Tables for
Volume H
Powder diffraction
Edited by C. J. Gilmore, J. A. Kaduk and H. Schenk

International Tables for Crystallography (2018). Vol. H, ch. 3.2, pp. 257-258

Section Absorption within a homogeneous sample

P. W. Stephensa*

aDepartment of Physics and Astronomy, Stony Brook University, Stony Brook, NY 11794–3800, USA
Correspondence e-mail: Absorption within a homogeneous sample

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In equations (3.2.2)[link] and (3.2.10)[link], it was assumed that the neither the incident nor the diffracted radiation is absorbed within the sample. However, it is generally the case that neutrons or X-rays are attenuated as they travel through any material, such that the fraction of original intensity surviving after a distance x is I(x)/I(0) = exp(−μx). Here μ is the linear absorption coefficient, which generally depends strongly on the composition of the sample and the X-ray wavelength.

X-ray attenuation coefficients for elements are often given as mass attenuation coefficients, μ/ρ, in units of cm2 g−1. These are available in various sources, such as International Tables for Crystallography, Volume C, Table[link] , or from internet resources such as . The exact positions of X-ray absorption edges can depend on the chemical environment of the atom, and so tabulated or computed atomic absorption coefficients are not entirely trustworthy within about ±100 eV of an absorption edge. The X-ray mass attenuation coefficient is related to the imaginary part of the atomic scattering factor [equation (3.2.6)[link]] as μm = 2reλf ′′/m, where m is the atomic mass. For a compound or other mixture of elements of total density ρ in which the (dimensionless) mass fraction of element i is [g_m^i], the X-ray linear absorption constant is given by[ \mu_{\rm X\hbox{-}ray} = \rho \textstyle\sum\limits_i g_m^i (\mu_m/\rho)^i. ]

In the case of neutrons used for powder diffraction, the absorption cross section is typically tabulated in barns (1 barn = 10−24 cm2); see Table in International Tables for Crystallography, Volume C or websites such as . The neutron absorption cross section is generally inversely proportional to velocity, and values are usually tabulated for neutrons with a speed of 2200 m s−1 (i.e., 25.3 meV kinetic energy, 1.80 Å wavelength). For neutrons, the equivalent expression with absorption cross sections depends on the number densities [g_n^i] (atoms/volume) of each element:[\mu_{\rm neutron} = \textstyle\sum\limits_i g_n^i (\sigma^i_{\rm abs} + \sigma^i_{\rm inc}).]

Considering absorption, the effective volume is given by an integral over the sample,[V_{\rm eff} = \textstyle\int {\rm d}^3\, {\bf R} \exp [- \mu (L_{\rm in} + L_{\rm out})],\eqno(3.2.15)]where Lin and Lout are the paths of the incident and diffracted radiation to the point R within the sample, respectively.

The simplest case is of a sample in the form of a flat plate, with a thickness significantly greater than 1/μ, in the dividing position so that the angles of incidence and diffraction from the plane of the sample are both equal to θ. Let the dimensions of the beam be W in the equatorial direction (in the diffraction plane) and H in the axial direction (parallel to the diffractometer axis, i.e., perpendicular to the scattering plane) as shown in Fig. 3.2.2[link]. Assume that the detector is large enough in the equatorial direction that all diffracted radiation is captured (i.e., no loss of signal due to parallax). Radiation scattered from a volume element a distance z below the surface of the sample will have a total path length 2z/sin θ; integration yields an effective sample volume of WH/2μ, independent of diffraction angle. The situation is more complicated if a flat sample is not optically deep or if the sample is shorter than W/sin θ; in either case, computation of the effective volume is straightforward, as long as the sample geometry is known with sufficient accuracy. The calculations are described in detail in Chapter 5.4[link] of this volume.

[Figure 3.2.2]

Figure 3.2.2 | top | pdf |

Sketch illustrating symmetrical diffraction from a flat sample.

Another important case is that of a cylindrical sample, such as a Lindemann capillary for X-rays or a can of appropriate material (often vanadium) for neutrons. The integration for Veff above is given in convenient form for computation by Ida (2010[link]), and is tabulated in International Tables for Crystallography, Volume C, Section[link] . In the case of monochromatic radiation (neutrons or X-rays), for a given absorption constant μ, Veff is an increasing function of diffraction angle 2θ so that higher-angle peaks will appear relatively stronger. Weak-to-moderate absorption from a cylindrical sample affects the observed intensities in essentially the same algebraic form as a negative effective Debye–Waller factor (Hewat, 1979[link]), leading to significantly underestimated (perhaps even negative) measured Debye–Waller factors. In the case of pulsed neutrons, the effect is compounded by the fact that the absorption cross section is generally inversely proportional to the neutron velocity; consequently, in a given detector, faster neutrons, representing larger |G|, will suffer less attenuation in the sample.


Hewat, A. W. (1979). Absorption corrections for neutron diffraction. Acta Cryst. A35, 248.Google Scholar
Ida, T. (2010). Efficiency in the calculation of absorption corrections for cylinders. J. Appl. Cryst. 43, 1124–1125.Google Scholar

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