# 1.7 A practical introduction to x-ray absorption spectroscopy  (Page 4/5)

 Page 4 / 5

## Transmission and fluorescence modes

X-ray Absorption measurements can be performed in several modes: transmission, fluorescence and electron yield; where the two first are the most common. The choice of the most appropriate mode to use in one experiment is a crucial decision.

The transmission mode is the most used because it only implies the measure of the X-ray flux before and after the beam passes the sample. Therefore, the adsorption coefficient is defined as [link] . Transmission experiments are standard for hard X-rays, because the use of soft X-rays implies the use the samples thinner than 1 μm . Also, this mode should be used for concentrated samples. The sample should have the right thickness and be uniform and free of pinholes.

${\mu }_{E}=\text{ln}\left(\frac{{I}_{0}}{I}\right)$

The fluorescence mode measures the incident flux I 0 and the fluorescence X-rays I f that are emitted following the X-ray absorption event. Usually the fluorescent detector is placed at 90° to the incident beam in the horizontal plane, with the sample at an angles, commonly 45°, with respect to the beam, because in that position there is not interference generated because of the initial X-ray flux ( I 0 ). The use of fluorescence mode is preferred for thicker samples or lower concentrations, even ppm concentrations or lower. For a highly concentrated sample, the fluorescence X-rays are reabsorbed by the absorber atoms in the sample, causing an attenuation of the fluorescence signal, it effect is named as self-absorption and is one of the most important concerns in the use of this mode.

## Uniformity

The samples should have a uniform distribution of the absorber atom, and have the correct absorption for the measurement. The X-ray beam typically probes a millimeter-size portion of the sample. This volume should be representative of the entire sample.

## Thickness.

For transmission mode samples, the thickness of the sample is really important. It supposes to be a sample with a given thickness, t , where the total adsorption of the atoms is less than 2.5 adsorption lengths, µ E t ≈ 2.5; and the partial absorption due to the absorber atoms is around one absorption length ∆ µ E t ≈ 1, which corresponds to the step edge.

The thickness to give ∆ µ E t = 1 is as [link] . where ρ is the compound density, n is the elemental stoichiometry, M is the atomic mass, σ E is the adsorption cross-section in barns/atom (1 barn = 10 -24 cm 2 ) tabulated in McMaster tables, and E + and E - are the just above and below the energy edge. This calculation can be accomplished using the free download software HEPHAESTUS.

$t=\frac{1}{\mathrm{\Delta \mu }}=\frac{1\text{.}\text{66}\sum _{i}{n}_{i}{M}_{i}}{\rho \sum _{i}{n}_{i}\left[{\sigma }_{i}\left({E}_{+}\right)-{\sigma }_{i}\left({E}_{-}\right)\right]}$

For non-concentrate samples, the total X-ray adsorption of the sample is the most important. It should be related to the area concentration of the sample (ρ t , in g/cm 2 ). The area concentration of the sample multiplied by the difference of the mass adsorption coefficient ( ∆µ E ) give the edge step, where a desired value to obtain a good measure is a edge step equal to one, (∆µ E /ρ)ρ t ≈ 1.

The difference of the mass adsorption coefficient is given by [link] , where ( µ E /ρ) i is the mass adsorption coefficient just above (E+) and below (E-) of the edge energy and f i is the mass fraction of the element i . Multiplying the area concentration, ρ t, for the cross-sectional area of the sample holder, amount of sample needed is known.

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