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XAFSmass is a powerful and helpful program that lets you easily optimise the dilution of your sample for transmission XAFS analysis. This guide will demonstrate how to use

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XAFSmass to estimate the appropriate dilution of a sample for transmission, achieving the desired total absorption and edge step.

Terminology

Absorption unit: One absorption unit represents the thickness of material sufficient to decrease in flux to 1/e (36.79%) of its initial value. e.g. A flux of 1E10 photons/second incident on a sample 1 absorption units thick would result in a flux of 3.68E9 photons after the sample, alternatively expressed, 1 photon in every 2.7 photons makes its way through the sample. Absorption units are a logarithmic scale. Thus a sample two absorption units thick will have a transmission of (1/e)*(1/e)  which is 13.53% or 1 photon in every 7.4.

Total absorption Td): The total attenuation of a beam of photons of a defined energy. All elements in a sample contribute to the total absorption. Total absorption is expressed in absorption units.

Edge step/absorptance step (Δμ): The magnitude of the change in absorption as a monochromatic beam is scanned across an absorption edge. Only atoms of the element corresponding to the absorption edge contribute to the edge step. The size of the edge step is proportional to the concentration of the element of interest in the sample. Edge step is expressed in absorption units.

Note

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MEX-1 can measure edge steps down to 0.1 without problem. Close to 1 is ideal, 0.01 means the sample is probably better suited to fluorescence-mode.

Optical thickness: In this guide, optical thickness refers to how absorbing a material is, expressed in absorption units. e.g. "My sample has an optical thickness of 2.3 absorption units". Optical thickness and total absorption can be used interchangeably.

XAFSmass: This guide employs XAFSmassQt, the modern version of XAFSmass. It has a much more powerful chemical formula parser than the older version of XAFSmass. Details can be found here: https://xafsmass.readthedocs.io/# It is beyond the scope of this guide to describe installation of XAFSmass. Using the old version of XAFSmass WILL NOT WORK with these instructions.

TotalAbs.pngImage AddedXAFSmass for wiki.pngImage Added

The problem

The problem to be solved when choosing an appropriate dilution for a transmission sample is to balance the total absorption, the edge step and an appropriate amount of material to make a pellet. In an ideal world, we want an edge step between 0.5 and 1.5 absorption units, whilst keeping the total absorption within reasonable limits, i.e. ideally below 1.5 absorption units, though perhaps relaxing that constraint to 2.5 absorption units or below for more difficult samples. There are two motivations for keeping total absorption at <2.5 absorption units:

  1. the avoidance of "thickness effects". The optically thicker a sample is, the more the measured energy-dependent absorption will be affected by imperfections in sample homogeneity. At a total absorption of 5 absorption units, 1 photon in 150 will make it through the sample. If regions of the sample illuminated by the beam have varying absorption, because your sample was poorly mixed, or worse, there are areas where the element of interest is absent, the measured absorption signal will be dominated by the contributions from those regions. More information can be found in the following papers:

    1. Stern, E. A. & Kim, K. (1981). Thickness effect on the extended-x-ray-absorption-fine-structure amplitude. Physical Review B23, 3781–3787.

    2. Heald, S. M. (1988).

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    1. EXAFS with synchrotron radiation. In: Koningsberger, D. C. & Prins, R. (eds) X-Ray Absorption: Principles, Applications, Techniques of EXAFS, SEXAFS and XANES. John Wiley & Sons,

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    1. 87–118.

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optimal signal to noise. See here for further explanation.

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    1. Goulon, J., Goulon-Ginet, C., Cortes, R. & Dubois, J. M. (1982). On experimental attenuation factors of the amplitude of the EXAFS oscillations in absorption, reflectivity and luminescence measurements. Journal de Physique 43, 539–548.

  1. optimal signal to noise. See here for further explanation.

In addition, we need a mass of material that is large enough to reliably press into pellets. For many materials, the optimal mass for a 7 mm pellet will be in the single figure milligram range, a mass too small to reliably press high quality pellets. Thus, we must dilute the material of interest with an inert dilutant, preferably a material that has low atomic number (and thus has low absorption) that also forms good pellets. The material of choice at the MEX beamline is microcrystalline cellulose, C6H10O5. A good mass for a 7 mm pellet is 40 mg of sample; for a 13 mm pellet, 90 mg is a good mass.

