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