC ν€ The disintegration energy is: E d = M X c 2 − M Y c 2 = EC γ + EC ν € €
- 41. 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 willcreate 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: * 123 53 I + e(atomic) → ( 123Te) + ν + γ − 52 €
- 42. Internal conversion - Auger electronsThis process refers to electrons that can be ejected from the electronic cloud as aconsequence 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: (hν )γ >> ( E L ) K ,L...From a practical point of view, it is impossible to differentiate β - decay electronsfrom those resulting from internal conversion. €They have, however, a characteristic emission energy (ray): ECCI = (hν )γ − ( E L ) K ,L... €
- 43. Raies characteristic rays interne IC des électrons de conversion Electrons- émis Nombre de β ⎛ ⎞ ⎜ EC − ⎟ ⎝ β ⎠ max emitted EC − βHere again, the electron vacancy will be filled by more superficial electrons. Thiswill create X rays emissions characteristics of the electronic transition.These fluorescence X rays can leave the atom (i.e. for heavy atoms) or ejectsuperficial 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.
- 44. V Conclusion: decay scheme of 99mTcThis illustrate some of the processes seen in the lecture… 99 m 43 Tc 6,007 heures ec1 γ1 ec3 γ3 β1− 142,681 keV 140,511 keV ec2 γ2 322,40 keV 99 Tc 43 99 Tc 43 β2− 2,14 × 105 ans β− γ2 γ3 89,80 keV 293,60 keV β3− γ1 99 44 Ru 0 keV Stable
- 45. Emission Electronic layer Energy (keV) Absolute yield (%) ec1 M, N 1.73 < E < 2.13 99.11 K 119.467 8.7 ec2 L 137.46 1.12 ICelectrons M, N 139.96 < E < 140.50 0.23 K 121.64 0.61 ec3 L 139.64 0.20 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
- 46. 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 physicspublishing - Bristol and Philadelphia (1993)- F. Lagoutine, N. Coursol et J. Legrand, Table de radionucléides, CEA/LMRI (1983)Acknowledgement: T Carlier, NM, Nantes University Hospital
- 47. Interactions radiation - matterIntroduction: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
- 48. High energy X and γ rays Interactions with matterIntroduction:In the medical ﬁeld, electromagnetic radiations: X and γ Radiology, Nuclear Medicine 20 to 500 keV Radiotherapy 1 to 100 MeVThe study of the interactions leads to: Basic rules for radioprotection Physics basis for radiotherapy Physics basis for detectors for medical imagingPlan of the study: Macroscopic (experimental) aspects µ and HVL Microscopic (atomic) phenomenons PE effect, Compton, pair production Energy ranges ConsequencesHere it’s the particle side of electromagnetic radiations that’s considered
- 49. I Macroscopic effects: { } Nt Na N0 N1 Np Nd Ns Detector Détecteur 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
- 50. A Attenuation coefficient:Relation between N0 and N1? N N0 N(x) N(x+dx) N(e) 0 x x+dx e x N(x) - N(x+dx) = dN = - µ N dx
- 51. N(x) - N(x+dx) = dN = - µ N dxThe 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 crossedThis leads to a ﬁrst order differential equation: dN = −µdx N Ln(N ) = −µx + C N Ln(N ) − Ln(N 0 ) = Ln( ) = −µx N0 Therefore: N(x) = N 0 e− µx € €
- 52. Consequences: N(x) = N 0 e− µx• 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) µ µm = ρ
- 53. 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: N0 N (HVL) = N 0 e−µHVL = −ln(2) = −µHVL 2 Ln(2) 0.693 HVL is expressed as a length (cm). HVL = = It is independent of N0. € µ µ€ 1 HVL decreases the photon number by a factor 2 2 HVL 4 3 HVL 8 10 HVL € 1000 (1024) n HVL 2n
- 54. 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 Ek E0 Eb
- 55. More likely to happen if E0 is close to Eb.Probability : null if E0 < Eb high if E0 > Eb low if E0 ≈ EbThis 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
- 56. Fluorescence photon emission Temps time 0 Ekc E E0 Ei E1 E El b Ei Ep Ea atome à Stable effet Photoelectric émission du photon Fluorescence létat basal photoélectrique de fluorescence atom Effect photon EThe atom is at an initial stable state (Ea <0)PE effect produces an ion (Ei <0), and Ei = Ea + EbIf 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
- 57. Temps time 0 E Ekc E0 Ei E1 E El b Ei Ep Ea Stable atome à effet Photoelectric émission du photon Fluorescence létat basal atom photoélectrique de fluorescence Effect photon EEp 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
- 58. But E1 cannot trigger another PE effect in an atom of the same kind!The direction of emission of the photoelectron is independentof 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.
