Photons, also called X-rays or y-rays, are electromagnetic radiation. Considered as particles, they travel with the speed of light c and they have zero rest mass and charge. The relationship between the energy of a photon, its wavelength A, and frequency is
There is no clear distinction between X-rays and y-rays. The term X-rays is applied generally to photons with E < 1 MeV. Gammas are the photons with
E > 1 MeV. In what follows, the terms photon, y, and X-ray will be used
interchangeably.
X-rays are generally produced by atomic transitions such as excitation and ionization. Gamma rays are emitted in nuclear transitions. Photons are also produced as bremsstrahlung, by accelerating or decelerating charged particles.
X-rays and y-rays emitted by atoms and nuclei are monoenergetic. Bremsstrah- lung has a continuous energy spectrum.
Figure 4.1
Heavy Light 8 r Ar Si
'- yV
Fission fragment
.5 Universal range-energy plot for E > 1. It allows de
Energy, keV
:termination of range in silicon for many heavy ions (Ref. 22).
Br 0 . 1 1 5
Light \ Flss,on 0 . i 1 5
~eavy! fragment 0.1 1 3
I 0 . 1 1 3
There is a long list of possible interactions of photons, but only the three most important ones will be discussed here: the photoelectric effect, Compton scattering, and pair production.
4.8.1 The Photoelectric Effect
The photoelectric effect is an interaction between a photon and a bound atomic electron. As a result of the interaction, the photon disappears and one of the atomic electrons is ejected as a free electron, called the photoelectron (Fig. 4.16).
The kinetic energy of the electron is
T = E, - Be (4.45)
where E, = energy of the photon
Be = binding energy of the electron
The probability of this interaction occurring is called the photoelectric cross section or photoelectric coeficient. Its calculation is beyond the scope of this book, but it is important to discuss the dependence of this coefficient on parameters such as E,, Z, and A. The equation giving the photoelectric coefficient may be written as
zn
T (m-') = aN-[I - O(Z)] (4.46)
E,"
where T = probability for photoelectric effect to occur per unit distance traveled by the photon
a = constant, independent of Z and E,
m , n = constants with a value of 3 to 5 (their value depends on E,; see Evans)
N, Z have been defined in Sec. 4.3.
The second term in brackets indicates correction terms of the first order in Z.
Figure 4.17 shows how the photoelectric coefficient changes as a function of E, and Z. Fig. 4.17 and Eq. 4.46 show that the photoelectric effect is more important for high-Z material, i.e., more probable in lead ( Z = 82) than in Al ( 2 = 13). It is also more important for E, = 10 keV than E, = 500 keV (for the same material). Using Eq. 4.46, one can obtain an estimate of the photoelectric coefficient of one element in terms of that of another. If one takes the ratio of T
for two elements, the result for photons of the same energy is
Figure 4.16 The photoelectric effect.
154 MEASUREMENT AND DETECTION OF RADIATION
where pi and A , are density and atomic weight, respectively, of the two elements, and T~ and 7, are given in m-'. If 7, and 7, are given in m2/kg, Eq.
4.47 takes the form
4.8.2 Compton Scattering or Compton Effect
The Compton eflect is a collision between a photon and a free electron. Of course, under normal circumstances, all the electrons in a medium are not free but bound. If the energy of the photon, however, is of the order of keV or more, while the binding energy of the electron is of the order of eV, the electron may be considered free.
The photon does not disappear after a Compton scattering. Only its direc- tion of motion and energy change (Fig. 4.18). The photon energy is reduced by a certain amount that is given to the electron. Therefore, conservation of energy gives (assuming the electron is stationary before the collision):
If Eq. 4.48 is used along with the conservation of momentum equations, the energy of the scattered photon as a function of the scattering angle 0 can be calculated. The result is (see Evans)
Using Eqs. 4.48 and 4.49, one obtains the kinetic energy of the electron:
(1 - cos 6)Ey/mc2
T = E
1 + (1 - c o s ~ ) ~ ~ / m c ~
A matter of great importance for radiation measurement is the maximum and minimum energy of the photon and the electron after the collision. The minimum energy of the scattered photon is obtained when 8 = T. This, of course, corresponds to the maximum energy of the electron. From Eq. 4.49,
E y ' , min =
E Y 1 + 2E,/mc2
Z photon energy and ( b ) atomic ( b ) number of the material.
