NDT.net • Dec 2004 • Vol. 12 No.12

## X rays linear attenuation coefficient in steel. I. Thickness dependence

V. Dorobantu
Physics Department, "Politehnica" University,
Timisoara, Romania, Tel/Fax: + 40 -256 - 472801

Corresponding Author Contact:
V. Dorobantu, Email: wdorobantzu@zappmobile.ro

### Abstract

The polychromatic nature of X rays involves a thickness dependence of the linear attenuation coefficient - µ - , and starting from these physical considerations an expression for µ is obtained and verified for X rays absorption in steel.

### 1. Introduction

X ray inspections in order to determine whether a steel structure is suitable for further use became a current practice. The steel structures, as well as other materials, may be damaged in a lot of ways: corrosion, erosion, deposits, mechanical, etc. Using X rays as non - destructive testing (NDT) agent, one can find where the defect is placed and (very important) what is its size, and for this purpose, the linear attenuation coefficient of X rays must be known.

### 2. Theoretical background

The essence of the X - ray technique is how much of the incident radiation, on certain material, will remain after traversing the material. Let us suppose we have a primary X-rays intensity I0 incident on the surface of a material. After crossing the material, with the thickness x, we have got intensity I, the connection between I0 and I being:

 I (X) = I0e µx (1)

where µ is called the linear attenuation factor, or absorption factor. Intensity I has the usual meaning, being the average of the Poynting vector S, namely

 I (X) = e0 c {Eµx} (2)

with e0, vacuum dielectric constant, c, light speed in vacuum, {Eµx} the average of the electric field squared. As one can see from the above expression for I, the entire process of absorption is contained in µ. Writing the refractive index [1] in the general form:

 n = n1 - i n2 (3)

where n1 and n2 are real numbers and i - imaginary number ( i2 = -1), only n2 is responsible for absorption.

 (4)

where E0 and w are the electric field and the frequency of the incident electromagnetic wave. Comparing expressions (1) and (4) we have µ.

 (5)

The aetiology of the X - rays' attenuation recalls, mainly, three aspects [2]:

a) photoelectric effect responsible for x - rays' collisions with bound electrons;
b) Compton effect due inelastic collisions between X rays and free electrons;
c) electron-positron pair production, when the energy of electromagnetic radiation is greater than the double of the electron rest energy (1.02 MeV).
> I shall not enter the details regarding the refractive index affecting the absorption, because it will be the subject of other paper.

### 2. Thickness dependence

As one can see from expression (5), µ is defined for a monochromatic radiation and formula (1) holds if this condition of monochromaticity is fulfilled. X rays, being the result of the deexcitations process of the highly excited states of an atom, are polychromatic radiations and because they have different energies, the interactions with the traversed material are also, different. It means that the process of absorption of X rays can be written as contributions of all waves from that beam:

 (6)

where wj is the weight of the j - X ray component from the total number n. Taking the logarithm in (6), we will get an important result: the linear attenuation coefficient depends on thickness x.

In the above expression m is the number of terms we take in the series expansion. Further, if we take another series expansion of the natural logarithm in (8) we get:

 (10)

where bk are coefficients depending on different µ's, and j is the number of terms we take in series expansion. Also, if we evaluate a1 from (9), we can get:

 (11)

One can see from the above expression, that the linear attenuation coefficient of electromagnetic waves, of any kind, decreases with traversed thickness x. The coefficients, µaverage and bk , can be determined from experimental data.

### 3. Experiments

Two reference blocks of steel have been studied: 168.3 mm outer diameter and 14 mm wall's thickness, and 273 mm outer diameter and 10 mm wall's thickness, each of them with inside and outside steps, inside and outside holes with depth of 10%, 20% and 50% from the remaining wall's thickness, as well as ground patch on the inside surface and flat area on the outside surface according to[3] . We have irradiated with X - rays the two reference blocks, first the halves of each cylinder on all holes, and patch, and flat area, then the entire tube. We have made measurements on halves of pipelines to have enough data in getting an empirical formula for µ. Also, we made measurements on a piece of massive cylinder steel with 10 cm diameter and with steps at 450 of different thicknesses. The voltage we have used was between 120 and 240 kV.

### 4. Results

We have tried to find an expression for attenuation factor µ as a function of the thickness traversed by X rays. The process of film's blackening can roughly be described as follows: the electromagnetic waves impinges the film and set free electrons from negative Br ions. These electrons are captured by electron traps, so the free positive Ag ions are attracted by these negative centers becoming neutral and blackening the film. So, the number of neutral Ag atoms formed in film is directly proportional with the number of electromagnetic waves incident on film. At its turn, the number of photons incident on film is that number of photons which have traversed the material to be studied, and according to formula (1) is exponentially decreased, and as a result the film's blackening density can be written as:

 D = D0 eµx (12)

where D-film blackening density, D0 a quantity depending on the current of X ray source, on the film to source distance, exposure time t, and some constants which are characteristic to the film we use. Because of the thickness dependence of µ, we have different µ's on the defect and near it. Measuring the film densities on the defect (D1) and just near it (D2), see Fig. 1, then the ratio of the two D's is a function of the dimension of the defect, as well as of the entire thickness traversed by electromagnetic waves.

The ratio:

 (13)

where Dfog is the fog density. For µ as a function of the thickness x, we can use the expression (11) but it seems to be unpractical to get the dimension of the defect by solving an algebraic equation of a degree higher than two. As a consequence, is desirable to get a simpler function for µ instead of a polynomial. Such a function it was proved to be:

The problem is to find a and b with a reasonable approximation, so to be able to get a good (trusting) results for the dimension of the defect. We have measured blackening densities of the AGFA D7 film for different material thicknesses ( 3 - 28 mm), different accelerating voltages (120 - 240 kV), and different dimensions of defects ( 0.5 - 5.5 mm). Knowing the dimensions of the defects, we have got the a and b coefficients for µ(x). Here are some results. I have to mention that the results refer to wide X rays beam, namely without collimation, which, in fact, is the usual situation we deal with. Accelerating voltage U = 140 kV.

 (17)

With this µ, the calculated dimensions of defects versus real dimensions looks like this:

 Fig.2

The errors are well within 10% Accelerating voltage, U = 200 kV.

 (18)

The calculated defects' dimensions versus real dimensions are:

 Fig.3

In both cases, as well as in general, the defect's dimension is calculated solving the second degree equation resulting from (16). In the general case the solution is:

 (19)

Here is a comparison of the experimental data [4],[5],our own, for µ, and the calculated, when U = 200 kV. The errors are within 6%.

 Fig.4

### References

1. Richard Feynman, " The Feynman Lectures on Physics Vol.2", Addison- Wesley Pub. Co.1966
2. Emilio Segré, " Nuclei and Particles", W.B. Benjamin, Inc., New York,1965
3. " Report of the 1st research coordination meeting on Validation of Protocols for Corrosion and Deposit Evaluation in Large Diameter Pipes by Radiography", Vienna, Nov. 2002