Tsoulfanidis N. Measurement and detection of radiation
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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.+
'~ables 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.
ENERGY LOSS
AND
PENETRATION OF RADIATION THROUGH
IMA'ITER
159
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
-
-
=
epPt
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
ENERGY
LOSS
AND
PENETRATION
OF
RADIATION THROUGH
MATTER
161
where
p,
=
total
mass
attenuation coefficient for a compound or a mixture
wi
=
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:
23
Na:
p
=
0.00546 m2/kg w
=
,
=
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 137~s source is kept in a large water vessel. What is the
energy deposited by the gammas in H20 at a distance 0.05 m from the source?
Answer
137~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
(Ed
=
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
ENERGY LOSS
AND
PENETRATION OF RADIATION THROUGH MA7TER
163
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, pr) expresses the effect of scattering as the photons
travel the distance r, it should not be surprising to expect
B(E, pr)
+
1
as
pr
-+
0.
Table
4.4
Types
of
Buildup Factors
Quantity of interest
Corresponding buildup factor
Flw
4
Number buildup factor
Energy deposited in medium
Energy deposition buildup factor
Dose (absorbed) Dose buildup factor
ENERGY
LOSS
AND
PENETRATION
OF
RADIATION
THROUGH
MA'ITER
165
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, pr)
=
1
+
a(~)pre~(~)~-~~
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)(KX
-
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
=
pr
=
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 pr
=
(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)
166
MEASUREMENT AND DETECIION OF RADIATION
4.9 INTERACTIONS OF NEUTRONS WITH MATTER
Neutrons, with protons, are the constituents of nuclei (see Sec. 3.4). Since a
neutron has no charge, it interacts with nuclei only through nuclear forces.
When it approaches a nucleus, it does not have to go through a Coulomb
barrier, as a charged particle does.
As
a result, the probability (cross section) for
nuclear interactions is higher for neutrons than for charged particles. This
section discusses the important characteristics of neutron interactions, with
emphasis given to neutron cross sections and calculation of interaction rates.
4.9.1 Types of Neutron Interactions
The interactions
of neutrons with nuclei are divided into two categories:
scattering and absorption.
Scattering.
In this type of interaction, the neutron interacts with a nucleus, but
both particles reappear after the reaction.
A
scattering collision is indicated as
an (n, n) reaction or as
Scattering may be elastic or inelastic. In elastic scattering, the total kinetic
energy of the two colliding particles is conserved. The kinetic energy is simply
redistributed between the two particles. In inelastic scattering, part of the
kinetic energy is given to the nucleus as an excitation energy. After the collision,
the excited nucleus will return to the ground state by emitting one or more
-prays.
Scattering reactions are responsible for neutron's slowing down in reactors.
Neutrons emitted in fission have an average energy of about
2
MeV. The
probability that neutrons will induce fission is much higher if the neutrons are
very slow-"thermal"-with kinetic energies of the order of eV. The fast
neutrons lose their kinetic energy as a result of scattering collisions with nuclei
of a "moderating" material, which is usually water or graphite.
Absorption.
If the interaction is an absorption, the neutron disappears, but one
or more other particles appear after the reaction takes place. Table 4.5 illus-
trates some examples of absorptive reactions.
4.9.2 Neutron Reaction Cross Sections
Consider a monoenergetic parallel beam of neutrons hitting a thin targett of
thickness
t
(Fig. 4.26). The number of reactions per second,
R,
taking place in
'A
thin target is one that does not appreciably attenuate the neutron beam (see
Eq.
4.80).
ENERGY
LOSS
AND
PENETRATION
OF
RADIATION
THROUGH
MATIER
167
Table
4.5
Absorptive Reactions
Reaction Name
~+;X+~-(Y
+P
(n,
p)
reaction
n
+
$X
+$I:Y
+
:He
(n,
a)
reaction
n
+$x+~-~x+
~n
(n,
2n)
reaction
n
+$x+*+;x+~
(n ,7)
reaction
n+$~+~l~,
+*,
+n+n+..-
z,
fkon
this target may be written as
neutrons per m2 s
R
(reactions/s)
=
hitting the target
probability of interaction
per n/m2 per nucleus
where
I,
a,
and
t
are shown in Fig.
4.26.
The parameter
a,
called the
cross
section,
has the following physical meaning:
a
(m2)
=
probability that an interaction will occur per target nucleus
per neutron per m2 hitting the target
The unit of
a
is the barn (b).
Since the nuclear radius is approximately
10-l5
to
10-l4
m,
1
b is approximately
equal to the cross-sectional area of a nucleus.
e
Target
(A,
Z)
Figure
4.26
A
parallel neutron beam
hitting a thin target:
a
=
area
of
target
struck by the beam.