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This is Round Two on the basic principles of interactions of photons with matter.

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Let us review the objectives of today’s lecture. We are going to focus on the various kinds of interaction coefficients. Then we will determine various units of path length, each path length associated with a different type of interaction coefficient.

Next we will identify the various interactions that photons undergo with matter and identify their dependences on energy, angle, and atomic number. We will go into the interactions in much more detail in subsequent lectures.

Finally, we will identify the implications of the various interactions in imaging and in radiation oncology.

Let me remind you of the course mantra “Follow the energy.”

We are going to use the variety of interaction coefficients to look at how energy is distributed when photons interact with matter.

•What happens to this energy?

•Where is it going?

•How much of the energy goes where?

This is the information we are going to get from the coefficients.

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Let me remind you again of the linear attenuation coefficient.

The linear attenuation coefficient is the quantity that we can measure directly. The linear attenuation coefficient is best interpreted as the fraction of photons that interact in the absorber per unit absorber thickness. That’s the definition of linear attenuation coefficient that we are going to use. It’s really the probability of an event occurring.

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We said that there are several other attenuation coefficients that we frequently use and that it’s important that you get a feel for each one of these.

The mass attenuation coefficient is the fraction of photons removed from the beam per density-weighted unit of absorber thickness. The density times the path length is the density-weighted path length, which is sometimes called the radiological path length. We are going to see the radiological path length in many calculations involving transport of photons through real patients. The mass attenuation coefficient is going to help us determine how energy is deposited in the patient per unit mass. I hope you remember the definition of dose from one of our earlier lectures. Recall that dose is energy deposited per unit mass. The mass attenuation coefficient is going to be our handle to connect photon fluence to energy deposited per unit mass, or dose. So that’s going to be a very important coefficient that we’re going to be using later on in the course.

We’re going to find that for certain processes, when we try to determine attenuation coefficient, we’re going to need to look atthe electron density-weighted path length. And when we do that the attenuation coefficient is then the electronic attenuation coefficient. So that the electronic attenuation coefficient is the mass attenuation coefficient divided by the number of electrons per unit mass. How do we determine the number of electrons per unit mass? It’s the number of atoms per mole, which is Avogadro’s number, times the number of electrons per atom, which is the atomic number, divided by the mass per mole, which is the mass number. The electron density-weighted path length is the physical density-weighted path length time Avogadro’s number times Z over A.

What is the value of Z over A? With the exception of hydrogen, for which Z over A is 1, the value of Z over A is close to ½ for every other material. For large atoms, Z over A is a bit less than ½, but for most atoms Z over A is very close to ½. Consequently, electron density for most materials is the mass density times a constant.

Finally, we are going to be looking at some interactions involving atoms, and some of these are characterized by an atomic attenuation coefficient and an atomic density. The atomic density is the electron density divided by the atomic number.

These are the four attenuation coefficients that we are going to be dealing with in this course.

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Let’s summarize, then. The electron density, or number of electrons per unit volume, is the mass density multiplied by N0, the number of electrons per unit mass. N0 is given by Avogadro’s number, the number of atoms per mole, multiplied by the atomic number, the number of electrons per atom, divided by the mass number, the mass per mole.

The atomic density, or number of atoms per unit volume, is the electron density divided by the atomic number, the number of electrons per atom.

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The linear attenuation coefficient is perhaps the most straightforward of the attenuation coefficients. We express thicknesses and path lengths in terms of physical length, and the linear attenuation coefficient is expressed as an inverse length. The linear attenuation coefficient is a quantity that can be measured directly.

However, some of these other coefficients may give us actually more insight into the interaction process.

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Let’s look at one example now, but we will look at this in a lot of detail later on.

Let’s take an interaction called Compton scatter. We find that the probability of Compton scatter is going to depend on the electron density, because the model we use to calculate Compton scatter approximates a material as a free electron gas. So, if we have a free electron gas, we are going to weight the thickness of the absorber, the path length, with the electron density to yield the electronic attenuation coefficient.

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Compton scatter depends on electron density. The electron density-weighted thickness is the physical density ρ, multiplied by the thickness x, multiplied by the number of electrons per unit mass N0. We said this was the physical density multiplied by the thickness multiplied by Avogadro’s number multiplied by the atomic number divided by the mass number.

However, Avogadro’s number is a constant and Z over A is approximately constant. So the electron density-weighted thickness is going to be roughly proportional to the physical density-weighted thickness.

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Electron density is proportional to physical density. Physical density is easy to measure. Mass divided by volume is the physical density.

We are going to be using these density-weighted coefficients for many applications. One example where we’re using the electronic attenuation coefficient is going to be when we calculate the attenuation coefficient for Compton scatter.