Diluting the material of interest with cellulose changes both the total absorption and the edge step. Thus, the problem to be solved is to choose a dilution that gives reasonable values for total absorption, edge step and mass of material to be pressed into a pellet. For many real-world samples, optimum values for all three parameters cannot be achieved. Thus, the task for the user is to use tools like XAFSmass to explore the parameter space to find the least worst compromise.

Equipment required

To make sample pellets, you will need the following equipment available in the Chemistry Lab.

image-20241113-061851.pngImage Added

image-20241113-062137.pngImage Added

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For a more sophisticated grinding and pellet pressing method, the automatic grinder and hydraulic press below can be used, but will require training from the Chemistry Lab technicians outside of beam time.

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Example 1 - tungsten metal powder

Using

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XAFSmass, enter W as the compound, µtd (total absorption) as 1.5, S (area of the pellet) as 0.3848451 (the area of a 7 mm diameter pellet expressed in cm2). Choose the energy from the drop down list to be the W L3 edge (+50 eV):

image-20240531-061305.pngImage Modified

XAFSMass reports that the edge step will be 0.942, in the middle of the ideal range of 0.5-1.5, but the mass required is very small, 2.46 mg. It is unlikely that it will be possible to successfully press a 7 mm pellet of this mass, particularly of a hard material like W. Thus, it will be necessary to dilute the W metal with cellulose. The python version of XAFSmass has a much more featureful parser than the older version, and allows us to easily express dilutions. Let's start with investigating a 50-50 mix of W metal and cellulose. Change the compound to W%50(C6H10O5) and click calculate:

image-20240531-061403.pngImage Modified

XAFSMass automatically converts our expression for dilution into a molecular formula, W0.882C6H10O5, and returns a slightly smaller edge step of 0.928 absorption units. However, the mass required is still far less than the optimal 40 mg for a 7 mm pellet. Lets try diluting the W to 5%, by entering W%5(C6H10O5):

image-20240531-061451.pngImage Modified

This result is much closer to what we want. The edge step is within our optimal range at 0.73 absorption units, and the mass required is 38.1 mg, close enough to 40 mg that a successful pellet will be possible. Enthusiastic users can optimise further and find that a 40 mg pellet requires a dilution of 4.68 wt% W, and gives an edge step of 0.718:

image-20240531-061754.pngImage Modified

Comparison with the "simple" method

A simpler way of using XAFSmass is to enter only the material formula and a µtd (total absorption). XAFSmass will then return a mass and edge step. That mass of material can then be mixed with a dilutant such as cellulose to a pre-defined final mass, known to make robust pellets, such as 40 mg for 7 mm OD pellets:

image-20240531-061825.pngImage Modified

In the example above, 2.46 mg of W would be mixed with 37.54 mg of cellulose. Let us include cellulose in the calculation. 2.46 mg of W diluted with 37.54 mg of cellulose corresponds to a concentration of 6.15 wt%. This can be entered into XAFSmass via the following formula W%6.15(C6H10O5):

image-20240531-062235.pngImage Modified

If we inspect the XAFSmass window, we can see that the mass required for the total absorption we specified (1.5 abs units) is 32.4 mg. This is not equivalent to the pellet we made above, and we can confirm that by looking at the absorptance step box, where is reports that the mass of W is 1.99 mg and the edge step is 0.764. To calculate 2.46 mg of W mixed with 37.54 mg of cellulose we must manually increase the value of µtd (total absorption) until the mass is 40.0 mg AND the absorptance step box report 2.46 mg W and an edge step of 0.942:

image-20240531-062445.pngImage Modified

After changing the value a few times we find that the total absorption of the W + cellulose mix is 1.85. This is a little higher than the 1.5 abs. units we specified in the "simple" calculation, and probably would not have too much impact on the final result.

However, this simple approach only works some of the time. The shortcoming is that the cellulose is not included in the total absorption calculation, and for edges at lower energies, the absorption contributed by the dilutant can be significant, and using this method will result in sample far too thick for transmission measurements.

