Uses of Powder Diffraction
Powder X-ray Diffraction

Brendan J. Kennedy
School of Chemistry
The University of Sydney

Three Unique Features ofSynchrotron Radiation•Intensity•Enables Rapid DataCollection
Kinetics
Unstable Compounds
Environmental Cells–Enables Focussing
Small Samples
Small areas/volumes•Energy Range•Enables Spectroscopy–Elemental Identification–Bonding Studies–Speciation•Enables Optimal Conditions–Environmental Cells–Selected Elements•Low Divergence
Enables High Resolution•Micro Beams•Small Volumes•Complex Materials

Qualitative Analysis
Identification of single-phasematerials
Identification of multiple
phases in microcrystalline
mixtures
Recognition of amorphousmaterials in partially crystalline
mixtures
Quantitative Analysis
Lattice Parameter
Determination
Phase Fraction Analysis

Structure Refinement
Rietveld Method
Structure Solution
Reciprocal Space Methods
Real Space Methods
Thermal expansion and Phase
Transitions

Peak Shape Analysis
Crystallite Size Distribution
Microstrain Analysis
Extended DefectConcentration

What is special about a crystal?
Solid phases are often crystalline, but need not be e.g. glass an “amorphous material”
CrystalGlass• Fractures into shards• Takes on any shape,depending on preparation• Properties do not varywith orientation.

Si• Cleaves along preferreddirections• Grows with welldeveloped crystal faces• Properties depend onorientation in which theyare measured.

Oxygen1

Crystal Structure

The Unit Cell

• CRYSTAL: Contains a periodical array of atoms/ions.
This can be represented by a simple lattice of points.

c• A group of atoms is associated with each lattice points.

β αa γ• LATTICE: An infinite array of points in space, in which
each point has identical surroundings to all others.
• CRYSTAL STRUCTURE: The periodic arrangement ofatoms in the crystal.

Example of 2D symmetry ina wallpaper patternbThe unit cell is a basic parallelopiped shaped block from which thewhole volume of the crystal may be built by repetition in 3 dimensions.
Any point in the unit cell may be specified with respect to the origin byparameters x, y, z measured parallel to the unit cell axes and expressed
as fractions.

Example of 2D symmetry in awallpaper pattern
To show symmetry:

To show symmetry:
1. Pick a point●

2. Find all equivalent points●●

1. Pick a point
2. Find all equivalent points
These points form a 2D lattice●Connecting 4 lattice points toform a parallelogram gives apossible unit cell(http://www.clarku.edu/~djoyce/wallpaper/)2

Example of 2D symmetry in awallpaper pattern
Connecting 4 lattice points toform a parallelogram gives apossible unit cell

Different unit cells can bechosen●●But some unit cells are
preferable for higher symmetry

PC Lattice

BCC Latticeα-Po is primitiveCubic
Identical atoms atcorners but nothingat the and body orface centers.
Lattice type P

α-Iron is BodyCentered Cubic
Identical atoms atcorners and bodycenter (nothing atface centers)
Lattice type I
Also Nb, Ta, Ba,Mo...

Bravis Lattice Conditions

Triclinic

1 (P)Monoclinic

2 (P, C)

a≠b≠c
α≠β≠γ
a≠b≠c
α = β = 90° ≠ γa≠b≠c
α = β = γ = 90°a=b≠cα = β = γ = 90°
a=b=cα = β = γ = 90°z●

● Unit cell – the basic unit that
repeats in every direction

Name

Orthorhombic 4 (P,F,I,A)

β αγcaxb

Tetragonal2 (P, I)

Cubic

3 (P, F,I)

Trigonal

1 (P)

a=b=c
α = β = γ < 120° ≠ 90°

1 (P)

a=b≠c
α = β = 90°
γ = 120°

yLattice parameters:
a, b, c; α, β, γHexagonal

PCIF

FCC Lattice

Sodium Chloride(NaCl) Na is muchsmaller than Cl
Face Centered Cubic
Rocksalt structure
Lattice type F
Also NaF, KBr,MgO….

