Download presentation

Presentation is loading. Please wait.

Published byKristin Howlett Modified about 1 year ago

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 JCPDF# is supposed to fit it ( Sodium Chloride Halite ) JCPDF#

5
It’s a single crystal 22 At °2 , Bragg’s law fulfilled for the (111) planes, producing a diffraction peak. The (200) planes would diffract at °2 ; however, they are not properly aligned to produce a diffraction peak The (222) planes are parallel to the (111) planes

6
A random polycrystalline sample that contains thousands of crystallites should exhibit all possible diffraction peaks 22 22 22 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
Salt Sprinkled on double stick tape What has Changed? NaCl Powder Samples It’s the same sample sprinkled on double stick tape but after sliding a glass slide across the sample Hint Typical Shape Of Crystals

8
Example 2 PZT

9
A Tetragonal PZT Lattice Parameters –a= Å –b= Å Sample Re-polished and Re-measured What happened to cause the peaks to shift?

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

11
Z-Displacements R Tetragonal PZT –a= –b= Disp 2θ2θ θ

12
Example 3 Nanocrystalline Materials

13
(deg.) Intensity (a.u.) Which of these diffraction patterns comes from a nanocrystalline material? 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 Hint: Why are the intensities different?

14
Crystallite Size Broadening Peak Width B(2 ) 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
(deg.) Intensity (a.u.) (deg.) Intensity (a.u.) Methods used to Define Peak Width 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

16
Williamson-Hull Plot y-interceptslope K≈0.94 Grain size broadening Grain size and strain broadening 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. Gaussian Lorentzian (Cauchy) 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. SrTiO 3 and CaTiO 3 2θ (Deg.) Assuming that they are both random powder samples what is the likely cause?

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

21
θ (Deg) Intensity(Counts) Two samples of Yttria stabilized Zirconia 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 ZrO 2 50% Y in ZrO 2 Why might the two patterns differ?

22
CeO 2 19 nm (deg.) Intensity (a.u.) ZrO 2 46nm Ce x Zr 1-x O 2 0

23
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 Kα1Kα1 Kα2Kα2

25
Diffraction Pattern Collected Where A Ni Filter Is Used To Remove K β K1K1 K2K2 What could this be? KK W L 1 Due to tungsten contamination

26
Wavelengths for X-Radiation are Sometimes Updated Copper Anodes Bearden (1967) Holzer et al. (1997) Cobalt Anodes Bearden (1967) Holzer et al. (1997) Cu K Å Å Co K Å Å Cu K Å Å Co K Å Å Cu K Å Å Co K Å Å Molybdenum Anodes Chromium Anodes Mo K Å Å Cr K Å Å Mo K Å Å Cr K Å Å Mo K Å Å Cr K 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 No peaks seen in a locked coupled 2θ scan of a crystalline material D8 Focus Why?

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

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

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 Pattern 1 2 3

34
Questions

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google