Consider a monochromatic beam of gamma rays. Let the radiation intensity at point K, recorded by the device in the absence of an absorber, be equal to I_{0 } (Fig. 19, a). If a very thin layer of substance with a thickness of Δx is placed between the radiation source and the device, then the radiation intensity at point K will change by a small amount and become equal to I (Fig. 19, b). A small change in intensity is denoted by

Fig.19. Attenuation of a narrow beam of radiation.

ΔI, so I_{0}– I= -ΔI^{1}. The sign Δ means that the value in question is very small.Measurements carried out with various absorbers and radiations of various energies showed that

-ΔI=μI Δx, (3.11)

I.e. When radiation passes through thin layers of the absorber, the change in radiation intensity is proportional to the thickness of the substance layer and its initial intensity (in the absence of an absorber), and also depends on the magnitude of μ, which is a function of the energy of the incident radiation and the kind of substance and is called the* linear attenuation coefficient.*

*1 Since the absorption of gamma rays in the air is negligible, it can be neglected. The minus sign at ΔI indicates that the intensity is weakening.*

The greater the energy of the incident photons (the rigidity of the radiation), the less attenuation, i.e. The attenuation coefficient decreases with increasing radiation. If Δx is measured in cm, then μ will be measured in cm^{-1}.

Let’s put Δx=1, then from (3.11) we get

(3.12)

Thus, *the linear attenuation coefficient characterizes the relative change in the intensity of radiation when passing through a layer of substance with a thickness equal to one.*

Dividing and multiplying the right part (3.11) by the density of the absorber ρ, we get

(3.13)

Let us now consider what the value Δxρ is.As is known, density p is the mass of a substance per unit volume; it is usually expressed in *g/cm ^{3}*, therefore, Δxρ represents the mass of a substance enclosed in a cylinder or parallelepiped with a base area equal to 1 cm

^{2}and a length of Δx cm; obviously, Δxρ is measured in units of

*g/cm*.If we put Δxρ=1, then from (3.13) it follows that

^{2}

(3.14)

μ/p is called the mass attenuation coefficient and characterizes the relative attenuation of the intensity of the radiation flux during the passage of a layer of substance with a mass of 1 g enclosed in a cylinder or parallelepiped with a base area equal to 1 cm^{2}.Accordingly, μ/p is measured in cm^{2}/g. Sometimes atomic µ_{A} or electronic µe attenuation coefficients are used; they characterize the relative change in the intensity of radiation when passing through a substance calculated for 1 atom or 1 electron. Accordingly, μ/p is measured in cm^{2}/g.Sometimes atomic µ_{A} or electronic µ_{e} attenuation coefficients are used; they characterize the relative change in the intensity of radiation when passing through a substance calculated for 1 atom or 1 electron. Thus,

and

where, N_{A} is the number of atoms in 1 cm^{3} of the absorber, and Z is the atomic number.

From the expression (3.11), by integration, it is possible to obtain the ratio between the radiation intensities I_{0} and I at a given point in the absence of an absorber and after passing through an absorber layer with a thickness of* d cm*, respectively. This relation has the form:

(3.15)

where μ is the linear attenuation coefficient.

If the radiation intensity I_{0}^{’} generated by a point source at a distance of 1 cm or 1m is known, then, according to (3.10) and (3.15), its intensity I^{’} at a distance of R cm or R m after passing through the absorber layer with a thickness of *d cm* is equal to

If the thickness of the absorber m is expressed in units of g/cm^{2}, then (3.15) will be written as follows:

(3.16)

The above expressions are valid only for a narrow beam of monochromatic radiation.

* The thickness of the absorber d*

Fig. 20. Attenuation of the radiation intensity depending on the thickness of the absorber.

a- linear scale; b- semi-logarithmic scale.

Thus, the attenuation of the intensity of the X-ray or gamma ray flux occurs exponentially. 20, *a* shows an exponential curve, or exponent, showing the nature of the change in the intensity of a narrow beam of monochromatic radiation depending on the thickness of the absorber. We see that when the radiation passes through an arbitrarily large thickness of the absorber, its intensity does not decrease to zero. This means that it is impossible to completely absorb X-rays or gamma rays, but you can only reduce their intensity to an arbitrarily small value.

If the logarithm of intensity — ln I is postponed on the ordinate axis, then the dependence of the intensity change on the thickness of the absorber layer will be expressed as a straight line (Fig. 20, b).

*The half-attenuation layer*. From fig.20, *a* it follows that as the thickness of the absorber increases, the initial radiation intensity I_{0} gradually decreases and with a certain thickness of the absorber equal to Δ_{0.5}, the intensity I becomes equal to I_{0}/2, i.e. It is reduced by half. *The thickness of the absorber, after passing through which the radiation intensity is halved, is called the half attenuation layer*, it is usually denoted by the symbol Δ_{0.5} and measured in cm. After passing through the absorber, the thickness of which is equal to two layers of attenuation, the radiation intensity decreases by 4 times, i.e. I= I_{0}/4 or I/I_{0}=1/4=(1/2)^{2}; after passing through three layers of semi-attenuation, the radiation intensity decreases by 8 times: I/ I_{0}=1/8=(1/2)^{3}. Thus, in order to reduce the radiation intensity by 2^{n} times (i.e. I/ I_{0} =(1/2)^{n}), it is necessary that the thickness of the absorber be equal to n layers of half attenuation.

