Interaction rayonnements/matière - 3 Rayons X & Gamma Jour 2 – Leçon 3

Objectif Discussion des modes d’interaction des photons avec la matière: Effet Photoélectrique Diffusion Compton Production de paires Apprendre davantage sur: Coefficients d’atténuation linéique et massique Épaisseur moitié

Effet Photoélectrique Quand un photon éjecte un électron, le processus est appelé “photoélectrique”. Il s'agit d'une interaction photoélectrons plutôt que d'une interaction de particules chargées. Le photon transfère toute son énergie à l’électron au cours d’un seul événement, l’électron éjecté conserve toute l’énergie qui lui a été communiquée à l’exception de l’énergie de liaison Puisqu’un électron a été éjecté, il y a création d’une vacance dans la couche K, lorsque cette vacance est remplie par un autre électron des couches supérieures, il y a émission d’un rayon X caractéristique

Effet Photoélectrique We have actually discussed the photoelectric effect previously without calling it by name. When we discussed ionization, we considered the possibility of a tightly bound electron being ejected from its orbit and a characteristic x-ray being produced when the vacancy created was filled by an electron from a higher orbit. Although we were primarily considering the effects of charged particle interactions, it was noted that electrons can be ejected by not only charged particles but also, neutrons and photons. When we consider the specific case of an incident photon ejecting an electron, we call the process the “photoelectric” effect. It is a photon-electron interaction rather than a charged particle interaction. The photon must transfer all its energy to the electron in a single cataclysmic event after which the photon ceases to exist and the ejected electron retains all the energy originally possessed by the photon except for the energy required to overcome the binding energy of the electron. Of course, since an electron has been ejected, a vacancy exists in the K shell and thus one or more characteristic x-rays will be produced as the vacancy is filled by electrons from higher orbits.

Effet Photoélectrique Ephotoélectron = E photon incident – Eénergie de liaison - Exemple: E photon incident = 80 keV Eénergie de liaison = 20 keV Ephotoélectron = 60 keV The simple equation for the energy transferred during the photoelectric interaction is given here. As an example, if the energy of the incoming photon is 80 keV and the binding energy of the electron in the K shell is 20 keV, then the energy carried away by the electron as it leaves the atom is 80-20 = 60 keV.

Effet Photoélectrique L’effet photoélectrique est prédominant pour: Des photons de faibles énergies : Des matériaux de numéro atomique “Z” élevé La probabilité est proportionnelle à: Z4 E3 The photoelectric effect is most likely to occur when the energy of the incident photon is relatively low but higher than the binding energy of the bound electrons. It is also most likely if the atomic number (Z) of the material upon which the photons are incident is large. The K shell binding energy for lead is approximately 88 keV so that it would be very effective at removing photons with energies just above 88 keV via the photoelectric effect. This makes lead a very good shield for low energy photon emitters. The cross section or likelihood for this interaction is directly proportional to the 4th power of the Z of the materials and inversely proportional to the cube of the energy. As the atomic number increases (Al to Cu to W to Pb) the photoelectric effect becomes more likely. Conversely, as the energy increases (10 keV to 100 keV to 1000 keV), the photoelectric effect becomes less likely.

Diffusion Compton La diffusion Compton se produit lorsque le photon X incident est dévié de sa trajectoire initiale par interaction avec un électron.  L'électron est éjecté de sa position orbitale et le photon X perd une fraction de son énergie en raison de cette interaction, mais il continue à se déplacer à travers le matériau le long d'une trajectoire modifiée. Compton scattering occurs when the incident x-ray photon is deflected from its original path by an interaction with an electron.  The electron is ejected from its orbital position and the x-ray photon loses energy because of the interaction but continues to travel through the material along an altered path.  Energy and momentum are conserved in this process.  The energy shift depends on the angle of scattering and not on the nature of the scattering medium.  Since the scattered x-ray photon has less energy, it has a longer wavelength and less penetrating than the incident photon.

Électron moins lié (Eie) Diffusion Compton Électron moins lié (Eie) Électron diffusé(Ese) photon incident (Eip)   photon diffusé(Esp)

Diffusion Compton l’échange d’énergie dépend de l'angle de diffusion et non pas de la nature du milieu de diffusion. Etant donné que le photon X diffusé a une énergie dégradée, il a une longueur d'onde plus longue et moins pénétrant que le photon incident.