Example 2 - Sc2O3

To demonstrate, let us compare the approach with scandium oxide, Sc2O3:

image-20240528-223257.pngImage Modified

Following the "simple" approach, we would mix 1.03 mg with 38.97 mg of cellulose to make a 7 mm pellet.

Comparison with the "simple" method

Let's see what happens when we include cellulose in the calculation. 1.03 mg of Sc2O3 in a 40 mg pellet equates to 2.575 wt%. Enter the formula (Sc2O3)%2.575(C6H10O5) into XAFSmass, and select the Sc K-edge:

image-20240528-223335.pngImage Modified

The mass shown is 10.8 mg, which is incorrect. Our task now is to manually increase the value for µtd (total absorption) until the mass of material is 40.0 mg AND the absorptance step for Sc matches that from when we did the calculation with Sc2O3 on its own (in this case, 1.29):

image-20240528-223358.pngImage Modified

Thus, if we used the "simple" method the resulting pellet would have a µtd (total absorption) of 5.585 absorption units. This sample is far too absorbing for a transmission measurement! The cellulose contributed 4.085 absorption units that were ignored by using the "simple" method. This demonstrates that it is important to include the dilutant in calculations. The new version of XAFSmass provides a convenient way of expressing dilution, so there is no reason to use the "simple" method any longer.

Example 3 - Fe3O4

The energy of the Fe K-edge is approximately halfway between Sc K and W L3. Let us compare the two methods again.

The simple method suggests 1.9 mg of Fe3O4 should be mixed with 38.1 mg cellulose:

image-20240528-223415.pngImage Modified

Include the cellulose in the calculation:

image-20240528-223433.pngImage Modified

The total absorption increase by 1.025 absorption units, and total absorption is 1.68 times higher if the cellulose is excluded.

Example 4 - CeO2

The energy of the Ce L3 edge is roughly halfway between Fe K and Sc K.

The simple method:

image-20240528-223501.pngImage Modified

Including cellulose:

image-20240528-223517.pngImage Modified

The total absorption has increased by 1.977 absorption units when cellulose is included in the calculation. This sample is too thick for transmission.

Example 5 - In-doped PbSe

Sometimes, the composition of a material makes it difficult, or impossible to make a good transmission sample that gives both a reasonable edge step and a reasonable total absorption

image-20240528-223547.pngImage Modifiedimage-20240528-223603.pngImage Modified

This material involves many heavy elements, so the results are dominated by the high total absorption. A total thickness of 8 absorption units is required to get an edge step of only 0.1 at In L3, and even then the mass is a tiny 2.78 mg. Diluting with cellulose will only increase the total absorption further. This material is better suited to measurement in fluorescence mode.

Example 6 - Fe in soil

For a complex natural sample, for which you know the composition in weight percent oxides, the workflow is a little more involved:

  • Convert to moles of oxides. A convenient molecular mass calculator can be found here.

  • Convert to moles of elements

  • Construct formula involving dilution with cellulose to enter into XAFSmass

An example composition, in weight percent oxides:

Al2O3

SiO2

K2O

CaO

TiO2

MnO

Fe2O3

sum

21.46

44.39

2.05

4.49

1.51

0.27

25.06

99.23

Convert to moles of oxides. The top row is the molecular mass of the oxide:

101.96

60.08

94.2

56.08

79.87

70.94

159.69

Al2O3

SiO2

K2O

CaO

TiO2

MnO

Fe2O3

0.210475

0.738848

0.021762

0.080064

0.018906

0.003806

0.156929

Convert to moles of the element based on stoichiometry:

Al

Si

K

Ca

Ti

Mn

Fe

O

0.420949

0.738848

0.043524

0.080064

0.018906

0.003806

0.313858

2.723352

Create a formula for XAFSmass, in this case diluted to 10 wt% in cellulose:

(Al0.421Si0.739K0.044Ca0.080Ti0.019Mn0.004Fe0.314O2.723)%10(C6H10O5)

Enter into XAFSmass:

image-20240528-223643.pngImage Modified

Your task as a user is to explore the total absorption-dilution space to find the optimum combination of dilution and total absorption that returns a reasonable edge step and a workable mass to make a pellet.

