1. Misinterpreting X-Ray Diffraction Results by Tom and Keith

2. X-ray How many of you have carried out x-ray diffraction?
How many of you have interpreted x-ray diffraction results?
Who is responsible for Bragg’s Law?

3. Example 1
Rock Salt

4. Why are peaks missing? The sample is made from Morton’s Salt
111
200
220
311
222
JCPDF#
The sample is made from Morton’s Salt
JCPDF# is supposed to fit it (Sodium Chloride Halite)

5. It’s a single crystal
111
200
220
311
222 2q
The (200) planes would diffract at °2q; however, they are not properly aligned to produce a diffraction peak
The (222) planes are parallel to the (111) planes.
At °2q, Bragg’s law fulfilled for the (111) planes, producing a diffraction peak.

6. A random polycrystalline sample that contains thousands of crystallites should exhibit all possible diffraction peaks
200
220
111
222
311 2q 2q 2q
For every set of planes, there will be a small percentage of crystallites that are properly oriented to diffract (the plane perpendicular bisects the incident and diffracted beams).
Basic assumptions of powder diffraction are that for every set of planes there is an equal number of crystallites that will diffract and that there is a statistically relevant number of crystallites, not just one or two.

7. Hint Typical Shape Of Crystals
Powder Samples
200
111
220
311
222
Salt Sprinkled on double stick tape
What has Changed?
NaCl
<100>
Hint Typical Shape Of Crystals
It’s the same sample sprinkled on double stick tape but after sliding a glass slide across the sample

8. Example 2
PZT

9. A Tetragonal PZT What happened to cause the peaks to shift?
Lattice Parameters
a= Å
b= Å
Sample Re-polished and Re-measured
011
110
111
002
200
What happened to cause the peaks to shift?

10. Change In Strain/Lattice Parameter?
A Tetragonal PZT
Lattice Parameters
a= Å
c= Å
Change In Strain/Lattice Parameter?
a=4.07A c=4.16A
101/110
002/200
111
Z-Displaced Fit
Disp.=1.5mm
D= 1.5 mm
a [A] 4.07
c [A] 4.16
Disp

11. Z-Displacements Tetragonal PZT R θ 2θ a=4.0215 b=4.1100 Disp 011 110
111
002
200 θ
Disp 2θ

12. Example 3
Nanocrystalline
Materials

13. Hint: Why are the intensities different?
Which of these diffraction patterns comes from a nanocrystalline material? 66 67 68 69 70 71 72 73 74 2 q
(deg.)
Intensity (a.u.)
Hint: Why are the intensities different?
These two diffraction patterns come from the exact same sample (silicon).
The apparent difference in peak broadening is due to the instrument optics, not due to specimen broadening
These diffraction patterns were produced from the exact same sample
The apparent peak broadening is due solely to the instrumentation
0.0015° slits vs. 1° slits

14. Crystallite Size Broadening
Peak Width B(2q) varies inversely with crystallite size
The constant of proportionality, K (the Scherrer constant) depends on the how the width is determined, the shape of the crystal, and the size distribution
the most common values for K are 0.94 (for FWHM of spherical crystals with cubic symmetry), 0.89 (for integral breadth of spherical crystals with cubic symmetry, and 1 (because 0.94 and 0.89 both round up to 1).
K actually varies from 0.62 to 2.08
For an excellent discussion of K, refer to JI Langford and AJC Wilson, “Scherrer after sixty years: A survey and some new results in the determination of crystallite size,” J. Appl. Cryst. 11 (1978) p
Remember:
Instrument contributions must be subtracted

15. Methods used to Define Peak Width
46.7
46.8
46.9
47.0
47.1
47.2
47.3
47.4
47.5
47.6
47.7
47.8
47.9 2 q
(deg.)
Intensity (a.u.)
Full Width at Half Maximum (FWHM)
the width of the diffraction peak, in radians, at a height half-way between background and the peak maximum
Integral Breadth
the total area under the peak divided by the peak height
the width of a rectangle having the same area and the same height as the peak
requires very careful evaluation of the tails of the peak and the background
FWHM
46.7
46.8
46.9
47.0
47.1
47.2
47.3
47.4
47.5
47.6
47.7
47.8
47.9 2 q
(deg.)
Intensity (a.u.)