e,T - freed electron Photon
- EY +e--B--- Y , E j - scattered photon Figure 4.18 The Compton effect.
and
The maximum energy of the scattered photon is obtained for 8 = 0, which essentially means that the collision did not take place. From Eqs. 4.49 and 4.50,
The conclusion to be drawn from Eq. 4.51 is that the minimum energy of the scattered photon is greater than zero. Therefore, in Compton scattering, it is impossible for all the eneqy of the incident photon to be given to the electron. The energy given to the electron will be dissipated in the material within a distance equal to the range of the electron. The scattered photon may escape.
Example 4.15 A 3-MeV photon interacts by Compton scattering. (a) What is the energy of the photon and the electron if the scattering angle of the photon is 90"? (b) What if the angle of scattering is 180"?
Answer (a) Using Eq. 4.49,
E , = = 0.437 MeV
1 + (1 - 0)3/0.511 T = 3 - 0.437 = 2.563 MeV (b) Using Eq. 4.51,
J
Eyl, min = = 0.235 MeV 1 + (2)3/0.511
T = 3 - 0.235 = 2.765 MeV
Example 4.16 What is the minimum energy of the y-ray after Compton scattering if the original photon energy is 0.511 MeV, 5 MeV, 10 MeV, or 100 MeV?
156 MEASUREMENT AND DETECTION OF RADIATION
Answer The results are shown in the table below (Eq. 4.51 has been used).
The probability that Compton scattering will occur is called the Compton coeficient or the Compton cross section. It is a complicated function of the photon energy, but it may be written in the form
where u = probability for Compton interaction to occur per unit distance f ( E J = a function of E,
If one writes the atom density N explicitly, Eq. 4.53 takes the form
In deriving Eq. 4.54, use has been made of the fact that for most materials, except hydrogen, A .= 2 2 to A = 2.62. According to Eq. 4.54, the probability for Compton scattering to occur is almost independent of the atomic number of the material. Figure 4.19 shows how u changes as a function of E, and Z. If the Compton cross section is known for one element, it can be calculated for any other by using Eq. 4.53 (for photons of the same energy):
where ul and a, are given in m-'. If a, and a, are given in m2/kg, Eq. 4.55 takes the form
Figure 4.19 Dependence of the Compton cross section on ( a ) photon energy and ( b ) atomic (b) number of the material.
4.8.3 Pair Production
Pair production is an interaction between a photon and a nucleus. As a result of the interaction, the photon disappears and an electron-positron pair appears (Fig. 4.20). Although the nucleus does not undergo any change as a result of this interaction, its presence is necessary for pair production to occur. A y-ray will not disappear in empty space by producing an electron-positron pair.+
Conservation of energy gives the following equation for the kinetic energy of the electron and the positron:
T e + T,+= E, - (mc21e- -(mc21e+ = E, - 1.022 MeV (4.56) The available kinetic energy is equal to the energy of the photon minus 1.022 MeV, which is necessary for the production of the two rest masses. Electron and positron share, for all practical purposes, the available kinetic energy, i.e.,
Te-= Te+= +(E, - 1.022 MeV) (4.57)
Pair production eliminates the original photon, but two photons are created when the positron annihilates (see Sec. 3.7.4). These annihilation gammas are important in constructing a shield for a positron source as well as for the detection of gammas (see Chap. 12).
The probability for pair production to occur, called the pair production coeficient or cross section is a complicated function of E, and Z (see Evans and Roy & Reed). It may be written in the form
where K is the probability for pair production to occur per unit distance traveled and f ( E,, Z ) is a function that changes slightly with Z and increases with E,.
Figure 4.21 shows how K changes with E, and 2. It is important to note that K has a threshold at 1.022 MeV and increases with E, and Z. Of the three coefficients ( r and a being the other two), K is the only one increasing with the energy of the photon.
