Transcript 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.

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?
JCPDF# 01-0994
200
220
111
222
311
• The sample is made from Morton’s Salt
• JCPDF# 01-0994 is supposed to fit it (Sodium Chloride Halite)
It’s a single crystal
200
220
111
222
311
2q
At 27.42 °2q, Bragg’s law
fulfilled for the (111) planes,
producing a diffraction peak.
The (200) planes would diffract at 31.82
°2q; however, they are not properly
aligned to produce a diffraction peak
The (222) planes are parallel to the (111)
planes.
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.
Powder Samples
200
NaCl
Hint
• Salt Sprinkled on
<100>
double stick tape
Typical Shape
Of Crystals
• What has Changed?
111
220
311 222
It’s the same sample sprinkled on double
stick tape but after sliding a glass slide
across the sample
Example 2
PZT
A Tetragonal PZT
• Lattice Parameters
– a=4.0215 Å
– b=4.1100 Å
Sample Re-polished
and Re-measured
011
110
111
200
002
What happened to cause the peaks to shift?
A Tetragonal PZT
• Lattice Parameters
Change In
Strain/Lattice Parameter?
a=4.07A
c=4.16A
– a=4.0215 Å
– c=4.1100 Å
101/110
111
002/200
Z-Displaced Fit
Disp.=1.5mm
Disp
Z-Displacements
• Tetragonal PZT
– a=4.0215
– b=4.1100
R
θ
011
Disp
2θ
110
111
200
002
d
d Actual
Disp  cos2 q

RDetector  sin q
d Actual 
d Measured
1
Disp cos2 q
RDetector sin q
Example 3
Nanocrystalline
Materials
Which of these diffraction patterns comes
from a nanocrystalline material?
Intensity (a.u.)
Hint: Why are the
intensities different?
66
67
68
69
70
71
72
73
74
2q (deg.)
• 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
Crystallite Size Broadening
0.94
B2q  
Sizecosq
• 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) p102-113.
• Remember:
– Instrument contributions must be subtracted
Methods used to Define Peak Width
• Full Width at Half Maximum
(FWHM)
– 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
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
2q (deg.)
Intensity (a.u.)
• Integral Breadth
Intensity (a.u.)
– the width of the diffraction
peak, in radians, at a height
half-way between background
and the peak maximum
FWHM
46.7
46.8
46.9
47.0
47.1
47.2
47.3
47.4
2q (deg.)
47.5
47.6
47.7
47.8
47.9
Williamson-Hull Plot
FWHM  cosq  
K 
 Strain  4  sinq 
Size
(FWHMobs-FWHMinst)
cos(q )
y-intercept
Gausian Peak Shape Assumed
slope
Grain size broadening
4 x sin(q)
K≈0.94
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)
BExp  BSize  BStrain  BInst
B
Exp

 BInst  BSize  BStrain
Gaussian
2
2
2
2
BExp
 BSize
 BStrain
 BInst
2
2
2
2
BExp
 BInst
 BSize
 BStrain


Integral Breadth (PV)
2
Exp
 Lorentzian Exp  Gaussian
Example 4
Crystal Structure
vs.
Chemistry
Two Perovskite Samples
• What are the differences?
•
Assuming that they are both random
powder samples what is the likely cause?
– 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
200
210
2θ (Deg.)
211
What is a structure factor?
What is a scattering factor?
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
Intensity(Counts)
•
45
50
55
2θ (Deg)
60
65
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
ZrO2
46nm
Intensity (a.u.)
CeO2
19 nm
45
46
CexZr1-xO2
0<x<1
47
48
49
2q (deg.)
50
51
52
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:
113
– It’s Alumina
– Cu source
– Detector has a single
channel analyzer
Kα1
Kα2
006
Diffraction Pattern Collected
Where A Ni Filter Is Used
To Remove Kβ
Ka1
Ka2
WhatWcould
La1
this
Due
tobe?
tungsten
contamination
K
E keV   h 
hc


6.02
  
Wavelengths for X-Radiation are
Sometimes Updated
Copper
Anodes
Bearden
(1967)
Holzer et al.
(1997)
Cobalt
Anodes
Bearden
(1967)
Holzer et al.
(1997)
Cu Ka1
1.54056Å
1.540598 Å
Co Ka1
1.788965Å
1.789010 Å
Cu Ka2
1.54439Å
1.544426 Å
Co Ka2
1.792850Å
1.792900 Å
Cu K
1.39220Å
1.392250 Å
Co K
1.62079Å
1.620830 Å
Cr Ka1
2.28970Å
2.289760 Å
Molybdenum
Anodes
Chromium
Anodes
Mo Ka1
0.709300Å
0.709319 Å
Mo Ka2
0.713590Å
0.713609 Å
Cr Ka2
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 XRay 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
D8 Focus
• No peaks seen in a locked
coupled 2θ scan of a
crystalline material
Why?
Bruker Diffractometer
with Area Detector
α
α = 35°
2θ=50° ω=25 °
Detector distance= 15 cm
After Crushing
The Unknown Sample
D8 Focus
JCPDF 75-0097
We now have two visible peaks
that index with CaF
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