Misinterpreting X-Ray Diffraction Results by Tom and Keith
description
Transcript of Misinterpreting X-Ray Diffraction Results by Tom and Keith
Misinterpreting X-Ray Diffraction Results
by Tom and Keith
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?
Example 1
Rock Salt
Why are peaks missing?
111
200
220
311
222
•The sample is made from Morton’s Salt
•JCPDF# 01-0994 is supposed to fit it (Sodium Chloride Halite)
JCPDF# 01-0994
It’s a single crystal
2
At 27.42 °2, Bragg’s law fulfilled for the (111) planes, producing a diffraction peak.
The (200) planes would diffract at 31.82 °2; however, they are not properly aligned to produce a diffraction peak
The (222) planes are parallel to the (111) planes.
111
200
220
311
222
A random polycrystalline sample that contains thousands of crystallites should exhibit all possible diffraction peaks
2 2 2
• 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.
111
200
220
311
222
• 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
<100>Hint
Typical Shape Of Crystals
200
111220 311 222
Example 2
PZT
A Tetragonal PZT• Lattice Parameters
– a=4.0215 Å– b=4.1100 Å
011
110
111
002200
Sample Re-polished and Re-measured
What happened to cause the peaks to shift?
A Tetragonal PZT
• Lattice Parameters– a=4.0215 Å– c=4.1100 Å
Z-Displaced Fit
Disp.=1.5mm
Change In Strain/Lattice Parameter?
101/110
111 002/200
Disp
a=4.07A c=4.16A
Z-Displacements
011
110
111
002200
sin
cos2
DetectorActual R
Disp
d
d
R• Tetragonal PZT
– a=4.0215– b=4.1100
sincos21
DetectorRDisp
MeasuredActual
dd
Disp 2θ
θ
Example 3
Nanocrystalline
Materials
66 67 68 69 70 71 72 73 74
2 (deg.)
Inte
ns
ity
(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?
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) p102-113.
• Remember: – Instrument contributions must be subtracted
cos
94.02
SizeB
46.746.846.947.047.147.247.347.447.547.647.747.847.9
2 (deg.)
Inte
nsity
(a.
u.)
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 (deg.)
Inte
nsity
(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
Williamson-Hull Plot
4 x sin()
(FW
HM
ob
s-F
WH
Min
st)
c
os
()
sin4cos
StrainSize
KFWHM
y-intercept slope
K≈0.94
Grain size broadeningGrain size and stra
in broadening
Gausian Peak Shape Assumed
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.
2222InstStrainSizeExp BBBB
InstStrainSizeExp BBBB
Gaussian
Lorentzian (Cauchy)
Integral Breadth (PV)
StrainSizeInstExp BBBB
2222StrainSizeInstExp BBBB
GaussianExpLorentzianExp 2
Example 4
Crystal Structure
vs.
Chemistry
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.
SrTiO3 and CaTiO3
2θ (Deg.)
Assuming that they are both random powder samples what is the likely cause?
200 210 211
What is a structure factor?
What is a scattering factor?
45 50 55 60 65
2θ (Deg)
Inte
nsi
ty(C
ou
nts
)
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 ZrO2
50% Y in ZrO2
Why might the two patterns differ?
CeOCeO22
19 nm19 nm
45 46 47 48 49 50 51 52
2 (deg.)
Inte
nsity
(a.
u.)
ZrOZrO22
46nm46nm
CexZr1-xO2
0<x<1
Polycrystalline films on Silicon
• Solid Solution Inhomogeneity– Variation in the composition of a solid solution can create a
distribution of d-spacing for a crystallographic plane
Why do the peaks broaden toward each other?
Example 5
Radiation from a copper source -
Is that enough information?
“Professor my peaks split!”
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
Diffraction Pattern Collected Where A Ni Filter Is Used
To Remove KβK1
K2
What could this be?
K
W L1
Due to tungsten contamination
02.6hchkeVE
Wavelengths for X-Radiation are Sometimes Updated
Copper
Anodes
Bearden
(1967)
Holzer et al.
(1997)
Cobalt
Anodes
Bearden
(1967)
Holzer et al.
(1997)
Cu K1 1.54056Å 1.540598 Å Co K1 1.788965Å 1.789010 Å
Cu K2 1.54439Å 1.544426 Å Co K2 1.792850Å 1.792900 Å
Cu K 1.39220Å 1.392250 Å Co K 1.62079Å 1.620830 Å
Molybdenum
Anodes
Chromium
Anodes
Mo K1 0.709300Å 0.709319 Å Cr K1 2.28970Å 2.289760 Å
Mo K2 0.713590Å 0.713609 Å Cr K2 2.293606Å 2.293663 Å
Mo K 0.632288Å 0.632305 Å Cr K 2.08487Å 2.084920 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?
Example 6
Unexpected Results From An Obviously Crystalline Sample
Unexpected Results From an Unknown Sample
• No peaks seen in a locked coupled 2θ scan of a crystalline material
D8 Focus
Why?
Bruker Diffractometer with Area Detector
2θ=50° ω=25 ° Detector distance= 15 cm
α
α = 35°
After Crushing The Unknown Sample
We now have two visible peaks that index with CaF
D8 FocusJCPDF 75-0097
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
Quiz
Match The Sample/Measurement Conditions With The Diffraction Pattern
1
2
3
Questions