Powder X
X--ray Diffraction

Brendan J. Kennedy
School of Chemistry
The University of Sydney

Uses ofo Powderowde Diffraction
c oQualitative Analysis
Identification of single-phaseg pmaterials
Identification of multiple phasesin microcrystalline mixtures
Recognition of amorphousmaterials in partially crystalline
mixtures
Quantitative Analysis
Lattice Parameter Determination
Phase Fraction Analysis
Peak Shapep AnalysisyCrystallite Size Distribution
Microstrain Analysis
Extended Defect Concentration

Structure Refinement
Rietveld Method
Structure Solution
Reciprocal
p
Space
p
Methods
Real Space Methods
Thermal expansion andPhase Transitions2

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

3What is specialpabout a crystal?
y
Solid phases are often crystalline, but need not be - e.g.
glasslan “amorphous“hmaterial”
t i l”
CrystalGlass• Fractures into shards• Takes on any shape,depending on preparation• Properties do not vary withorientation.

Si

• Cl
Cleaves alonglpreferredfddirectionsp• Grows with well developed
crystal faces• Properties depend onorientation in which they aremeasured.

Oxygen4

Crystaly
Structure• CRYSTAL: Contains a periodical array of atoms/ions. This canbe represented by a simple lattice of points.
points• A group off atomst
i associatedis
i t d with
ith eachh latticel tti points.
i t• LATTICELATTICE: An
A iinfinite
fi it array off points
i t ini space, in
i whichhi h eachh
point has identical surroundings to all others.
• CRYSTAL STRUCTURE: The periodic arrangement of atoms
in the crystal.
crystal5The Unit Cell
c
 
a 
b
The unit cell is a basic parallelopiped shaped block from which thewhole volume of the crystal may be built by repetition in 3
dimensions.
dimensions
Any point in the unit cell may be specified with respect to the originpx,, y, z measured parallelp
to the unit cell axes andbyy parameters
expressed as fractions.
6

Example of 2D symmetry
in a wallpaper pattern
Too sshowow sy
symmetry:e y:
1. Pick a point
2. Find all equivalentpoints(http://www.clarku.edu/~djoyce/wallpaper/)7

Example of 2D symmetry ina wallpaper pattern
T showTo
h symmetry:t
1 Pick a point
1.
2. Find all equivalent points
These points form a 2D lattice
Connecting 4 lattice points toform a parallelogram gives apossible unit cell8

Example of 2D symmetry in awallpaper pattern
ConnectingC
ti 4 latticel tti points
i t totform a parallelogram gives apossible unit cell●

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

Differentt unit
Diff
it cells
ll can bebchosen●But some unit cells are
preferable for higher symmetry
●

9

z

a
x

b

Bravis Lattice Conditions

Triclinic

1 ((P))

Monoclinic

2 (P, C)

abc

abc
 =  = 90  abc =  =  = 90
a=bc =  =  = 90
a=b=c =  =  = 90

Orthorhombic 4 (P,F,I,A)(P F I A)

 


c

Name

y

Tetragonal

2 (P, I)

Cubic

3 (P, F,I)

Trigonal

1 (P)

a=b=c
 =  =  < 120  90

1 (P)

a=bc
 =  = 90
 = 120

Lattice parameters:
a, b, c; 
Hexagonal

P

C

I

F

10

PC Lattice

BCC Lattice-Po is primitiveCubic
Identical atoms at
corners but nothingat the and body orface centers
centers.
Lattice type P

-Iron is BodyCentered Cubic
Identical atoms at
corners and body
center (nothing at
face centers)
Lattice type I
Also Nb, Ta, Ba, Mo...

FCC Lattice

Sodium Chloride(NaCl) Na is muchsmaller than ClF
Face
Centered
C t
d Cubic
C bi
Rocksalt structure
Lattice type F
Also NaF, KBr,MgO….
11

12

Diffraction of XX-ray
y Waves• Diffraction: When light passes sharp edges orgoes through narrow slits the rays areproduce fringes
g of lightg and dark
deflected and pbands.
Diffraction grating and helium-neon laser