Example 7 - metal alloy in carbon

In this example, we take advantage of the powerful parser in the python/Qt version of XAFSmass. A common type of sample measured at XAFS beamlines is a metal alloy dispersed in a matrix such as carbon. In this example, the composition is 4 wt% Pt, 26 wt. % Ni and the remainder is carbon. We can enter that into XAFSmass as the formula Pt%4Ni%26C. Helpfully, XAFSmass will immediately recalculate that for us into moles: Pt0.00352Ni0.076C. We can then take this molecular formula, and use it to explore dilution with cellulose. For example a 50 wt. % mix of the alloy+carbon material and cellulose, we can enter the following into XAFSmass: (Pt0.00352Ni0.076C)%50(C6H10O5). When you are unfamiliar with a material, it can frequently be helpful to construct a table of results of XAFSmass calculations to allow you to identify the optimum dilution. The table below shows the results of such an examination for the material Pt%4Ni%26C (7.0 mm OD pellet, diluted with cellulose):

Material

Molecular formula

dilution (wt. %)

dilutant

formula for XAFSmass

total absorption

edge step

mass of 7 mm OD pellet

Pt%4Ni%26C

Pt0.00352Ni0.076C

0

Pt0.00352Ni0.076C

2

1.58

7.99

Pt%4Ni%26C

Pt0.00352Ni0.076C

90

C6H10O5

(Pt0.00352Ni0.076C)%90(C6H10O5)

2

1.56

8.81

Pt%4Ni%26C

Pt0.00352Ni0.076C

80

C6H10O5

(Pt0.00352Ni0.076C)%80(C6H10O5)

2

1.55

9.83

Pt%4Ni%26C

Pt0.00352Ni0.076C

70

C6H10O5

(Pt0.00352Ni0.076C)%70(C6H10O5)

2

1.53

11.1

Pt%4Ni%26C

Pt0.00352Ni0.076C

60

C6H10O5

(Pt0.00352Ni0.076C)%60(C6H10O5)

2

1.51

12.7

Pt%4Ni%26C

Pt0.00352Ni0.076C

50

C6H10O5

(Pt0.00352Ni0.076C)%50(C6H10O5)

2

1.48

15

Pt%4Ni%26C

Pt0.00352Ni0.076C

40

C6H10O5

(Pt0.00352Ni0.076C)%40(C6H10O5)

2

1.43

18.1

Pt%4Ni%26C

Pt0.00352Ni0.076C

30

C6H10O5

(Pt0.00352Ni0.076C)%30(C6H10O5)

2

1.36

23

Pt%4Ni%26C

Pt0.00352Ni0.076C

20

C6H10O5

(Pt0.00352Ni0.076C)%20(C6H10O5)

2

1.24

31.4

Pt%4Ni%26C

Pt0.00352Ni0.076C

10

C6H10O5

(Pt0.00352Ni0.076C)%10(C6H10O5)

2

0.98

49.7

Pt%4Ni%26C

Pt0.00352Ni0.076C

5

C6H10O5

(Pt0.00352Ni0.076C)%5(C6H10O5)

2

0.98

49.7

Pt%4Ni%26C

Pt0.00352Ni0.076C

0

Pt0.00352Ni0.076C

1.5

1.18

5.99

Pt%4Ni%26C

Pt0.00352Ni0.076C

90

C6H10O5

(Pt0.00352Ni0.076C)%90(C6H10O5)

1.5

1.17

6.61

Pt%4Ni%26C

Pt0.00352Ni0.076C

80

C6H10O5

(Pt0.00352Ni0.076C)%80(C6H10O5)

1.5

1.16

7.37

Pt%4Ni%26C

Pt0.00352Ni0.076C

70

C6H10O5

(Pt0.00352Ni0.076C)%70(C6H10O5)

1.5

1.15

8.3

Pt%4Ni%26C

Pt0.00352Ni0.076C

60

C6H10O5

(Pt0.00352Ni0.076C)%60(C6H10O5)

1.5

1.13

9.55

Pt%4Ni%26C

Pt0.00352Ni0.076C

50

C6H10O5

(Pt0.00352Ni0.076C)%50(C6H10O5)