3

Young’s Double Slit Experiment• Thomas Young first
demonstrated interference in lightwaves from two sources in 1801
• Light is incident on a screen witha narrow slit, So• The light waves emerging fromthis slit arrive at a second screenthat contains two narrow, parallelslits, S1 and S2• The narrow slits, S1 and S2 act assources of waves• The waves emerging from theslits originate from the same
wave front and therefore arealways in phase

Interference Patterns• Constructive interference occurs
at the center point• The two waves travel the same
distance– Therefore, they arrive in phase• The upper wave travels one
wavelength farther than the lower
wave– Therefore, they arrive in phase• A bright fringe occurs

Resulting Interference Pattern• The light from the two slits form a visible pattern ona screen• The pattern consists of a series of bright and darkparallel bands called fringes• Constructive interference occurs where a brightfringe occurs• Destructive interference results in a dark fringeInterference Patterns• The upper wave travels
one-half of a wavelength
farther than the lowerwave• The trough of the bottom
wave overlaps the crest of
the upper wave (180°phase shift)• This is destructive
interference– A dark fringe occurs

4Interference Equations• The path difference, δ, is found from the tan triangle• δ = r2 – r1 = d sin θ

Diffraction of X-ray Waves• Diffraction: When light passes sharpedges or goes through narrow slits therays are deflected and produce fringes oflight and dark bands.
Diffraction grating and helium-neon laser• For a bright fringe, produced by constructive interference, the path
difference must be either zero or some integral multiple of of thewavelength• δ = d sin θbright = m λ– m = 0, ±1, ±2, …
– m is called the order number

Bragg’s Law
Incident radiation“Reflected” radiation

Lattice Planes
12Pθ
SdθRQ• It is possible to describe certain directions andplanes with respect to the crystal lattice using a
set of integers referred to as Miller Indicies

Transmitted radiation

Beam “2” travels the extra distance SQR

nλ = SQ + QR
= d hkl sin θ + d hkl sin θ= 2d hkl sin θ
But not all planes result in diffraction !!!

5Crystallographic Directions AndPlanes

Examples of Miller IndicesaLattice Directions
Individual directions: [uvw]Symmetry-related directions: <uvw>
Miller Indices:1. Find the intercepts on the axes in terms of the lattice
constant a, b, c
2. Take the reciprocals of these numbers, reduce to thethree integers having the same ratio
(hkl)
hkl)
Set of symmetrysymmetry-related planes: {hkl
{hkl}
hkl}

(100)
yxz

Families of Planes• different families of planes– d-spacing between (400) planes is 1/4 that of the (100) spacing.
– The (300) plane does not contain atoms and so is not observed

1h2 + k 2 + l 2=2d hkl
a2

(200)

(400)

For cubic system with a = 4.0 A

ad100 = 4.0(100)(110)

Lattice Spacing• Miller indices describe the orientation of a family of planes– the spacing between adjacent planes in a family is referred to as a“d-spacing”(200)

d110 = 2.828

(300)

6

Single Crystal vs Powder

Powder – A Polycrystalline Mass
All orientations ofcrystallites possible
Single crystal reciprocal lattice- smeared into spherical shells

McCusker

Powder Diffraction
MonochromatorθSourceθ
λ=2dsinθ
Detector(200)

Sample2750
2500

2θNaCl - 0.809 Å(220)2250
200017501500(222)

1250
1000
500(311)750(111)

Intensity• By varying the angle θ, the Bragg's Law conditions aresatisfied by different d-spacings in polycrystalline
materials.
• Plotting the angular positions and intensities of theresultant diffracted peaks produces a pattern which ischaracteristic of the sample.

25001520

2530

Two Theta (degrees)7

Information Contained in a
Diffraction PatternOn
M
1.
e0
C
1.
r0
S
8.
a0C

15000

3

Intensity

10000
5000
0100

120

33O
n
M
1.
e0C.8
r0S.1
a0CO
nM.1e0C.1r0S.8a0C

If the pathlength between
rays 1 and 2 differs by λ, thepath length between rays 1and 3 will differ by λ/2 anddestructive interference in(b) will lead to no diffracted
intensity

Intensity

2000
1000
029.8

30.0

30.2

30.4

Two Theta

29.2 29.4 29.6 29.8 30.0 30.2
Two Theta500

α−Fe - Im3m

NaCl Fm3m
FCC

BCC
(220)

Centering and Absences(200)

Changes in symmetry and microstrain
upon chemical substitution can beestablished by examination of thepatterns(110)

Peak Shapes & Widths
Crystallite Size (2-200 nm)
Non-uniform microstrain
Extended Defects (stacking faults,etc.)