'pair production may take place in the field of an electron. The probability for that to happen is much smaller and the threshold for the gamma energy is 4mc2 = 2.04 MeV.
E7 = 0.51 1 MeV
Figure 4.20 Pair production. The gamma disappears and a positron-electron pair is created. Two 0.511-MeV photons are produced when the positron annihilates.
158 MEASUREMENT AND DETECTION OF RADIATION
If the pair production cross section is known for one element, an estimate of its value can be obtained for any other element by using Eq. 4.58 (for photons of the same energy).
where K~ and K, are given in m-l. If K, and K, are given in m2/kg, Eq. 4.59 takes the form
4.8.4 Total Photon Attenuation Coefficient
When a photon travels through matter, it may interact through any of the three major ways discussed earlier. (For pair production, E, > 1.022 MeV.) There are other interactions, but they are not mentioned here because they are not important in the detection of gammas.
Figure 4.22 shows the relative importance of the three interactions as E, and Z change. Consider a photon with E = 0.1 MeV. If this particle travels in carbon ( Z = 6), the Compton effect is the predominant mechanism by which this photon interacts. If the same photon travels in iodine ( Z = 531, the photoelectric interaction prevails. For a y of 1 MeV, the Compton effect predominates regardless of Z. If a photon of 10 MeV travels in carbon, it will interact mostly through Compton scattering. The same photon moving in iodine will interact mainly through pair production.
The total probability for interaction p , called the total linear attenuation coefficient, is equal to the sum of the three probabilities:
Physically, p is the probability of interaction per unit distance.
There are tables that give p for all the elements, for many photon energies.+
' ~ a b l e s of mass attenuation coefficients are given in App. D.
Figure 4.21 Dependence of the pair production cross section on
1.022 MeV (a) photon energy and ( b ) atomic
( b ) number of the material.
Photoelectric
predominant I
E?, MeV
Figure 4.22 The relative importance of the three major gamma interactions (from The Atomic Nucleus by R. D. Evans. CopyrightO1972 by McGraw-Hill. Used with the permission of McGraw-Hill Book Company).
Most of the tables provide p in units of m2/kg (or cm2/g), because in these units the density of the material does not have to be specified. If p is given in m2/kg (or ~ m ~ / ~ ) , it is called the total mass attenuation coeficient. The relation- ship between linear and mass coefficients is
Figure 4.23 shows the individual coefficients as well as the total mass attenuation coefficient for lead, as a function of photon energy. The total mass attenuation coefficient shows a minimum because as E increases, r decreases, K
increases, and cr does not change appreciably. However, the minimum of p
does not fall at the same energy for all elements. For lead, p shows a minimum at E, - 3.5 MeV; for aluminum, the minimum is at 20 MeV; and for NaI, the minimum is at 5 MeV.
If a 'parallel beam of monoenergetic photons with intensity I(0) strikes a target of thickness t (Fig. 4.24), the number of photons, Z(t), emerging without having interacted in the target is given by
The probability that a photon will traverse thickness t without an interaction is number transmitted I(0)e-fit
- - = e p P t
number incident I(0)
160 MEASUREMENT AND DETECTION OF RADIATION
\ Total mass sttenuatior coefficient
-. -.
Compton ( 0 ) -.
Pair
\ .
\' '.
:' '. '.
: , Photo '..
Ey, MeV
Figure 4.23 Mass attenuation coefficients for lead ( Z = 82, p = 11.35 X lo3 kg/m3).
Based on this probability, the average distance between two successive interactions, called the mean free path (mfp) (A), is given by
Thus, the mean free path is simply the inverse of the total linear attenuation coefficient. If p = 10 m-I for a certain y-ray traveling in a certain medium, then the distance between two successive interactions of this gamma in that medium is A = 1/p = 1/10 m = 0.10 m.
The total mass attenuation coefficient for a compound or a mixture is calculated by the same method used for (dE/dx), in Sec. 4.5. It is easy to show (see Prob. 4.15) that
where p, = total mass attenuation coefficient for a compound or a mixture w i = weight fraction of ith element in the compound
pi = total mass attenuation coefficient of ith element
Example 4.17 What is the total mass attenuation coefficient for 1.25-MeV gammas in NaI?