Le photon est transformé en deux particules Création de paires Le photon est transformé en deux particules (énergie en masse) électron (-) positron (+) L'énergie au repos d'un électron positif ou négatif est 0,511 MeV

Création de paires Pour créer les deux particules il faut un minimum d’énergie de 2 x 0.511 MeV = 1.02 MeV If, however, the photon has an energy of more than 1.02 MeV, the excess energy is given to the particles as kinetic energy which permits them to travel at some velocity away from the point of creation. These are then charged particles which undergo the typical charged particle interactions losing energy during each interaction. Si le photon a une énergie exacte de 1.02 MeV, toute celle-ci est utilisée pour créer l'énergie de masse au repos des deux particules de sorte que les particules restent au repos et se recombinent par la suite.

Création de paires Un positron ne peut pas exister au repos. Il se combine avec un électron. Les deux particules s’annihilent en se transformant en énergie. At some point, the positron has lost all of its kinetic energy and comes to rest. It then combines with a nearby electron also at rest. The two particles annihilate each other, disappear and are replaced by photons which are pure energy. As always, the laws of conservation of energy and momentum must be satisfied. The energy law is satisfied by the conversion of the rest mass energy of the particle to a photon of equal energy. However, the conservation of momentum can only be satisfied if the photons travel in opposite directions, essentially canceling each other out since the two particles had no momentum, each being at rest when they combined. This is in contrast to the pair production interaction which created the electron and the positron. The photon which initiated the event possessed some momentum since it was traveling in some specific direction. If the photon possessed any excess energy beyond 1.02 MeV, the kinetic energy transferred to the particles must cause them to travel in such a manner as to equal the momentum of the initial photon. They are able to travel at some angle to each other as long as the horizontal components of the two particles, when summed, equals the forward momentum of the initial photon, and the vertical components cancel.

Création de paires Si le photon incident a une énergie  1.02 MeV, l’excès d’énergie est transformé en énergie cinétique, énergie qui leur permet de se déplacer à une vitesse donnée à partir du point de création. Ceux-ci subissent ensuite des interactions typiques de particules chargées en perdant de l'énergie lors de chaque interaction. If the photon has an energy of exactly1.02 MeV, all of it is used to create the rest mass energy of the two particles so that the particles remain at rest and soon recombine. If, however, the photon has an energy of more than 1.02 MeV, the excess energy is given to the particles as kinetic energy which permits them to travel at some velocity away from the point of creation. These are then charged particles which undergo the typical charged particle interactions losing energy during each interaction.

Interactions des Photons L'effet de diffusion Compton se produit sur toutes les énergies, mais des pics se trouvent dans la gamme d’énergie moyenne. Bien que l'événement ne ne se produit qu’au-dessus de 1,02 MeV il domine tant que l'énergie des photons augmente. WATER Combined Probability Compton As can be seen in this graphic, for very low energy photons, the predominant interaction is Photoelectric. The Compton Scattering effect occurs over all energies but peaks in the midrange while the Pair Production event occurs only above 1.02 MeV and dominates as the photon energy increases. Graphs such as these exist for each material with which a photon interacts. This graph represents photon interactions in water. The curves shift depending on the material but the basic distribution of where each type of interaction dominates is the same for all. The vertical axis actually represents the mass attenuation coefficient which is a measure of the effectiveness of the material to stop photons. This concept will be discussed in a future lesson. Pair Production Photoelectric Photon Energy (MeV)

Atténuation & Absorption Quand les photons interagissent avec la matière trois choses peuvent se produire. Le photon peut être: Transmis à travers le milieu non affecté Diffusé dans une direction différente de celle du photon incident Absorbé par le matériau de telle sorte qu'aucun photon ne diffuse

Attenuation & Absorption L'atténuation du faisceau de photons peut être considérée comme une combinaison de diffusion et d'absorption. Attenuation = Diffusé + Absorbé Si les photons sont diffusés ou absorbés, ils ne vont pas loin dans la direction de la cible visée.