1200

200600(113)(111)

100(222)

Intensity(211)

300(200)

Intensity400• We can extend these types of calculation
to include other modes of lattice centering.
They all lead to systematic absences

00
20

30

20

4030

Simple (Primitive)

Reflections that must
be absentnone

Monoclinic
a = 5.45 b = 5.25 c = 5.00 Åo
β= 100Pm3m

Orthorhombic
a = 5.45 b = 5.25 c = 5.00 ÅIm3m

Base (C) centeredh and k mixedTetragonal
a = 5.45 c = 5.00 ÅFm3m

Body (I) centered(h+k+l) odd

1015

20

2530

35

40

4550cubica = 5.45 Å55

Two Theta

Face (F) centeredh, k and l mixed40

Two Theta (degrees)

Two Theta (degrees)

Bravais lattice

(511)

80

(420)

60(422)

40

3000

(331)

20
4000

Peak Intensities
Unit Cell Contents
Point Symmetry
Quantitative Phase Fractions• The positions of the atoms in a unit cell determine the intensities ofthe reflections• Consider diffraction from (100) planes in(400)

Peak Positions
Crystal System
Space Group Symmetry
Unit Cell DimensionsQualitative Phase Identification

Centering and Absences

Influence of centering

1015202530354045505560657075

Two Theta

Influence of symmetry

8Multiplicity

Diffraction Patterns• For high symmetry materials the Bragg angles and d-spacings fordifferent reflections may be equivalent to one anotherFor example (100), (010), (001) etc are equivalent in a cubic
material•

In a powder, all planes with the same d-spacing contribute to thescattered intensity at a given Bragg angle• The number of planes that are symmetry equivalent is referred to as
the multiplicity and its appears as a multiplicative term in powder
diffraction intensity calculations• The multiplicity of a reflection depends upon the symmetry of thecrystal
Multiplicity of {100} for cubic is 6, but for tetragonal it would
only be 4 as (100) and (001) are not equivalent

Need for High Q
50000

Bi2InNbO7 RT
0.79 Å X-rays

orderedmodel

disorderedmodel

Intensity

30000

20000

5658606264565860

Need for High Resolution

There are many more reflections at higher Q.
Therefore, most of the structural information isat higher Q

40000

62• Spacing of peaks depends on size of unit celland the space group.
• The bigger the unit cell and/or the lower thesymmetry the more diffraction peaks areobserved.
• Intensity of peaks depends on (amongst otherthings) the arrangement of the atoms in the unitcell.
• For two materials that had identical unit cells, thepeak positions would be IDENTICAL, however
their intensities would be DIFFERENT.

64

10000

Differentiating Braggs Law gives theresolution as:∆d ∆λ ∆θ=+dλ tan θ

λ = 2d sin θ

03040506070

Source80Two θ (Degrees)

Refinement of structure gave
unusual displacement parameters
for the Bi cations, indicative ofcation disorder. The patterns
could only be adequately fitted byincluding 6-fold disorder of the Bi.
This involves a displacement alongthe (1 -1 0) direction

AtomBiIn/NbO(1)O(2)
BiSitexyz

Model 1. Ordered Bi. Rp 4.08 Rwp 6.07%
16d00.25
0.75
16c000Biso
2.74(6)3.00(8)48f
0.350(3)
0.125
0.125
7.6(7)
8b0.375
0.375
0.375
7.6(7)
Model 2. Disordered Bi Rp 3.09 Rwp 3.93%
96h
00.2249(1) 0.7751(1) 0.96(7)In/Nb16c0000.61(3)

O(1)
O(2)

48f8b0.322(1)
0.375

0.125
0.375

0.125
0.375

1.7(2)
1.7(2)

SampleSource +Instrument

Resolution
In Powder Diffraction it typically refers
to the width of the peaks.
In Single Crystal Diffraction it typicallyrefers to the minimum d-space studied.

Rutile
HRPD 1.49 ÅIntensity

20MRPD 1.66 Å

2040

60

80

100

120

140

Two Theta (Degrees)

Both definitions are relevant.

9Peak Overlap

The Solution - Rietveld
k p2yicalc = y iback + ∑ ∑ G ikp I 2k

• Powder Diffraction
patterns are a one
dimensional
representation of athree dimensional
structure.
• Often peaks due toindividual Braggreflections overlap

p k = k1p

•
•
•
•
•

yic the net intensity calculated at point i in the pattern,yiback is the background intensity,
Gik is a normalised peak profile function,
Ik is the intensity of the kth Bragg reflection,
k1 ... k2 are the reflections contributing intensity topoint i,• the superscript p corresponds to the possible phasespresent in the sample.