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Another process that we will study is pair production. We are going to see that pair production involves the atomic nucleus. So the attenuation coefficient is going to involve the atomic density. We weight the thickness with the atomic density to obtain an atomic attenuation coefficient. We will see that a little later on.

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We very often are going to be dealing with molecules, but to calculate an attenuation coefficient we typically need an atomic number. So there are times when we will have to calculate an effective atomic number. The effective atomic number is an empirical quantity and is given as a generalized weighted mean of the atomic numbers of each component element. We raise the atomic number of each of the components to some empirical power m, then multiply it by the fraction of electrons per gram of each component, add them together, and take the mth root.

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Here are some values of the parameter m. For the photoelectric effect, m is in the range 3 to 4, for the Compton effect, m is zero, and for pair production, m is equal to 1. Notice that the effective atomic number for a compound is going to be different depending on the particular interaction that is taking place.

In many cases we will have different photon interactions occurring simultaneously, the relative probability of occurrence of each interaction being a function of the photon energy. We could have both pair production and Compton taking place. We could have Compton scatter and photoelectric effect taking place. So the effective atomic number of compounds is going to be a fairly complicated quantity.

One of the reasons we need to look at effective atomic number is that the basis for most of our measurements is water. The human being is mainly water, so we are going to try to do measurements in water. But sometimes we can’t do our measurements in water; we have to use what’s called water-equivalent materials. What this says is that a material that’s water-equivalent at one energy at which, say, photoelectric effect is the only interaction, may not necessarily be water-equivalent at a higher energy where there is a combination of photoelectric effect and Compton effect. And it may not be water-equivalent at an even higher energy where there may be a different ratio between Compton and photoelectric effects. The only way for the material to be truly water-equivalent is if its elemental composition is the same as that of water. In that case, it would be water. We find that different materials are water-equivalent at different energies.

Same thing with air-equivalence; we need to worry about air-equivalence when we are designing an ionization chamber. No material is really air-equivalent at all energies. Because of this fact, when we calibrate an ionization chamber, we need to calibrate at specific energies. So if you send your ionization chamber to the Accredited Dosimetry Calibration Laboratory to get a calibration, you are going to find that what you are going to ask them is to calibrate the ionization chamber at different energies. The reason that you have got to calibrate at different energies is that fact that materials are not water-equivalent or air-equivalent at all energies.

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Let’s do some calculations and try to work out a problem.

We want to find the mass attenuation coefficient for aluminum at cobalt-60 energy and then determine the corresponding linear, electronic, and atomic attenuation coefficients.

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To do the first part is a simple table lookup. At the end of Johns & Cunningham, there are many tables, which can be very useful for this type of calculation. Table A-4e, in particular, provides the radiological properties of aluminum. Other tables provide the radiological properties of water, carbon, etc.

A Google search of “mass attenuation coefficient aluminum” will take you to another source, physics reference data from NIST.

You next need to know the energy of cobalt-60 gamma rays. Use the mean energy of the two gammas emitted by the decay of cobalt-60. That’s a number you will eventually have memorized; it is 1.25 MeV.

We look at the table, and we find that for a photon energy of 1.25 MeV, the mass attenuation coefficient is 0.0549 cm2/g. That’s the answer to the first part of the problem. It’s very straightforward, just table look up, but you have to know what table to use and what energy to use.

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Now, for the second part of the problem, we need to find the other coefficients.

Normally, you would find that it is the mass attenuation coefficient that is tabulated, but often when we want to calculate attenuation effects, we need the linear attenuation coefficient. To determine the linear attenuation coefficient, we multiply the mass attenuation coefficient by the mass density ρ. This is a number we can find in the same table in Johns & Cunningham and we find it is 2.699 103 kg/m3. We convert the mass attenuation coefficient of 5.49 10-2 cm2/g to 5.49 10-3

m2/kg and multiply it by 2.699 103 kg/m3 to obtain 14.81 m-1, which is the desired linear attenuation coefficient.

OK, now what is the half-value layer for aluminum? We divide the linear attenuation coefficient into 0.693 to obtain the half-value layer. 0.693 divided by 14.81 is 0.0468 m, or 4.68 cm. Now we know how much aluminum is required to attenuate the cobalt beam to half its initial intensity. Almost 5 cm of aluminum.

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The next calculation is that of the electronic attenuation coefficient. To determine the electronic attenuation coefficient, we take the mass attenuation coefficient and divide it by the electron density. Earlier we found the electron density to be equal to Avogadro’s number multiplied by Z over A. 6.023 1023 atoms per mole is Avogadro’s number, multiply that by 13 electrons per atom and divide by 27 g per mole to get the number of electrons per g. Divide that number into the mass attenuation coefficient and multiply to convert kg to g. We find that the electronic attenuation coefficient is 1.893 10-29 m2/electron.