Attenuation & Absorption Source de radioactive Détecteur b When photon radiation (gammas or x-rays) are incident upon a material there are 3 possible interactions as ware discussed in a previous lesson. These are the photoelectric effect, compton scattering and Pair Production. Lets look at each interaction. In the photoelectric effect, the incident photon is completely eliminated and an electron is ejected from the atom. The freed electron can then undergo charged particle interactions: it can be deflected thus producing Bremsstrahlung x-rays, or it can ionize other electrons possibly resulting in the emission of characteristic x-rays. In the compton interaction, the initial photon ceases to exist, however, a new, lower energy photon is produced which travels in a direction which differs from that of the initial photon. It can even go backwards. A free electron is also produced. The new photon can itself undergo a photon interaction and the free electron can undergo charged particle interactions. In the pair production interaction, provided it has sufficient energy, the incident photon disappears and is replaced by 2 charged particles which can undergo charged particle interactions. When the positron finally comes to rest, it annihilates with an electron and produces two photons traveling in opposite directions. This photon can undergo the photoelectric effect and compton scattering. In each case, the incident photon is “absorbed” by the material. Of those not absorbed, we are only interested in the photons which are created by the initial event, not those produced by charged particle interactions or photon interactions by secondary photons where a secondary photon is the one produced by the initial photon. In simple terms, we are only looking at the initial photons produced by the source (the “primary”) and the photons created by the initial photon interaction (the “secondary”). The “primary” photon will either be absorbed or unaffected by passage through the material as indicated by “b”. The “secondary” photons may be absorbed such as “d” in the slide or they may be attenuated such as “a” and “c” which means that their direction has been changed so that they are not detected in the forward direction. They were not absorbed but they are no longer traveling in the same direction as the incident photon. If we consider a beam of photons rather that a single photon, all of the primary and secondary photons that reach the “detector” which is located along the path of the incident photon would be considered unaffected or “unattenuated” even if some of them were scattered but ended up striking the detector. Those photons which do not strike the detector are considered attenuated. So it is clear that photons which are absorbed are also attenuated since they did not reach the detector. Absorbed photons are a special case of attenuated photons. The photon that initiated this chain reaction was “absorbed” by the material. d c

Atténuation 90% 90% 90% 90% 100 90 81 73 66 Each material has a specific attenuation coefficient which indicates how effective it is attenuating photons. I did not say “stopping” photons because that would imply we are only interested in absorption. When we speak of attenuation we acknowledge that some photons may pass through the material but at an angle such that they do not travel along the path of the primary beam. The attenuation coefficient for each material depends on the energy of the photons. The higher the energy, the more likely they are to pass through. Remember that the probability of the photelectric effect which involved total absorption of the initial photon decreases as the energy of the photon increases. Assume that we have a monoenergetic beam of photons (they all have the same energy). This beam strikes a given thickness of some material which permits 90% of the photons of that energy to pass through and attenuates 10%. If we start with 100 photons we would have 90 after passing through the material. Now suppose we have another piece of the same material with the same thickness. Since the coefficient is only based on the energy and that hasn’t changed, the second piece will also pass 90% of the photons. However, there are only 90 photons left after the initial beam passed through the first piece. So the number of photons that pass through the second piece is 90% of 90 photons or 81 photons. We can repeat this process and we see that for each identical piece of material, 90% of whatever is incident on that piece passes through and 10% is attenuated, removed from the primary path.

I = Io e Atténuation Exponentielle -  x This constant fractional attenuation is expressed as follows: The rate of change of the intensity of the beam as a function of the thickness of the material is a constant fraction of the intensity. The equation for this relationship is shown in the slide. The negative sign indicates that the intensity “I” is decreasing as the thickness “x” increases.  represents the fractional linear attenuation coefficient. This coefficient is a function of “E” the energy of the photons and “m” thickness of the material.  Représente le coefficient d’atténuation linéique, son unité est cm-1.

Epaisseur moitié Une épaisseur moitié d’un matériau donné ne laisse passer que 50% ou ½ du rayonnement incident. Une deuxième épaisseur moitié de ce matériau ne laissera passer que la moitié du rayonnement incident (déjà réduit d’un facteur 2) c’est à dire le quart du rayonnement initial (½ x ½). Si “n” épaisseurs moitiés est utilisée, (½)n du rayonnement initial sera transmis, “n” peut être n’importe quel nombre.