1.5

1.11

11.2

Pt%4Ni%26C

Pt0.00352Ni0.076C

40

C6H10O5

(Pt0.00352Ni0.076C)%40(C6H10O5)

1.5

1.07

13.6

Pt%4Ni%26C

Pt0.00352Ni0.076C

30

C6H10O5

(Pt0.00352Ni0.076C)%30(C6H10O5)

1.5

1.02

17.2

Pt%4Ni%26C

Pt0.00352Ni0.076C

20

C6H10O5

(Pt0.00352Ni0.076C)%20(C6H10O5)

1.5

0.931

23.5

Pt%4Ni%26C

Pt0.00352Ni0.076C

10

C6H10O5

(Pt0.00352Ni0.076C)%10(C6H10O5)

1.5

0.735

37.3

Pt%4Ni%26C

Pt0.00352Ni0.076C

5

C6H10O5

(Pt0.00352Ni0.076C)%5(C6H10O5)

1.5

0.517

52.5

Pt%4Ni%26C

Pt0.00352Ni0.076C

0

Pt0.00352Ni0.076C

2.5

1.97

9.99

Pt%4Ni%26C

Pt0.00352Ni0.076C

90

C6H10O5

(Pt0.00352Ni0.076C)%90(C6H10O5)

2.5

1.95

11

Pt%4Ni%26C

Pt0.00352Ni0.076C

80

C6H10O5

(Pt0.00352Ni0.076C)%80(C6H10O5)

2.5

1.94

12.3

Pt%4Ni%26C

Pt0.00352Ni0.076C

70

C6H10O5

(Pt0.00352Ni0.076C)%70(C6H10O5)

2.5

1.91

13.8

Pt%4Ni%26C

Pt0.00352Ni0.076C

60

C6H10O5

(Pt0.00352Ni0.076C)%60(C6H10O5)

2.5

1.88

15.9

Pt%4Ni%26C

Pt0.00352Ni0.076C

50

C6H10O5

(Pt0.00352Ni0.076C)%50(C6H10O5)

2.5

1.84

18.7

Pt%4Ni%26C

Pt0.00352Ni0.076C

40

C6H10O5

(Pt0.00352Ni0.076C)%40(C6H10O5)

2.5

1.79

22.7

Pt%4Ni%26C

Pt0.00352Ni0.076C

30

C6H10O5

(Pt0.00352Ni0.076C)%30(C6H10O5)

2.5

1.7

28.7

Pt%4Ni%26C

Pt0.00352Ni0.076C

20

C6H10O5

(Pt0.00352Ni0.076C)%20(C6H10O5)

2.5

1.55

39.2

Pt%4Ni%26C

Pt0.00352Ni0.076C

10

C6H10O5

(Pt0.00352Ni0.076C)%10(C6H10O5)

2.5

1.22

62.1

Pt%4Ni%26C

Pt0.00352Ni0.076C

5

C6H10O5

(Pt0.00352Ni0.076C)%5(C6H10O5)

2.5

0.862

87.5

Plotting the results:

image-20240528-223729.pngImage Modified

The intersection of grey dotted lines marks our target of 40 mg and an edge step of 1.0 absorption units. Each of the dots on the coloured lines represents a different diltution (you can inspect the values for mass and edge step in the table to locate each dilution on the plot). The results suggest that a total absorption of a little less than 2.0, i.e. ~1.85 and a dilution with cellulose to between 10 and 20 wt. % will give the optimum result. However, we do not have to stick to making only 40 mg pellets, nor do we have to stick to edge steps of 1.0. Depending on your material and your skill making pellets, lower masses will also work. As an accomplished presser of pellets, I would attempt to make a pellet with only 20 mg of material and a dilution of 20 wt % and a total absorption of 1.5 abs. units. Alternatively, for this sample, I could make a 40 mg pellet with a total absorption of 1.5 abs units, and accept an edge step of ~0.7 which will give excellent data at MEX1.

Think of 40 mg as a good starting point for a mass of a pellet that should almost always work, however if you find you can successfully make lower mass pellets, take advantage of the extra flexibility it offers. Similarly, whilst an edge step of 1.0 is ideal, MEX1 can return good data at edge steps as low as 0.1, (or lower for the more adventurous/less discerning user).