I don’t think we really have a good feel for that sort of number, certainly not as good as for the linear attenuation coefficient of 14.81 m-1. I think it is important to have a feel for the order of magnitude, which is on the order of 10-29 m2/electron.

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Finally, how do we determine the atomic attenuation coefficient? We multiply the electronic attenuation coefficient by the number of electrons in the atom, which is the atomic number. 1.893 10-29 m2/electron multiplied by 13 electron/atom gives us for the atomic attenuation coefficient a value of 2.500 10-28 m2/atom.

We could do this calculation for whatever energy we want, and for whatever material for which we can obtain data.

Be aware that if you ever see this type of calculation on an examination, you will not be required to memorize tables of attenuation coefficients. I might give you values for one attenuation coefficient for a particular material and expect you know how to obtain another.

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Let’s return to the linear attenuation coefficient. The linear attenuation coefficient is a measure of probability. It’s a measure of the probability of an interaction per unit path length, or another way of describing it would be to say it is the fraction of the photons in a beam attenuated per unit path length.

Let’s look at a photon interaction. What could happen to the energy in the photon? Some of the energy of the incident photon could be transferred to the kinetic energy of a secondary electron. The energy that’s transferred to the kinetic energy of an electron is called the kinetic energy released per unit mass or kerma. Remember that from an earlier lecture.

So some energy can be transferred to kinetic energy of a secondary electron and some energy could be transferred to a scattered photon depending on the particular interaction.

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Let’s look in particular at the energy that is transferred to electrons. This energy transfer is a stochastic process. For any particular photon interaction at a given energy we have no idea how much energy is going to be transferred to the electron. What we do know is the probability of energy transfer. We know that if we have a large number of photons coming in, there is going to be a mean energy transferred to the electrons. We shall denote this mean energy transfer to kinetic energy of the electrons as Etr with a bar over it.

We can then define a quantity called the energy transfer coefficient. The energy transfer coefficient is the mean fraction of energy transferred to kinetic energy of electrons per unit path length. So, if we start out with a beam which has 10 MeV photons and 5 MeV of energy is transferred to kinetic energy of the electrons per centimeter of path length, the energy transfer coefficient is the fraction of energy transferred, 5 divided by 10, or one half, per cm of path length. Energy transfer coefficients are also tabulated. If we have energy transfer coefficients and we know the energy of the incident photon, we can determine the energy transferred to the kinetic energy of the secondary electrons per unit path length. This quantity may be important if we are trying to follow the energy. Remember, we are following the energy. If some energy is being transferred to electrons, we want to determine how much. We can do this using the energy transfer coefficient.

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We have this kinetic energy that’s transferred to the electrons. What can happen to this kinetic energy? Some of this kinetic energy can be re-radiated as Bremsstrahlung. We know that electrons that are moving can interact downstream with nuclei. These interactions will cause the electrons to deflect and give off Bremsstrahlung. Recall that we spoke about this earlier when we discussed the production of x-rays when electrons struck a target.

The remainder of the energy, that which is not re-radiated, is absorbed. The amount of energy that is absorbed is called the collision kerma.

So, the incident photon energy goes into scattered photon energy plus kerma, that is, kinetic energy of electrons. This kerma goes into Bremsstrahlung, which are re-radiated photons, plus collision kerma. Once we know the collision kerma, we can talk about the mean energy absorbed and an energy absorption coefficient, which is the mean fraction of incident photon energy absorbed per unit path length.

As we have said several times already, follow the energy. Incident photon energy goes into scattered photon energy plus kinetic energy of electrons. The fraction of energy that goes into kinetic energy of charged particles gives us the energy transfer coefficient. This kinetic energy of charged particles further is either re-radiated as Bremsstrahlung or is absorbed locally. The amount of energy absorbed locally is the collision kerma, and the fraction of energy that is absorbed locally gives us the energy absorption coefficient.

Now think about this: When we talk about radiation dose, we’re looking at energy absorbed per unit mass. The mass energy absorption coefficient is going to be the handle that enables us to connect the incident photon fluence to the radiation dose. So, roughly speaking, fluence multiplied by mass energy absorption coefficient is going to give us radiation dose.

We’ll deal with this a lot more later in the course when we talk about cavity theory.

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Let’s look at the energy transfer coefficient before we look at the energy absorption coefficient. The energy transfer coefficient, which is the mean fraction of energy transferred to charged particles per unit path length, is the linear attenuation coefficient times the fraction of energy. Fraction of energy is the mean energy transferred to kinetic energy of charged particles divided by the incident photon energy. So Etr over h is the mean fraction of photon energy transferred to charged particles. If we want to determine the amount of energy transferred to charged particles, we take the energy transfer coefficient divided by the linear attenuation coefficient and multiply it by the incident photon energy. We just solve for Etr.

