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**Interaction of radiation with matter - 3**

X and Gamma Rays Day 2 – Lecture 3

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**Objective To discuss photon interactions including: To learn about:**

Photoelectric effect Compton Scattering Pair Production To learn about: Linear and mass attenuation coefficients Half value layer

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Photoelectric Effect When an incident photon ejects an electron, the process is called “photoelectric” effect. It is a photon-electron interaction rather than a charged particle interaction. The photon transfers 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. 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.

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Photoelectric Effect 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.

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Photoelectric Effect Ephotoelectron = Eincident photon – Ebinding energy - Example: Eincident photon = 80 keV Ebinding energy = 20 keV Ephotoelectron = 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 = 60 keV.

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**Photoelectric Effect The photoelectric effect is predominant for:**

Low energy photons but higher than BE of the electrons High atomic number “Z” materials Probability is proportional to: 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.

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Compton Scattering 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. 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.

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**Compton Scattering scattered electron (Ese) incident photon (Eip)**

loosely bound electron (Eie) scattered electron (Ese) incident photon (Eip) scattered photon (Esp)

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Compton Scattering The energy change 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.

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**Photon converted into two particles**

Pair Production Photon converted into two particles (energy into mass) electron (-) positron (+) The rest mass energy of a positive or negative electron is MeV

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Pair Production To create these two particles requires a minimum energy of 2 x 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. 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.

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Pair Production A positron cannot exist at rest. It combines with an electron. The two particles annihilate each other converting mass back into energy. 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.

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Pair Production If, the incident the photon has an energy 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 then undergo the typical charged particle interactions losing energy during each 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.

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Photon Interactions 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. 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)

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**Attenuation vs Absorption**

When photons interact with matter three things can occur. The photon may be: Transmitted through the material unaffected Scattered in a different direction from that traveled by the incident photon Absorbed by the material such that no photon emerges

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**Attenuation vs Absorption**

Attenuation of the photon beam can be considered a combination of scattering and absorption. Attenuation = Scattered + Absorbed If the photons are scattered or absorbed, they are no longer traveling in the direction of the intended target.

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**Attenuation vs Absorption**

Radiation Source Detector 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

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Attenuation 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.

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**I = Io e Exponential Attenuation - 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. represents the fractional linear attenuation coefficient unit is per cm.

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Half Value Layer A half value layer of any material will permit only 50% or ½ of the incident radiation to pass. A second half value layer will permit ½ of the incident radiation (already reduced by ½) to pass so that only ¼ of the initial radiation (½ x ½) is permitted to pass. If “n” half value layers are used, (½)n of the initial radiation is permitted to pass. “n” may be any number.

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**Half Value Layer - Example**

The half value layer (HVL) of a material is 2 cm. A researcher has a piece of the material which is 7 cm thick. What fraction of the initial radiation will pass through the piece?

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**Half Value Layer - Example**

Determine the number of “n” half value layers used 7 cm 2 cm HVL = 3.5 HVL = n (½)n of the initial radiation is permitted to pass. (½)3.5 = (use a calculator yx) Self Check – the answer must be between: (½)3 = 1/8 = and (½)4 = 1/16 =

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**Half Value Layer - Example**

The half value layer of material “A” is 2 cm and the half value layer of material “B” is 5 cm. A researcher has a piece of some material which is composed of 3 cm of “A” and 4 cm of “B”. What fraction of the initial radiation will pass through the piece?

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**Half Value Layer - Example**

The half value layer of material “A” is 2 cm and the half value layer of material “B” is 5 cm. A researcher has a piece of some material which is composed of 3 cm of “A” and 4 cm of “B”. What fraction of the initial radiation will pass through the piece? “A”: cm = 1.5 HVL = n (2 cm/HVL) “B”: cm = 0.8 HVL = n (5 cm/HVL) [(½)1.5 ] x [(½)0.8 ] = x = 0.203 A B 35 20 100 35% 57%

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**Half Value Layer - Example**

The initial intensity is It is desired to reduce the intensity to 12. How many HVL do we need? You don’t need to know anything about the material. A half value layer of ANY material passes ½.

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**Half Value Layer - Example**

The initial intensity is It is desired to reduce the intensity to 12. How many HVL do we need? Going from 192 to 12 means that the initial intensity is reduced by a factor of 192/12 = 16. Or we could say that the final intensity is 1/16 of the initial. How many HVL do we need? (½)n = 1/ or 2n = 16 This one is easy. Since 24 is 16, we need 4 HVL

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**Half Value Layer Given a specific material:**

For monoenergetic radiation, the HVL never changes. For polyenergetic x-ray beams, the HVL increases as more material is inserted into the beam

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**Half Value Layer mono- energetic poly- energetic***

1000 500 250 125 62 HVL mono- energetic E1 E1 E1 E1 E1 poly- energetic* 1000 500 300 200 155 HVL 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 * Effective energy of the initial polyenergetic beam is the same as the energy of the monoenergetic beam above

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**Attenuation Coefficients**

There are two types of attenuation coefficients: Linear Attenuation Coefficient (LAC) provides a measure of the fractional attenuation per unit length of material traversed Mass Attenuation Coefficient (MAC) provides a measure of the fractional attenuation per unit mass of material encountered

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**Linear Attenuation Coefficient**

I = Io e (- x) when x = HVL, then I = (½)Io (½)Io = Io e (- HVL) and HVL are functions of the energy of the photon radiation and the material through which it passes ½ = e (- HVL) ln(½) = ln(e (- HVL)) ln(½) = (- HVL) ln(2) = ( HVL) LAC = M,E = ln 2 HVL M,E ln(2) HVL =

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** = x Mass Attenuation Coefficient**

The relationship between LAC and MAC is: LAC = MAC x density = 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

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**Mass Attenuation Coefficient**

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

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**Attenuation Equations**

To express the attenuation of radiation as it passes through some material we can use either of two equations: I = Io e (- x) or when x = HVL 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.

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**Sample Problem #1 Use I = Io (½) n or I = Io e (- x)**

The dose rate is reduced from 300 mSv/hr to 100 mSv/hr using 5 cm of some material. The material has a mass attenuation coefficient of 0.2 cm2/g. What is the density of the material? Use I = Io (½) n or I = Io e (- x)

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**Solution to Sample Problem #1**

(½)n = 100/300 = 1/ or ln(½)n = ln(1/3) n = ln(1/3)/ln(½) = / = HVL 5 cm/1.585 HVL = 3.2 cm = HVL LAC = ln(2)/HVL = (µ/) x = MAC x = = = = 1.09 g/cm3 Using 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 The actual HVL is determined to be 3.2 cm. ln(2) HVL MAC 0.693 3.2 cm 0.2 cm2/g 0.217 cm-1 0.2 cm2/g

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Sample Problem #2 An x-ray beam is evaluated by sequentially placing thicknesses of aluminum in the beam path and measuring the amount of radiation transmitted. The results are: Al (mm) (mR/hr) Al (mm) (mR/hr) Determine the effective energy of the radiation emitted by this x-ray unit.

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**Solution to Sample Problem #2**

HVL approximately 3.9 mm = 0.39 cm µ = ln(2)/HVL = 0.693/0.39 cm = 1.78 cm-1 for aluminum = 2.7 g/cm3 MAC = LAC/ = 1.78 cm-1/2.7 g/cm3 = 0.66 cm2/g Looking up the MAC for aluminum yields an effective energy somewhere between 35 and 40 keV

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**Where to Get More 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)

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