The Answers
Rp =∑ y iobs − yicalc∑ yiobs

10000∑ w ( y − y ) = ∑ w y

1/2

2• The weighted ProfileR• The expected Profile
RR wp

iiobs
i

∑ w (yχ =2i2iobs

8000

Anatase
6000

4000

Rutile

2000

1/2

020

40

60

80

Two Theta (degrees)− y icalc )

2iobs

N-P

• The Goodness of fiticalc

 N-P R exp = 2 ∑ w i y iobs Anatase + Rutile

12000

Intensity (a.u.)• The Profile R

Phase Analysis

 R wp = R exp 2• Where a mixture of different phases is present,
the resultant diffraction pattern is formed byaddition of the individual patterns.
• The intensity of the peaks is proportional to theamount of the phase present.

10Quantitative Phase Analysis• Bragg scattering is proportional to N/V where N is thenumber of unit cells and V the unit cell volume.
There for the weight of a phase in the beam is:
WP =An Example• Synchrotron X-ray Diffraction pattern for
SrRuO3( SZMV ) P∑ (SMPV )i

70000

SrRuO3 Observed

60000

i

50000
40000

IntensityS - the scale factor
Z the number of formula unites per unit cell
M the molecular weight of the formula unit
I is the index running over all phases

30000
20000
10000
010203040Angle (2θ) degrees• Hence SZVM is proportional to the weight of thediffracting sample

The background

Peak Shapes– Incoherent– Temperature diffuse• Other materials– Specimen holder– air etc

6000

4000

2000

10203040Angle (2θ) degrees• Background can beeither fitted or
estimated.
• Here the capillary isa feature.

• Different
Diffractometers havedifferent peak shapes.
• The most widely
function is a pseudoVoigt (mixed Gaussianand Lorentzian).
G ik = γ

1/20[]

gamma0
gamma05
gamma1
Intensity

Intensity• Fluorescentradiation from thesample• Diffraction from thecontinuousspectrum• Diffuse scattering

Angle

1/2
1
1/2-1CC
1 + C0 X ik2 + (1 + γ)
Hk π
Hkπ

[exp - C1X ik2

H2 = Utan2θ+ Vtanθ + W]

NeutronBragg-Brentano
Synchrotron

0.4• The width of peaks isusually not constant.

FWHM (degrees)

8000

0.20.0020406080100120140

Two Theta (degrees)

11Peak Asymmetry• The fit to a single
phase sample looks
good

60000
50000

R p 2.42%
R wp 3.85%

40000

Intensity

asym
perfect

Intensity

2

χ 254

30000
20000
10000
0

9.65

9.70

9.75

9.80

9.85

Angle

Intensity

asym
perfect

0

35.50

35.55

35.60

35.65

35.70

35.75

35.80

Angle

A Common Problem• If the structural model is wrong then the mostcommon response of Rietveld programs is to:– broaden the peaks,– Increase the displacement parameters,• The former is most noticeable at high angles
where intensity is lowest.
• Due to absorption of the X-rays powder X-ray
diffraction often yields poor displacement
parameters

10• BUT…...

15202530354045

Angle (2θ) degrees• The detail of the fit
is not satisfactory the model is missing
something!

Rp 2.42%
Rwp 3.85%
2

χ 254

Intensity

9.60

354045Angle (2θ) degrees

An Alternate Model• The high angle
splitting is wellmodeled by atetragonal model - but
this overestimatessome intensities.

Rp 2.25%
Rwp 3.27%

100002χ 182.9

Intensity• Beam Divergence canresults in asymmetricpeaks at low angles.
• Results from not
integrating over theentire Debye cone.

The Simple Structural Model

0354045Angle (2θ) degrees• The Truth lies somewhere in the middle

12Strengths and Limitations ofPowder X-ray Diffraction

The finished Product
60000

SrRuO3 828 K λ= 0.700 ÅRP 1.79%
Rwp 2.22%

50000

Strengths2

χ 84.5Intensity (au)40000

Limitations

30000
20000
10000
0102030

40

5060

70

8090

Two Theta (degrees)• The sample contains a mixture of bothphases!

• Non-destructive – small
amount of sample• Relatively rapid• Identification of compounds /phases – not just elements• Quantification of concentrationof phases – (sometimes)• Classically for powders, butsolids possible too• Gives information regarding
crystallinity, strain, crystallite
size, and orientation• Bulk technique – generally –unless a microfocus source isused• Not a “stand-alone” technique– often need chemical data
• Complicated appearance• multiphase materials –identification /quantification
can be difficult

Experiment Design Issues
What Wavelength?
• Absorption is your enemy!
• Short Wavelengths are best! BUT….
• Consider required resolution. And…
• Avoid Absorption Edges.
What Size Capillary?
• Small capillaries reduce absorption AND (witharea detectors) improve resolution.
• BUT reduce amount of material.

13