13Young’sYoung 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 phase14

Resulting Interference Pattern• The light from the two slits form a visible pattern ona screen• The pattern consists of a series of bright and dark
pparallel bands called ffringesg• Constructive interference occurs where a brightfringe occurs
15• Destructive interference results in a dark fringeInterference Patterns• Constructive interference occurs
at the center point• The two waves travel the same
distance– Therefore,, theyy arrive in pphase• The
Th upper wave travelstl one
wavelength farther than the lower
wave– Therefore, they arrive in phase• A bright fringe occurs
16Interference Patterns• The upper wave travels
one-half of a wavelength
farther than the lower
wave• The trough of the bottom
wave overlaps the crest of
the upper wave (180
phase shift)• Thishi isi destructivediinterference– A dark fringe occurs
17Interference Equationsq• The path difference, δ, is found from the tan triangle• δ = r2 – r1 = d sin θ••

For a bright fringe, produced by constructive interference, the pathg multiple
p of of thedifference must be either zero or some integral
wavelengthδ = d sin θbright = m λ– m = 0,0 ±1,
±1 ±2,±2 …
– m is called the order number
18

Bragg’s Law
Incident radiation“Reflected” radiation

1
2

P


S

d


R

Q

Transmitted radiation

Beam “2” travels the extra distance SQRn  SQ  QR
 d hkl sin   d hkl sin  2d hkl sin 
But not all planes result in diffraction !!!

19

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

20Crystallographic Directions And
Planes
Lattice Directions
Individual directions: [uvw]Symmetry-related directions: <uvw>
Miller Indices:
1 Find the intercepts on the axes in terms of the lattice
1.
constant a, b, c
2 Take the reciprocals of these numbers,
2.
numbers reduce to thethree integers having the same ratio(hkl)
21
Set of symmetry-related planes: {hkl}Examples
p of Miller Indices
a

(100)

(200)

y
x
z

(110)

22

Families of Planes• Miller indices describe the orientation of a family of planes– the spacingp
g between adjacentjplanesp
in a familyy is referred to as a “dspacing”• 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

(100)

(200)

(400)

(300)
23

Lattice Spacing1h2  k 2  l 2

2
2
d hkl
a

For cubic system with a = 4.0 A

ad100 = 4.0

d110 = 2.828

24

Single Crystal vs Powder

McCusker25

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

26

• B
By varyingi theh anglel , theh Bragg's
B' L
Law conditions
di iaresatisfied by different d-spacings in polycrystalline materials.
• Plotting the angular positions and intensities of the resultant
diffracted peaks produces a pattern which is characteristic of
the sample.

27

Powder Diffraction
MonochromatorθSourceθ
λ=2dsinθ
D
Detector(200)

Sample2750
2500

NaCl - 0.809 Å
(220)

2250
2000
1750
1500

(222)

1250
1000
500

(311)

750

(111)

Intens
sity

2θ

250
0
15

20

25

30

Two Theta (degrees)28Intensity• The intensity of the diffracted beam can be given by:2
I  SLP Fhkl
• S is a scale factor that is dependent on the intensity of thesource and the detector efficiency.
• LP is a combination of the Lorentz and polarisation factors
which for a non-polarised X-ray beam is given by:LP 

1  cos 2 2sin 2   cos29

The Structure Factor Fhkl
• For a reflection from a plane hkl. The structure
f t Fhkl is:factori

Fhkl 

uniti  cell
ll



i 1


sin 2  hkl 
f i  exp  Bi
 exp[2ihkl  xi ]

2




• xI = (x( iyizi) are theth fractionalf ti l coordinates
di t forf atom
t
i in
i the
th
cell
• hkl are the Miller indices for the reflection• Bi is the Debye-Waller displacement (thermal) factor for
atom (in Å2)30• fi. Is the scattering factor.

Information Contained in a
Diffraction Pattern
a0
C

15000

3
O
n
M
1
.
0
e
C
1
.
0
r
S
8
.

10000

Inttensity

Peak Positions
Crystal System
Space Group Symmetry
Unit Cell Dimensions
Qualitative Phase Identification

5000
0

20

40

100

120

Intensity

3000
2000
1000
0
30.2

Two Theta

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

80

3
O
n
M
1
.
0
e
C
8
.
0
r
S
1
.
0
a
C

3

1
.