16. Williamson-Hull Plot y-intercept slope K≈0.94
Grain size and strain broadening
Grain size broadening
K≈0.94
Gausian Peak Shape Assumed

17. Dealing With Different Integral Breadth/FWHM Contributions Contributions
Lorentzian and Gaussian Peak shapes are treated differently
B=FWHM or β in these equations
Williamson-Hall plots are constructed from for both the Lorentzian and Gaussian peak widths.
The crystallite size is extracted from the Lorentzian W-H plot and the strain is taken to be a combination of the Lorentzian and Gaussian strain terms.
Lorentzian (Cauchy)
Gaussian
Integral Breadth (PV)

18. Example 4
Crystal Structure
vs.
Chemistry

19. Two Perovskite Samples
What are the differences?
Peak intensity
d-spacing
Peak intensities can be strongly affected by changes in electron density due to the substitution of atoms with large differences in Z, like Ca for Sr.
Assuming that they are both random powder samples what is the likely cause?
SrTiO3 and
CaTiO3
Ca Z=20; Sr Z=38
Zr Z=40; Y Z=39
Plays into why having an un-textured sample is important
200
210
211
2θ (Deg.)

20. What is a structure factor? What is a scattering factor?

21. Two samples of Yttria stabilized Zirconia
Why might the two patterns differ?
Substitutional Doping can change bond distances, reflected by a change in unit cell lattice parameters
The change in peak intensity due to substitution of atoms with similar Z is much more subtle and may be insignificant
10% Y in ZrO2
50% Y in ZrO2 45 50 55 60 65
2θ (Deg)
Intensity(Counts)
Ca Z=20; Sr Z=38
Zr Z=40; Y Z=39
Plays into why having an un-textured sample is important

22. Polycrystalline films on Silicon
Why do the peaks broaden toward each other?
Solid Solution Inhomogeneity
Variation in the composition of a solid solution can create a distribution of d-spacing for a crystallographic plane
CeO2
19 nm 45 46 47 48 49 50 51 52 2 q
(deg.)
Intensity (a.u.)
ZrO2
46nm
CexZr1-xO2
0<x<1

23. Is that enough information?
Example 5
Radiation from a copper source -
Is that enough information?
“Professor my peaks split!”

24. Why does this sample second set of peaks at higher 2θ values?
Hints:
It’s Alumina
Cu source
Detector has a single channel analyzer
006
113
Kα1
Kα2

25. Diffraction Pattern Collected Where A Ni Filter Is Used To Remove Kβ
Ka1
Ka2
What could this be?
W La1
Due to tungsten contamination
K alpha 1 and K alpha 2 overlap heavily at low angles and are easier to discriminate at high angles. Kb

26. Wavelengths for X-Radiation are Sometimes Updated
Copper
Anodes
Bearden
(1967)
Holzer et al.
(1997)
Cobalt
Cu Ka1 Å Å
Co Ka1 Å Å
Cu Ka2 Å Å
Co Ka2 Å Å
Cu Kb Å Å
Co Kb Å Å
Molybdenum
Chromium
Mo Ka1 Å Å
Cr Ka1 Å Å
Mo Ka2 Å Å
Cr Ka2 Å Å
Mo Kb Å Å
Cr Kb Å Å
Often quoted values from Cullity (1956) and Bearden, Rev. Mod. Phys. 39 (1967) are incorrect.
Values from Bearden (1967) are reprinted in international Tables for X-Ray Crystallography and most XRD textbooks.
Most recent values are from Hölzer et al. Phys. Rev. A 56 (1997)
Has your XRD analysis software been updated?

27. Example 6 Unexpected Results From An Obviously Crystalline Sample

28. Unexpected Results From an Unknown Sample
D8 Focus
No peaks seen in a locked coupled 2θ scan of a crystalline material
Why?

29. Bruker Diffractometer with Area Detectorα
α = 35°
2θ=50° ω=25 ° Detector distance= 15 cm

30. After Crushing The Unknown Sample
D8 Focus
JCPDF
We now have two visible peaks that index with CaF

31. 2D (Area) Diffraction allows us to image complete or incomplete (spotty) Debye diffraction rings
Polycrystalline thin film on a single crystal substrate
Mixture of fine and coarse grains in a metallic alloy
Conventional linear diffraction patterns can easily miss information about single crystal or coarse grained materials

32. Quiz

33. Match The Sample/Measurement Conditions With The Diffraction Pattern1 2 3

34. Questions