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The amount of energy transferred to charged particles is equal to the fraction of energy transferred to charged particles per unit path length times the energy carried by the beam times the path length. So now we know how much energy is actually transferred to charged particles.

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The next quantity we want to determine is the amount of energy actually absorbed by the charged particles. We use the same analysis used previously for energy transfer. The energy absorption coefficient, the mean fraction of energy absorbed per unit path length, is equal to mean energy absorbed divided by the incident photon energy multiplied by the linear attenuation coefficient.

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Energy absorbed is the energy absorption coefficient, the fraction of energy absorbed per unit path length, multiplied by the energy carried by the photon beam times the path length. Again μen and μtr are quantities that are tabulated, so if we know the incident energy fluence, we can determine how much energy is transferred to the charged particles and how much energy is absorbed by the charged particles.

This is really an exercise in accounting. We just have to know where the energy is going and where to place it, and that’s it.

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We now know that the energy transferred to electrons is either absorbed, or it is re-radiated as Bremsstrahlung. If we let g be the fraction of energy in the charged particles that is lost to Bremsstrahlung, then the energy absorption coefficient is equal to the energy transfer coefficient times (1-g). At low energies the amount of energy lost to Bremsstrahlung is going to be low, so the energy absorption coefficient will be approximately equal to the energy transfer coefficient. Recall that Bremsstrahlung production is inefficient at low energies. At high energies more Bremsstrahlung is produced, so the energy absorption coefficient will be somewhat less than the energy transfer coefficient. You’ll see that in a table shortly.

So, we now have the linear attenuation coefficient, the energy transfer coefficient, and the energy absorption coefficient.

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Now, we can combine that information with some of the other coefficients to give us mass, electronic, and atomic energy absorption and energy transfer coefficients. That gives us a total of 12 different types of coefficients to make matters even more confusing.

To try to keep matters straight, you can combine linear, mass, electronic, and atomic with attenuation, energy transfer, or energy absorption. Any one of the first four can be combined with any one of the last three to give you a coefficient.

I will point out that the most frequently used coefficient of all of them is the mass energy absorption coefficient. This coefficient gives the most direct connection between photon fluence and absorbed dose. We will be very concerned with that when we talk about dosimetry later in the course and later in your studies.

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Let’s put some quantities together.

Here are some coefficients for carbon. Do not worry, you are not expected to memorize this table.

First column is photon energy, starting with a very low energy and increasing in powers of 10 through energies we encounter in radiation medicine, from 10 keV to 100 MeV.

First, notice what happens to the mean energy transfer. The mean energy transfer starts out for 10 keV photons at 8.65 keV, which is around 85% of the energy of the photons. The mean energy transfer then only rises to 14 keV when the photon energy rises to 100 keV, so the fraction of energy transferred goes down to 14%. It then rises to 44% for 1 MeV photons, 73% for 10 MeV photons, and 96% for 100 MeV photons. What this is saying is that at very low energies and at very high energies, most of the energy of the incident photon gets transferred to kinetic energy of the electrons, but in the range of 100 keV to 1 MeV photons, a much lower fraction of energy gets transferred to electrons, with a larger amount of energy going into scattered photons. So photon scatter is important in intermediate energies.

Now, let’s compare energy absorbed to energy transferred. Keep in mind that the energy not absorbed is re-radiated as Bremsstrahlung. Notice, as we would expect, at low energies, all the electron energy is absorbed, and essentially none is re-radiated. At higher energies, such as 10 MeV, and even more so at 100 MeV, a large amount of Bremsstrahlung is produced.

We take μ over ρ and weight it by the various fractions of energy transferred and absorbed to obtain the mass energy transfer coefficient and mass energy absorption coefficient.

First note that at low energies the mass attenuation coefficient, μ over ρ, tails off quite quickly with increasing energy, but the curve flattens out at higher energies. During the first power of 10 in energy, we go down by a factor of 20 in μ over ρ, while over the next power of 10, we go down by only a factor of about 2.5, the next power of 10 we go down by a factor of 3, and the last power of 10, we barely go down.

We see that something complicated is going on, but don’t worry about that; we’re going to look at that in a lot of detail later on in the course. You should be aware that there are different processes going on and different processes have different energy dependences.

Let’s compare μtr over ρ to μen over ρ and we see that this is consistent with the observation that at low energies, almost all the kinetic energy is absorbed, but at higher energies, some energy is re-radiated.

Now we can use these coefficients for solving problems. We’re not going to do so now, but some problem sets will have you solve problems about energy transfer and energy absorption.

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