- 59. 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. Temps time 0 E0 Ekc E Ei E E’c Ebl E k Ei Ep+Eq Ea Stable atome à Photoelectric effet émissionelectron Auger délectron atom létat basal photoélectrique Effect Auger emission EIts kinetic energy is: E’c = (Eb - Ep) - Eq
- 60. 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 + EqMost probable event for light elements (C, N, O)Biologic effect equivalent to that of PE electronsImportance 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 adetector)
- 61. 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) Particle cross section: direction de propagation du photon Photon propagation direction BDisk of surface σpe Around each particle, In a plane ⊥ photon propagation. AThis 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!
- 62. Relation between cross section and attenuation coefﬁcient: It is possible to show that µpe = σpeP With P the particle number by volume unit NZ m ε With: P = ρ and: σ pe = A En ερNZ m Thus: µ pe =€ AE n € 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).
- 63. 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 imagingA 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... Trajectoire Electron de lélectron trajectory percuté Free Electron electron Photon incident (E) (E) Incident photon libre φ θ Scattered Photon photon diffusé (E)
- 64. RQ : Free electrons? • Metals • Outer shell electrons, with Eb << E These electrons can interact by Compton effectThere are relationship between: • Incident photon energy • Scattered photon energy • Deﬂexion angleEnergy conservation: E + mc 2 = E ʹ′ + p 2 c 2 + m 2 c 4E : Incident photon energyE’ : Scattered photon energy €mc2 : Electron rest energyc : Celerity of light in vacuumUnder the root square : scattered electron energy and momentum after the collision
- 65. Conservation of momentum: 1) Along the propagation axe of the incident photon: E E ʹ′ = cosθ + p cos φ c c 2) Perpendicular to the propagation axe of the incident photon: E ʹ′ 0= sin θ − p sin φ € c p2 can be calculated through these equations: 2 2 p 2 c 2 = ( E − E ʹ′ cosθ ) + ( E ʹ′ sin θ ) = E 2 + E ʹ′2 − 2EE ʹ′ cosθ € 2 p 2 c 2 = ( E − E ʹ′ + mc 2 ) − m 2 c 4 = E 2 + E ʹ′2 − 2EE ʹ′ + 2mc 2 ( E − E ʹ′)€ mc 2 ⎛ E ⎞ E Which gives: cos θ = 1− ⎜ −1⎟ And if: α= E ⎝ E ʹ′ ⎠ mc 2 € E ʹ′ ν ʹ′ λ 1 We get: = = = This relation binds E and E’ with θ, E ν λ ʹ′ 1+ α (1− cosθ ) scattered angle, and E’ ≤ E (normal) € €€
- 66. 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: σ C = σ 0 fKN (α )Whereσ0 is the Thompson cross section: 8 πr02 σ0 =r0 = 2.818 10-15 m is the electron radius 3And 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
- 67. Relation between cross section and attenuation coefﬁcient: NZOne ﬁnds: µC = σ C ρ ASince 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.
- 68. 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 chargeIf 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
- 69. Summary of X or gamma interactions
- 70. Charged particles matter interactionsIntroduction:Charged particle arriving in the neighbourhood of an atom: • interactions with the nucleus • interactions electronsIn all cases, the particle looses kinetic energy...(Neutrons are not charged particles and won’t be considered here).
- 71. 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 ionisations on 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.
- 72. 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 WRQ: Charged particle beams can be completely stopped...
- 73. 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).
- 74. 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/µmTrajectories 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)
- 75. 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
- 76. 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/µmStraight trajectories E(keV ) L(cm) ≈ LET (keV /µm)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...
- 77. Summary:

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