Epaisseur moitié - Exemple L’épaisseur moitié d’un matériau(x1/2) est 2 cm. Un chercheur a une plaque de ce matériau d’une épaisseur de 7 cm. Quelle est la fraction du rayonnement initial qui va passer à travers cette plaque?

Epaisseur moitié - Exemple On détermine la valeur du nombre n “n” l’épaisseur moitié utilisée (n = nombre d’épaisseurs moitié) 7 cm 2 cm = 3.5 = n (½)n du rayonnement initial est autorisé à passer (½)3.5 = 0.0883 (utiliser une calculatrice yx) La solution doit être entre: (½)3 = 1/8 = 0.125 et (½)4 = 1/16 = 0.0625

Epaisseur moitié - Exemple L’épaisseur moitié du matériau “A” est 2 cm et l’épaisseur moitié d’un autre matériau “B” est 5 cm.. Un chercheur a un morceau d'une matière qui est composée de 3 cm de "A" et 4 cm de "B". Quelle fraction du rayonnement initial va passer à travers la pièce?

Epaisseur moitié - Exemple L’épaisseur moitié du matériau “A” est 2 cm et l’épaisseur moitié du matériau “B” et 5 cm. Un chercheur a un morceau d'une matière qui est composée de 3 cm de "A" et 4 cm de "B". Quelle fraction du rayonnement initial va passer à travers la pièce? “A”: 3 cm = 1.5 x1/2 = n (2 cm/x1/2) “B”: 4 cm = 0.8 x1/2 = n (5 cm/ x1/2) [(½)1.5 ] x [(½)0.8 ] = 0.354 x 0.574 = 0.203 A B 35 20 100 35% 57%

Epaisseur moitié - Exemple L’intensité initiale est 192. On veut la réduire à une intensité de 12. On aura besoin de combien de x1/2? Vous n’aurez besoin d’aucune information au sujet du matériau. L’épaisseur moitié de n’importe quel matériau laisse passer ½.

Epaisseur moitié - Exemple L’intensité initiale est 192. On veut la réduire à une intensité de 12. On aura besoin de combien de x1/2? Partant de 192 à 12 ça veut dire que l’intensité initiale est réduite d’un facteur 192/12 = 16. Ou on peut dire que l’intensité finale est 1/16 de l’intensité initiale. De combien d’épaisseur moitiés on aura besoin? (½)n = 1/16 ou 2n = 16 Celui ci est facile. puisque 24 est 16, nous avons besoin de 4x1/2

Epaisseur moitié - Exemple On donne un matériau spécifique: Pour un rayonnement mono-énergitique , l’épaisseur moitié ne change jamais. Pour un faisceau de rayons x multi-énergies, l’épaisseur moitié (x1/2) augmente avec l’épaisseur du matériau inséré dans le faisceau

Epaisseur moitié x1/2 x1/2 mono- énergétique poly- énergétique* 1000 500 250 125 62 x1/2 mono- énergétique E1 E1 E1 E1 E1 poly- énergétique* 1000 500 300 200 155 x1/2 For a monoenergetic beam, the HVL never changes so that ½ of the beam is attenuated for each HVL through which it passes as in the first graphic above. If however, the beam is polyenergetic and initially has the same effective energy as the monoenergetic beam then we can assume that the polyenergetic beam has some photons which are lower in energy than the monoenergetic beam and some that are higher. As this polyenergetic beam passes through the first piece of material it is reduced by ½ since it behaves like a monoenergetic beam of that energy with that particular HVL. Now however, as the beam emerges from the first piece, the lower energy photons have been eliminated and the remaining higher energy photons produce a beam which has a higher effective energy. As a result, the HVL has changed. It now requires more material to reduce the beam by ½. Since we are using the same amount of material for each piece we see that the beam is not reduced by 1/2 . More penetrates than is expected. After 4 pieces there is more than twice as many photons than was the case for the monoenergetic beam. Thus if the hardening of a polyenergetic beam is not taken into consideration, the shielding may be underestimated. In this case more than 4 HVLs is required to reduce the initial beam to 1/16 of its initial value. E1 E2 E3 E4 E5 * L'énergie effective du faisceau poly-énergétique initiale est la même que l'énergie du faisceau mono-énergétique de dessus