30.0

60

O
n
M

e0
C

29.8

1
.
0

r
S
8
.
a0
C

Peak Intensities
Unit Cell Contents
Point Symmetry
Quantitative Phase Fractions
Q

4000

30.4

29.2 29.4 29.6 29.8 30.0 30.2
Two Theta

Changes in symmetry and microstrain uponchemical substitution can be established byexamination of the patterns

31Centering and Absences••

The positions of the atoms in a unit cell determine the intensities of thereflections
Consider diffraction from (100) planes in

If the pathlength between
rays 1 and 2 differs by λ,the path length between
rays 1 and 3 will differ by
λ/2 and destructive
interferencee e e ce in(b) will lead to nodiffracted intensity
32

Centeringg and Absences• We can extend these types of calculation toi l d otherincludeh modesd off latticel i centering.
iTheyhall lead to systematic absences
Bravais lattice
Simple (Primitive)

Reflections that must beabsentnone

Base (C) centeredh and k mixed

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

Face (F) centeredh, k and l mixed

33

NaCl Fm3m
FCC

(200)

Fe - Im3m

(511)

(422)

(113)

(111)

100

600

22)
(22

Intensity

(211)

200

(200)

Intensity

300

(331)

1200

400

(420)

(220)

BCC

(400)

(110)

500

0

0
20

30

20

40

30

40

Two Theta (degrees)

Two Theta (degrees)

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

Orthorhombic
a = 5.45 b = 5.25 c = 5.00 Å

Im3m

Tetragonal
a = 5.45 c = 5.00 Å

F 3
Fm3m
10

15

20

25

30

35

40

45

50

cubic
a = 5.45 Å
55

Two Theta

Influence of centering

10

15

20

25

30

35

40

45

50

55

60

65

70

75

Two Theta

Influence of symmetry 34

Multiplicitypy• 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 toas theh multiplicityl i li i andd its
i appears as a multiplicativel i li i term inipowder diffraction intensity calculations• The multiplicity of a reflection depends upon the symmetry of
the crystal
M lti li it off {100} for
Multiplicity
f cubic
bi is
i 6,
6 but
b t forf tetragonalt t
l it
would only be 4 as (100) and (001) are not equivalent35

Diffraction Patterns• Spacing of peaks depends on size of unit cell and thespace group.
• The bigger the unit cell and/or the lower thesymmetryyy the more diffraction ppeaks are observed.
• Intensity of peaks depends on (amongst other things)
the arrangement of the atoms in the unit cell.
• For two materials that had identical unit cells, thepeak positions would be IDENTICALIDENTICAL, however theirintensities would be DIFFERENT.
36

Need for High
g Q
50000

Bi2InNbO7 RT
0.79 Å X-rays

orderedmodel

disordered
model

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

40000

Inttensity

30000

20000

56

58

60

62

64

56

58

60

62

64

10000

0

20

30

40

50

60

70

80

Two  (Degrees)

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

Atom
Bi
In/Nb
O(1)
O(2)
Bi

Site

x

y

z

Model 1. Ordered Bi. Rp 4.08 Rwp 6.07%
16d
0
0.25
0.75
16c

0

0

0

Biso
2.74(6)3.00(8)48f
0 350(3)0.350(3)
0 125
0.125
0 125
0.125
7 6(7)
7.6(7)
8b
0.375
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)

I /Nb
In/Nb

16
16c

0

0

0

0 61(3)0.61(3)

O(1)
O(2)

48f
8b

0.322(1)
0.375

0.125
0.375

0.125
0.375

1.7(2)
1.7(2)

37

Atomic scattering factors as fcore and fvalence
C

Fe2+

38

Need for High Resolution

SourceSample  2d sin 
Source +Instrument

Resolution
In Powder Diffraction it typically refers to
the width of the peaks
peaks.
In Single Crystal Diffraction it typicallyrefersf tto theth minimum
i id-spacedstudied.
t di dRutile
HRPD 1.49 Å

Intensityd  d tan Differentiating Braggs Law gives theresolution as:

MRPD 1.66 Å

20

40

60

80

100

120

Two Theta (Degrees)

Both definitions are relevant.

140

39

Peak Overlap• Powder Diffraction
patterns are a one
dimensional
representation of athree dimensional
structure.
• Often peaks due toindividual Braggreflections overlap
40Phase Analysis
Anatase + Rutile

12000

a.u.)
Intensity (a

10000

8000

Anatase
6000

4000

Rutile

2000

0
20

40

60

80

Two Theta ((degrees)g)• Where a mixture of different phases is present, theresultant diffraction ppattern is formed byy addition of
the individual patterns.
• The intensity of the peaks is proportional to theamount of the phase present.
41

Quantitative Phase Analysis
Qy• Bragg scattering is proportional to N/V where N is thenumber of unit cells and V the unit cell volume. There forthe weight of a phase in the beam is:( SZMV ) P
WP  (SMPV )ii