Knowing the half-attenuation layer, it can be quite simple without using the formula (3.15), calculate what thickness of the absorber is necessary to attenuate the radiation by a certain number of times. For example, in order to attenuate the radiation by N times (i.e. I/ I_{0}=1/ N), one should take as many layers of semi-attenuation as many times the number 2 is a factor in the number N.

It is easy to show that the following relationship exists between the linear attenuation coefficient μ and the half attenuation layer Δ_{0.5}:

(3.17)

and vice versa,

(3.18)

Substituting the value of μ from (3.18), the formula (3.15) can be rewritten in the following form:

(3.19)

We derive the ratio (3.17). From the definition it follows that if the thickness of the absorber is d = Δ_{0.5}, then I = I_{0}/2.Substituting these values in (3.15), we get

or

Prologarithm the right and left parts using natural logarithms (the base of natural logarithms is the number e):

Since ln1=0, and ln e=1, we have

where from

*Example.* Determine how many layers of half attenuation should be taken so that the radiation intensity decreases by a factor of 100 (I/ I_{0} = 1/100).

To do this, you need to determine how many times the number 2 is a factor in the number 100. This is done as follows:

Therefore, the number 2 is about 6.5 times a factor of 100, and, therefore, it is necessary that the thickness of the absorber is equal to about 6.5 layers of half attenuation.

More generally, this can be written as follows. It follows from (3.18) that

or

Suppose that d=n·Δ_{0.5} , then

hence the number of half-attenuation layers

In our example, I/ I_{0}=1/100, ln I/I_{0}= – 4.605; representing we get

*A layer of tenfold weakening.* The thickness of the absorber, after passing through which the radiation intensity decreases by 10 times, i.e. I = I_{0}/10, is called a tenfold attenuation layer and is denoted Δ_{0,1} (see Fig. 20). This value is convenient to use when calculating protection for sources of high activity, the radiation intensity of which must be weakened hundreds or thousands of times.

It follows from the definition that after passing an absorber whose thickness is equal to one layer of tenfold attenuation, the radiation intensity decreases by 10 times, i.e. I/ I_{0} = 0.1; after passing an absorber whose thickness is equal to two layers of tenfold attenuation, the radiation intensity decreases by 100 times, I/ I_{0} = 0.01 = 1/10^{2}, etc. Thus, in order to attenuate the radiation by a factor of 10n [I/ I_{0}= (1/10)^{n}], it is necessary that the thickness of the absorber is equal to n layers of tenfold attenuation.

In the same way as for the half-attenuation layer, it can be shown that there is the following relationship between the tenfold attenuation layer and the linear attenuation coefficient:

(3.20)

Thus, the formula (3.15) can be written in the following form:

(3.21)

*The nature of attenuation of non-monochromatic radiation. *Any non – monochromatic radiation can be considered as monochromatic with some effective energy E_{eff} and an effective attenuation coefficient µ_{eff} .When passing through a substance , the radiation intensity of such monochromatic radiation is weakened as follows

Figure 21 Changes in the attenuation coefficient of a narrow beam of gamma radiation Ir^{192} and Ra depending on the thickness of the lead absorber.

the same as the intensity of the non-monochromatic radiation under consideration. Consequently, for a narrow beam of non-monochromatic radiation, the same attenuation law is valid as for monochromatic radiation, i.e.

(3.22)

The effective attenuation coefficient µ_{eff} of non-monochromatic radiation, in contrast to the attenuation coefficient of monochromatic radiation μ, depends not only on the type of substance of the absorber and the radiation energy, but also on the thickness of the absorber. This is due to the fact that as the various components of the spectrum of non-monochromatic radiation pass through the absorber, they will be absorbed differently; therefore, the spectral composition of the radiation will change with a change in the thickness of the absorber. Since low-energy photons weaken more strongly, the proportion of high-energy photons in the spectrum will increase with an increase in the thickness of the absorber, i.e. the radiation hardness will increase, and the µ_{eff} will decrease accordingly. As an example, FIG. 21 shows the nature of the change in the effective attenuation coefficient µ_{eff }of a narrow beam of gamma rays Ir^{192} and Ra depending on the thickness of the lead absorber.

If the radiation intensities of each component I_{01}, I_{02}, etc. are known for the linear spectrum, then the radiation intensity I after each passage of the absorber with thickness d can be calculated by the formula

(3.23)

where, μ_{1}, μ_{2}, etc. are, respectively, the linear attenuation coefficients of each component of the spectrum.

When radiation passes through a substance, its intensity decreases. Consequently, radiation interacts with matter, transferring part of its energy to it.

This transfer of energy from radiation to matter is due to five different elementary processes:

- photoelectric absorption.
- coherent scattering.
- incoherent scattering.
- the formation of pairs.
- nuclear photo effect.

In the first three cases, photons interact with the electrons of the atoms of the substance through which the radiation passes; the formation of pairs and the nucleated photoelectric effect represent the processes of interaction of photons with nuclei. The probability of each of these processes depends on the spectral composition of the radiation, the atomic number and the density of the absorbing medium.

From the book “Protection from X-rays and gamma rays”

A.V. Bibergal, U.Y. Margulis, E.I. Vorobyov.

Medgiz 1955