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The attenuation of gamma radiation (photons) by an absorber is qualitatively dif-ferent from that of either alpha or beta radiation. Whereas both these corpuscular radiations have definite ranges in matter and therefore can be completely stopped, gamma radiation can only be reduced in intensity by increasingly thicker absorbers;

it cannot be completely absorbed. If gamma-ray attenuation measurements are made under conditions ofgood geometry, that is, with a well-collimated, narrow beam of radi-ation, as shown in Figure 5-12, and if the data are plotted on semilog paper, a straight line results, as shown in Figure 5-13, if the gamma rays are monoenergetic. If the

Figure 5-12. Measuring attenuation of gamma rays under conditions of good geometry. Ideally, the beam should be well collimated and the source should be as far away as possible from the detector;

the absorber should be midway between the source and the detector and should be thin enough so that the likelihood of a second scattering of a photon already scattered by the absorber is negligible;

and there should be no scattering material in the vicinity of the detector.

Figure 5-13. Attenuation of gamma rays under conditions of good geometry. The solid lines are the attenuation curves for 0.662-MeV (monoenergetic) gamma rays. The dotted line is the attenuation curve for a heterochromatic beam.

gamma-ray beam is heterochromatic, a curve results, as shown by the dotted line in Figure 5-13. The equation of the straight line in Figure 5-13 is

lnI = −μt+lnI0 (5.26a)

or ln I

I0

= −μt. (5.26b)

Taking the inverse logs of both sides of Eq. (5.26b), we have, I

I0

=e^−μt, (5.27)

where

I0=gamma-ray intensity at zero absorber thickness, t =absorber thickness,

I =gamma-ray intensity transmitted through an absorber of thicknesst, e =base of the natural logarithm system, and

μ=slope of the absorption curve=the attenuation coefficient.

Since the exponent in an exponential equation must be dimensionless, μ and t must be in reciprocal dimensions, that is, if the absorber thickness is measured in centimeters, then the attenuation coefficient is called thelinear attenuation coefficient, μl, and it must have dimensions of “per cm.” Ift is in g/cm², then the absorption coefficient is called themass attenuation coefficient,μm, and it must have dimensions of per g/cm²or cm²/g. The numerical relationship betweenμlandμm, for a material whose density isρg/cm³, is given by the equation

μlcm⁻¹=μm

cm²

g ×ρ g

cm³. (5.28)

The attenuation coefficient is defined as the fractional decrease, or attenuation of the gamma-ray beam intensity per unit thickness of absorber, as defined by the equation below:

limit

t→0

I/I

t = −μ, (5.29)

where I/I is the fraction of the gamma-ray beam attenuated by an absorber of thicknesst. The attenuation coefficient thus defined is sometimes called thetotal attenuation coefficient.Values of the attenuation coefficients for several materials are given in Table 5-3.

For some purposes, it is useful to use the atomic attenuation coefficient,μa. The atomic attenuation coefficient is the fraction of an incident gamma-ray beam that is attenuated by a single atom. Another way of saying the same thing is that the atomic attenuation coefficient is the probability that an absorber atom will interact with one of the photons in the beam. The atomic attenuation coefficient may be defined by the equation

μacm² = μ1

1 cm N atoms

cm³

, (5.30)

QUANTUM ENERGY (MeV)

ρ,(g/cm³) 0.1 0.15 0.2 0.3 0.5 0.8 1.0 1.5 2 3 5 8 10

C 2.25 0.335 0.301 0.274 0.238 0.196 0.159 0.143 0.117 0.100 0.080 0.061 0.048 0.044

Al 2.7 0.435 0.362 0.324 0.278 0.227 0.185 0.166 0.135 0.117 0.096 0.076 0.065 0.062

Fe 7.9 2.72 1.445 1.090 0.838 0.655 0.525 0.470 0.383 0.335 0.285 0.247 0.233 0.232

Cu 8.9 3.80 1.830 1.309 0.960 0.730 0.581 0.520 0.424 0.372 0.318 0.281 0.270 0.271

Pb 11.3 59.7 20.8 10.15 4.02 1.64 0.945 0.771 0.579 0.516 0.476 0.482 0.518 0.552

Air 1.29×10⁻³ 1.95×10⁻⁴ 1.73×10⁻⁴ 1.59×10⁻⁴ 1.37×10⁻⁴ 1.12×10⁻⁴ 9.12×10⁻⁵ 8.45×10⁻⁵ 6.67×10⁻⁵ 5.75×10⁻⁵ 4.6×10⁻⁵ 3.54×10⁻⁵ 2.84×10⁻⁵ 2.61×10⁻⁵

H₂O 1 0.167 0.149 0.136 0.118 0.097 0.079 0.071 0.056 0.049 0.040 0.030 0.024 0.022

Concrete^a 2.35 0.397 0.326 0.291 0.251 0.204 0.166 0.149 0.122 0.105 0.085 0.067 0.057 0.054

aOrdinary concrete of the following composition: 0.56%H, 49.56% O, 31.35%Si, 4.56%Al, 8.26%Ca, 1.22%Fe, 0.24%Mg, 1.71%Na, 1.92%K, 0.12%S.

Source: From White G. X-ray Attenuation Coefficients. Washington, DC: US Government Printing Office; 1952. NBS Report 1003.

where N is the number of absorber atoms per cm³. Note that the dimensions of μaare cm², the units of area. For this reason, the atomic attenuation coefficient is almost always referred to as the cross sectionof the absorber. The unit in which the cross section is specified is thebarn, b.

1 b=10⁻²⁴cm².

The atomic attenuation coefficient is also called themicroscopic cross section and is symbolized byσ, while the linear attenuation coefficient is often called themacroscopic cross sectionand is given by the symbol. This nomenclature is almost always used in dealing with neutrons. Equation (5.30) can thus be written as

cm⁻¹=σ cm²

atom×Natoms

cm³ . (5.31)

Using the relationship given in Eq. (5.31), Eq. (5.27) may be rewritten as I

I0

=e^−μat =e^−σNt. (5.32)

The linear attenuation coefficient for a mixture of materials or an alloy is given by μl=μa1×N1+μa2×N2+ · · · =

n n=1

μan×Nn, (5.33)

where

μn=atomic coefficient of thenth element and N_n=number of atoms per cm³ of thenth element.

The numerical values forμahave been published for many elements and for a wide range of quantum energies.^*With the aid of atomic cross sections and Eq. (5.33), we can compute the attenuation coefficients of compounds or alloys containing several different elements.