Answer For this compound, the following data apply:
Na: p = 0.00546 m2/kg w = , 23 = 0.153
Using Eq. 4.64,
p (NaI) = 0.00546(0.153) + 0.00502(0.847) = 0.00509 m2/kg = 0.0509 The density of NaI is 3.67 x lo3 kg/m3; hence,
p (m-') = 0.00509 m3/kg(3.67 x lo3 kg/m3) = 18.567 m-' = 0.187 cm-' 4.8.5 Photon Energy Absorption Coefficient
When a photon has an interaction, only part of its energy is absorbed by the medium at the point where the interaction took place. Energy given by the photon to electrons and positrons is considered absorbed at the point of interaction because the range of these charged particles is short. However, X-rays, Compton-scattered photons, or annihilation gammas may escape. The fraction of photon energy that escapes is important when one wants to calculate heat generated due to gamma absorption in shielding materials or gamma radiation dose to humans (see Chap. 16). The gamma energy deposited in any material is calculated with the help of an energy absorption coefficient defined in the following way.
The gamma energy absorption coeficient is, in general, that part of the total attenuation coefficient that, when multiplied by the gamma energy, will give the energy deposited at the point of interaction. Equation 4.60 gives the total attenuation coefficient. The energy absorption coeficient pa ist
'A more detailed definition of the energy absorption coefficient is given by Chilton et al.
Incident photon beam
' 0
Figure 4.24 The intensity of the transmitted beam (only particles that did not interact) decreases exponen- tially with material thickness.
162 MEASUREMENT AND DETECTION OF RADIATION
where T, is the average energy of the Compton electron and pa may be a linear or mass energy absorption coefficient, depending on the units (see Sec.
4.8.4).
In writing Eq. 4.65, it is assumed that
1. If photoelectric effect or pair production takes place, all the energy of the gamma is deposited there.
2. If Compton scattering occurs, only the energy of the electron is absorbed.
The Compton-scattered gamma escapes.
In the case of photoelectric effect, assumption (1) is valid. For pair produc- tion, however, it is questionable because only the energy E, - 1.022 MeV is given to the electron-positron pair. The rest of the energy, equal to 1.022 MeV, is taken by the two annihilation gammas, and it may not be deposited in the medium. There are cases when Eq. 4.65 is modified to account for this effect."
Gamma absorption coefficients, as defined by Eq. 4.65, are given in App. D.
Example 4.18 A 1Ci 1 3 7 ~ s source is kept in a large water vessel. What is the energy deposited by the gammas in H 2 0 at a distance 0.05 m from the source?
Answer 1 3 7 ~ s emits a 0.662-MeV gamma. The mass absorption coefficient for this photon in water is (App. D) 0.00327 m2/kg. The total mass attenuation coefficient is 0.00862 m2/kg. The energy deposited at a distance of 0.05 m from the source is ( E d = 4p, E,)
4.8.6 Buildup Factors
Consider a point isotropic monoenergetic gamma source at a distance r from a detector, as shown in Fig. 4.25, with a shield of thickness t between source and detector. The total gamma beam hitting the detector consists of two compo- nents.
1. The unscattered beam consists of those photons that go through the shield without any interaction. If the source strength is S(-y/s), the intensity of the unscattered beam or the unscattered photon flux is given by the simple
Source
Pair Unscattered photon
Annihilation photon
Figure 4.25 If a point isotropic source is placed behind a shield of thickness t , both scattered and unscattered photons will hit the detector.
and exact expression
2. The scattered beam consists of scattered incident photons and others generated through interactions in the shield (e.g., X-rays and annihilation gammas). The calculation of the scattered beam is not trivial, and there is no simple expression like Eq. 4.62 representing it.
