4 Why are peaks missing? The sample is made from Morton’s Salt 111200220311222JCPDF#The sample is made from Morton’s SaltJCPDF# is supposed to fit it (Sodium Chloride Halite)
5 It’s a single crystal1112002203112222qThe (200) planes would diffract at °2q; however, they are not properly aligned to produce a diffraction peakThe (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 peaks2002201112223112q2q2qFor 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 Samples200111220311222Salt Sprinkled on double stick tapeWhat has Changed?NaCl<100>Hint Typical Shape Of CrystalsIt’s the same sample sprinkled on double stick tape but after sliding a glass slide across the sample
13 Hint: Why are the intensities different? Which of these diffraction patterns comes from a nanocrystalline material?6667686970717273742q(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 broadeningThese diffraction patterns were produced from the exact same sampleThe apparent peak broadening is due solely to the instrumentation0.0015° slits vs. 1° slits
14 Crystallite Size Broadening Peak Width B(2q) varies inversely with crystallite sizeThe constant of proportionality, K (the Scherrer constant) depends on the how the width is determined, the shape of the crystal, and the size distributionthe 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 2.08For 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) pRemember:Instrument contributions must be subtracted
15 Methods used to Define Peak Width 46.746.846.947.047.147.247.347.447.547.647.747.847.92q(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 maximumIntegral Breadththe total area under the peak divided by the peak heightthe width of a rectangle having the same area and the same height as the peakrequires very careful evaluation of the tails of the peak and the backgroundFWHM46.746.846.947.047.147.247.347.447.547.647.747.847.92q(deg.)Intensity (a.u.)
17 Dealing With Different Integral Breadth/FWHM Contributions Contributions Lorentzian and Gaussian Peak shapes are treated differentlyB=FWHM or β in these equationsWilliamson-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)GaussianIntegral Breadth (PV)
19 Two Perovskite Samples What are the differences?Peak intensityd-spacingPeak 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 andCaTiO3Ca Z=20; Sr Z=38Zr Z=40; Y Z=39Plays into why having an un-textured sample is important2002102112θ (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 parametersThe change in peak intensity due to substitution of atoms with similar Z is much more subtle and may be insignificant10% Y in ZrO250% Y in ZrO245505560652θ (Deg)Intensity(Counts)Ca Z=20; Sr Z=38Zr Z=40; Y Z=39Plays into why having an un-textured sample is important
22 Polycrystalline films on Silicon Why do the peaks broaden toward each other?Solid Solution InhomogeneityVariation in the composition of a solid solution can create a distribution of d-spacing for a crystallographic planeCeO219 nm45464748495051522q(deg.)Intensity (a.u.)ZrO246nmCexZr1-xO20<x<1
23 Is that enough information? Example 5Radiation 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 AluminaCu sourceDetector has a single channel analyzer006113Kα1Kα2
25 Diffraction Pattern Collected Where A Ni Filter Is Used To Remove Kβ Ka1Ka2What could this be?W La1Due to tungsten contaminationK 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 CopperAnodesBearden(1967)Holzer et al.(1997)CobaltCu Ka1ÅÅCo Ka1ÅÅCu Ka2ÅÅCo Ka2ÅÅCu KbÅÅCo KbÅÅMolybdenumChromiumMo Ka1ÅÅCr Ka1ÅÅMo Ka2ÅÅCr Ka2ÅÅMo KbÅÅCr KbÅÅOften 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 FocusNo peaks seen in a locked coupled 2θ scan of a crystalline materialWhy?
29 Bruker Diffractometer with Area Detector αα = 35°2θ=50° ω=25 ° Detector distance= 15 cm
30 After Crushing The Unknown Sample D8 FocusJCPDFWe 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 substrateMixture of fine and coarse grains in a metallic alloyConventional linear diffraction patterns can easily miss information about single crystal or coarse grained materials