Tutorial: Attenuation of X-Rays By Matter - In this X-ray tutorial, the authors attempt to answer the frequently asked question, "How deep do the X-rays penetrate my sample?" - Spectroscopy
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Tutorial: Attenuation of X-Rays By Matter
In this X-ray tutorial, the authors attempt to answer the frequently asked question, "How deep do the X-rays penetrate my sample?"
 Sep 1, 2005 Spectroscopy Volume 20, Issue 9, pp. 22-25

X-rays are attenuated as they pass through matter. That is, the intensity of an X-ray beam decreases the farther it penetrates into matter. Basically, each interaction of an X-ray photon with an atom of the material removes an X-ray from the beam, decreasing its intensity.

The amount of decrease in intensity of the X-ray beam depends upon two factors:

• The depth of penetration (x) or thickness
• A characteristic of the material called its "absorption coefficient" (A).

The intensity decreases exponentially with the distance traveled, or

I = I 0exp (– Ax)

where I 0 is the initial X-ray beam intensity. Note that this exponential decay of photon intensity applies in the optical region of the electromagnetic spectrum as well. In this region, it is known as the Beer–Lambert law.

The quantity A is the linear absorption coefficient. The quantity usually encountered in tabulations of material properties is the mass absorption coefficient (μ). These two coefficients are related by the density of the material (ρ) as μ = A/ρ.

Attenuation Length

An interesting application of this equation is to determine the depth of penetration of X-rays. The attenuation length is defined as the depth into the material where the intensity of the X-rays has decreased to about 37% (1/e) of the value at the surface. That is, I = (1/e)I 0, or I/I 0 = 1/e. [Recall that e, sometimes called Euler's number or Napier's constant, is the base of natural logarithms, or e ≈ 2.7183.] Then, substituting into Equation 1, we get

(I/I 0) = exp (–μρx)
ln (1/e) = (–μρx)
–1 = (–μρx)
x = 1/(μρ)

This also is referred to as the "mean free path" of the X-rays.

 Table 1
For example, given a 109Cd radioisotope source of X-rays with an energy of 22.1 keV, how far do these penetrate a piece of pure iron, (atomic number 26)? From Table I, we find the mass absorption coefficient for iron at 22.1 keV is μ = 18.2 cm2/g. The density of iron ρ = 7.86 g/cm3. Plugging in the numbers, we find x = 0.007 cm = 0.07 mm = 70 μm. A comparison of this depth for the same incoming X-ray energy both for lighter and heavier elements is shown in Table I.