Coefficients d’Atténuation Il y a deux types de coefficient d’atténuation: Coefficient atténuation linéique (CAL) fournit une mesure de la fraction d'atténuation par unité de longueur du matériau traversé Coefficient atténuation massique (CAM) fournit une mesure de la fraction atténuation par unité de masse du matériau traversé

Coefficient d’atténuation linéique (CAL) I = Io e (- x) quand x = x1/2, donc I = (½)Io (½)Io = Io e (-  x1/2 )  et x1/2 sont fonction de l’énergie du photon x et de la nature du matériau qu’il traverse ½ = e (-  x1/2) ln(½) = ln(e (-  x1/2)) ln(½) = (- x1/2) ln(2) = ( x1/2)  = ln 2 x1/2 ln(2) x1/2 = 

   = x  Coefficient d’attenuation massique (CAM) La relation entre (CAL) et (CAM) est: CAL = CAM . densité    = x   is the linear attenuation coefficient, dimension of 1/cm (or cm-1). In most tables you will find the mass attenuation coefficient which is / and has dimensions of (1/cm)/(g/cm3) which dimensionally equals cm2/g 1 = cm2 x g cm g cm3

Coefficient d’atténuation massique Photon Energy Material The mass attenuation coefficient for 100 keV photons passing through lead is 5.4 cm2/gm.

Equations d’atténuation Pour exprimer l’atténuation du faisceau de photons qui passe à travers un matériau, on peut utiliser l’une des équations ci-dessous: I = Io e (- x) ou quand x = x1/2 I = Io (½) n The exponential decay equation is typically written using the linear attenuation coefficient. However, as shown here, it can also be written as a function of the HVL.

Exemple #1 Utiliser I = Io (½) n ou I = Io e (- x) Le débit de dose est réduit de 300 mSv/h à 100 mSv/h en utilisant un matériau de 5 cm d’épaisseur. Ce matériau a un coefficient d’atténuation massique de 0.2 cm2/g. Quelle est la densité du matériau utilisé? Utiliser I = Io (½) n ou I = Io e (- x)

Solution #1 (½)n = 100/300 = 1/3 ou ln(½)n = ln(1/3) n = ln(1/3)/ln(½) = -1.0986/-0.693 = 1.585 x1/2 5 cm/1.585 x1/2 = 3.2 cm = x1/2 CAL = ln(2)/CAL = (µ/) .  = CAM .   = = = = 1.09 g/cm3 En utilisant I = Io (½) n Since 5 cm reduced the intensity from 300 to 100, that amount of material is greater than the HVL which would have reduced the intensity to 150. The actual HVL is determined to be 3.2 cm. ln(2) x1/2 CAM 0.693 3.2 cm 0.2 cm2/g 0.217 cm-1 0.2 cm2/g

Exemple #2 Un faisceau de rayons X est évalué en plaçant successivement des épaisseurs de l'aluminium dans le trajet du faisceau et en mesurant la quantité de rayonnement transmis. Les résultats sont les suivants: Al (mm) (mR/hr) Al (mm) (mR/hr) 0 350 4 170 1 290 5 150 2 240 6 140 3 200 10 100 Déterminer l’énergie effective des rayons x émis par cette unité.

Solution #2 X1/2 approximativement 3.9 mm = 0.39 cm µ = ln(2)/X1/2 = 0.693/0.39 cm = 1.78 cm-1  de l’Alu = 2.7 g/cm3 CAM = CAL/ = 1.78 cm-1/2.7 g/cm3 = 0.66 cm2/g Regardons sur le tableau le CAM pour l’Alu l’énergie effective se trouve entre 35 et 40 keV

Où trouver plus d’information Cember, H., Johnson, T. E, Introduction to Health Physics, 4th Edition, McGraw-Hill, New York (2009) International Atomic Energy Agency, Postgraduate Educational Course in Radiation Protection and the Safety of Radiation Sources (PGEC), Training Course Series 18, IAEA, Vienna (2002)