The total flux hitting the detector is
Obviously, for the calculation of the correct energy deposition by gammas, either for the determination of heating rate in a certain material or the dose rate to individuals, the total flux should be used. Experience has shown that rather than calculating the total flux using Eq. 4.67, there are advantages to writing the total flux in the form
where B is a buildup factor, defined and computed in such a way that Eq. 4.68 gives the correct total flux. Combining Eqs. 4.67 and 4.68, one obtains
How will B be determined? Equation 4.69 will be used, of course, but that means one has to determine the scattered flux. Then where is the advantage of using B? The advantage comes from the fact that B values for a relatively small number of cases can be computed and tabulated and then, by interpolation, one can obtain the total flux using Eq. 4.68 for several other problems. In other words, the use of the buildup factor proceeds in two steps.
164 MEASUREMENT AND DETECTION OF RADIATION
1. Buildup factor values are tabulated for many cases.
2. The appropriate value of B that applies to a case under study is chosen and used in Eq. 4.68 to obtain the total flux.
In general, the buildup factor depends on the energy of the photon, on the mean free paths traveled by the photon in the shield, on the geometry of the source (parallel beam or point isotropic), and on the geometry of the attenuating medium (finite, infinite, slab, etc.).
The formal definition of B upon which its calculation is based is quantity of interest due to total flux B(E' pr) = quantity of interest due to unscattered flux
Quantities of interest and corresponding buildup factors are shown in Table 4.4 The mathematical formulas for the buildup factors are (assuming a monoen- ergetic, E,, point isotropic source) as follows:
Number buildup factor:
Energy deposition buildup factor:
Dose buildup factor:
In Eqs. 4.70-4.72, the photon flux +(r, E ) is a function of space r and energy E, even though all photons start from the same point with the same energy E,. Since B(E, p r ) expresses the effect of scattering as the photons travel the distance r, it should not be surprising to expect B(E, p r ) + 1 as p r -+ 0.
Table 4.4 Types of Buildup Factors
Quantity of interest Corresponding buildup factor
F l w 4 Number buildup factor
Energy deposited in medium Energy deposition buildup factor
Dose (absorbed) Dose buildup factor
Note that the only difference between energy and dose buildup factors is the type of gamma absorption coefficient used. For energy deposition, one uses the absorption coefficient for the medium in which energy deposition is calcu- lated; for dose calculations, one uses the absorption coefficient in tissue.
Extensive calculations of buildup factors have been performed,26-31 and the results have been tabulated for several gamma energies, media, and distances.
In addition, attempts have been made to derive empirical analytic equations.
Two of the most useful formulas are as follows:
Berger formula:
B ( E , p r ) = 1 + a ( ~ ) p r e ~ ( ~ ) ~ - ~ ~ Taylor formula:
The constants a(E), b(E), A(E), a,(E), a2(E) have been determined by fitting the results of calculations to these analytic expressions. Appendix E provides some values for the Berger formula constants. The best equations for the gamma buildup factor representation are based on the so-called "geometric progression" (G-P)32 form. The G-P function has the form
B ( E , X ) = 1 + ( b - l ) ( K X - 1)/(K - 1) K + 1
= l + ( b - 1 ) x K = 1 (4.75)
tanh[(x/X,) - 21 - tanh(-2)
K ( x ) = a" + d (4.76)
1 - tanh( -2) where x = p r = distance traveled in mean free paths
b = value of B for x = 1
K = multiplication factor per mean free paths a , b, c , d, X, = parameters that depend on E
Extensive tables of these constants are given in Ref. 31. The use of the buildup factor is shown in Ex. 4.19. More examples are provided in Chap. 16 in connection with dose-rate calculations.
Example 4.19 A 1-Ci 137Cs source is kept in a large water tank. What is the energy deposition by the Cs gammas at a distance of 0.5 m from the source?
Answer Using the data of Ex. 4.18, the distance traveled by the 0.662-MeV photons in water is p r = (0.00862 m2/kg)(0.5 mX103 kg/m3) = 4.31 mean free path. From Ref. 32, the energy deposition buildup factor is B(0.662,4.31) = 13.5.
The energy deposition is MeV 3.7 x 10''
( kgs ) = 4 ~ ( 0 . 5 ) ~
eC4.31(0.00327)(0.662)13.5 = 4.62 X lo6 MeV/(kg s)