S - the scale factor
Z theth numberb off formulafl unites
it per unit
it cellll
M the molecular weight of the formula unit
I iis theth indexi d runningi over allll phasesh• H
Hence SZVM iis proportional
i l to theh weight
i h off thehdiffracting sample

42

The Solution - Rietveld
k p2

y icalc  y iback    G ikp I 2k
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,reflection
k1 ... k2 are the reflections contributing intensity to point i,
the superscript p corresponds to the possible phases present
in the sample.
43

Why is wavelength important?

samplesampleVi ibl light
Visibleli ht

X rays
X-rays

To penetrate a sample you need a wavelength of similar, or smaller, magnitude44

An Examplep• Synchrotron X
X-ray
ray Diffraction pattern for
SrRuO370000

SrRuO3 Observed

60000
50000

Intensity

40000
30000
20000
10000
0
10

20

30

40

Angle (2) degrees

45

The backgroundg
Intensity• Fluorescentradiation from thesample• Diffraction from thecontinuous
spectrum• Diffuse scattering– Incoherent– Temperature diffuse• Other materials– SpecimenSihholderld– air etc

8000

6000

4000

2000

10

20

30

40

Angle (2) degrees• Background can beeither fitted or
estimated.
• Here the capillary is afeature.
46

Peak Shapesp• Different
Diffractometers havedifferent peak shapes.
• The most widely
function is a pseudoVoigt (mixed Gaussian
anddL
Lorentzian).
t i )



Intensity

gamma0
gamma05
gamma1

Angle





1/2
1
C1/2
C
2
G ik  γ 0 1  C 0 X ik  (1  γ) 1 1/2 exp - C1X ik2
Hk π
Hk π

NeutronBragg-Brentano
Synchrotron

H2 = Utan2+ Vtan + W

FWHM (degree
es)

0.4

• The width of peaks is
usually not constant.



0.2

0.0
0

20

40

60

80

Two Theta (degrees)

100

120

140

47

Peak Asymmetry
asym
perfect

Intensity

9.60

9.65

9.70

9.75

9.80

9.85

Angle

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

0

35.50

35.55

35.60

35.65

35.70

35.75

35.80

Angle

48

The Simple Structural Model
60000
50000

Rp 2.42%
Rwp 3.85%

40000

Intensity• The fit to a single
phase sample looks
good

2

 254

30000
20000
10000
0

• BUT…...

15

20

25

30

35

40

45

Angle (2) degrees

Rp 2.42%
Rwp 3.85%
2

 254

In
ntensity• The
Th detaild il off the
h fit
fi is
i
not satisfactory - themodeld l isi missingi isomething!

10

35

40

45

Angle
g ((2) degreesg49

A Common Problem• If the structural model is wrong then the mostcommon response off Rietveld
Ri t ld programs iis tto:
– broaden the peaks,– Increase the displacement parameters,• The former is most noticeable at high angles
where intensity is lowest.
pof the X-rays
y p
powder X-rayy• Due to absorption
diffraction often yields poor displacementpparameters
50

An Alternate ModelRp 2.25%
Rwp 3.27%
3 27%

10000

2

 182.9

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

0
35

40

45

Angle (2) degrees• The Truth lies somewhere in the middle
51

The finished Product
60000

SrRuO3 828 K = 0
0.700
700 Å
RP 1.79%
Rwp 2.22%

50000

2

 84.5Inteensity (au)

40000
30000
20000
10000
0

10

20

30

40

50

60

70

80

90

Two Theta (degrees)• The sample contains a mixture of both phases!
52

Strengths and Limitations of
Powder X
X--ray Diffraction
Strengths
•
•
•
•
•
•

Non destructive – small
Non-destructive
amount of sample
Relatively rapid
de t cat o of
o compounds
co pou ds /Identification
phases – not just elements
Quantification of concentrationof phases – (sometimes)
Classically for powders, but
solids possible too
Gives information regarding
crystallinity,t lli it strain,t i crystallite
t llitsize, and orientation

Limitations••••

Bulk techniqueq –ggenerallyy–unless a microfocus source isused
Not a “stand-alone” techniqueq– often need chemical data
Complicated appearance
multiphase materials –identification /quantification
can be difficult
53

Experimentp
Designg Issues
What Wavelength?
g• Absorption is your enemy!
• Short Wavelengths are best! BUT
BUT….
• Consider required resolution. And…
• Avoid Absorption Edges.
What Size Capillary?
• Small capillaries reduce absorption AND (witharea detectors) improve resolution.
• BUT reducedamountt off material.
t i l54