Crystal structure analysis

C t l St t An l iCrystal Structure Analysis
X-ray Diffraction
Electron DiffractionElectron Diffraction
Neutron Diffraction
Essence of diffraction: Bragg DiffractionEssence of diffraction Bragg Diffraction
Reading: Basic West 3
West 5West 5
A/M 5-6
G/S 3 217

REFERENCES
Elements of Modern X-ray Physics, by Jens Als-Nielsen and Des McMorrow,
John Wiley & Sons, Ltd., 2001 (Modern x-ray physics & new developments)
X-ray Diffraction, by B.E. Warren, General Publishing Company, 1969, 1990
(Classic X-ray physics book)
Elements of X ray Diffraction 2nd Ed by B D Cullity Addison Wesley 1978Elements of X-ray Diffraction,2nd Ed., by B.D. Cullity, Addison-Wesley, 1978
(Covers most techniques used in traditional materials characterization)
High Resolution X-ray Diffractometry and Topography, by D. Keith Bowen
and Brian K. Tanner, Taylor & Francis, Ltd., 1998 (Semiconductors and thin
film analysis)
Modern Aspects of Small-Angle Scattering by H Brumberger Editor KluwerModern Aspects of Small Angle Scattering, by H. Brumberger, Editor, Kluwer
Academic Publishers, 1993 (SAXS techniques)
Principles of Protein X-ray Crystallography, by Jan Drenth, Springer, 1994
(C t ll h )(Crystallography)
218

SCATTERING
Scattering is the process in which waves or particles are forced to deviate from a
X-rays scatter by interaction with the electron density of a material.
Neutrons are scattered by nuclei and by any magnetic moments in a sample
g p p
straight trajectory because of scattering centers in the propagation medium.
Neutrons are scattered by nuclei and by any magnetic moments in a sample.
Electrons are scattered by electric/magnetic fields.
 p' p q E' E hMomentum transfer: Energy change:
Elastic (E’ = E)
p p q E E h Momentum transfer: Energy change:
El i i• Rayleigh (λ >> d )
E pcFor X-rays:
Elastic scattering geometry• Rayleigh (λ >> dobject)
• Mie (λ ≈ dobject)
• Geometric (λ << dobject)
• Thompson (X-rays)
Inelastic (E’ ≠ E)
Thompson (X rays)
• Compton (photons + electrons)
q 2 sin
2
p 


• Compton (photons + electrons)
• Brillouin (photons + quasiparticles)
• Raman (photons + molecular vib./rot.)

COMPTON SCATTERING
Compton (1923) measured intensity of scattered X-rays
Compton
Compton (1923) measured intensity of scattered X-rays
from solid target, as function of wavelength for different
angles. He won the 1927 Nobel prize.
X-ray source
Crystal
(selects
Collimator
(selects angle)
wavelength)
θ
Graphite
Target
θ
Result: peak in scattered radiation
hift t l l th th
Detector
shifts to longer wavelength than
source. Amount depends on θ (but
not on the target material). A. H. Compton. Phys. Rev. 22, 409 (1923).

COMPTON SCATTERING
Cl i l i t ill ti l t ti fi ld ill ti i itiClassical picture: oscillating electromagnetic field causes oscillations in positions
of charged particles, which re-radiate in all directions at same frequency and
wavelength as incident radiation (Thompson scattering).
Change in wavelength of scattered light is completely unexpected classicallyChange in wavelength of scattered light is completely unexpected classically
Compton’s explanation: “billiard ball” collisions between particles of light
Oscillating
electron
Incident light wave Emitted light wave
Compton s explanation: billiard ball collisions between particles of light
(X-ray photons) and electrons in the material
pBefore After
θ
 p
Incoming photon
p
scattered photon
ep
Electron
p
scattered electron

B f Aft
COMPTON SCATTERING
θ
 pBefore After
Incoming photon
scattered photon
θ
ep
Electron
p
scattered electron
Conservation of energy Conservation of momentum
 
1/22 2 2 2 4
e e eh m c h p c m c     ˆ
e
h
 

  p i p p e e e e 

h
From this Compton derived the change in wavelength
 
 
1 cos
1 cos 0
e
c
h
m c
  
 
  
   c
12
Compton wavelength 2.4 10 mc
e
h
m c
 
   
222

COMPTON SCATTERING
Note that there is also an
unshifted peak at each angleunshifted peak at each angle.
This comes from a collision
between the X-ray photon andbetween the X-ray photon and
the nucleus of the atom.
h
 1 cos 0
N
h
m c
     
N em msincesince
223

COMPTON SCATTERING
C t ib t t l b k d iContributes to general background noise
DiffDiffuse
background from
Compton
emission byemission by
gamma rays in
a PET scan.
Fluorodeoxyglucose (18F)
224

X-RAY SCATTERING
X λ (i Å) 12400/E (i V)X-rays:
• 100 eV (“soft”) – 100 keV (“hard”) photons
• 12,400 eV X-rays have wavelengths of 1 Å,
λ (in Å) = 12400/E (in eV)
, y g ,
somewhat smaller than interatomic distances in solids
Diffraction from crystals!
R t
id l diff i (θ 5°)
elastic (Thompson, ∆E = 0)
Roentgen
1901 Nobel
• wide-angle diffraction (θ > 5°)
• small-angle diffraction (θ close to 0°)
• X-ray reflectivity (films)
inelastic (∆E ≠ 0)
• Compton X-ray scattering
• resonant inelastic X ray scattering (RIXS)• resonant inelastic X-ray scattering (RIXS)
• X-ray Raman scattering
First X-ray: 1895
225

DIFFRACTION
Diffraction refers to the apparent bending of waves around small objects and theDiffraction refers to the apparent bending of waves around small objects and the
spreading out of waves past small apertures.
In our context, diffraction is the scattering of a coherent wave by the atoms in a
t l A diff ti tt lt f i t f f th tt dcrystal. A diffraction pattern results from interference of the scattered waves.
Refraction is the change in the direction of a wave due to a change in its speed.
C l diff iCrystal diffraction
I. Real space description (Bragg)
II Momentum (k) space descriptionII. Momentum (k) space description
(von Laue)
diffraction of plane waves
W. L. BraggW. H. Bragg von Laue226

OPTICAL INTERFERENCE
δ λ
perfectly in phase:
δ = nλ, n = 0, 1, 2, …
δ: phase difference
n: order
perfectly out of phase:
δ = nλ, n = 1/2, 3/2, …
perfectly out of phase:

BRAGG’S LAW OF DIFFRACTION
When a collimated beam of X-rays strikes pair of parallel lattice planes in a crystalWhen a collimated beam of X rays strikes pair of parallel lattice planes in a crystal,
each atom acts as a scattering center and emits a secondary wave.
 All of the secondary waves interfere with each other to produce the diffracted beam
B id d i l i t iti h t diff tiBragg provided a simple, intuitive approach to diffraction:
• Regard crystal as parallel planes of atoms separated by distance d
• Assume specular reflection of X-rays from any given planep y y g p
→ Peaks in the intensity of scattered radiation will occur when rays
from successive planes interfere constructively
2Θ 228

BRAGG’S LAW OF DIFFRACTION
N k is bs d l ss th diti f st ti i t fNo peak is observed unless the condition for constructive interference
(δ = nλ, with n an integer) is precisely met:
AC sind 
ACB 2 sind 
ACBn 
2 id 2 sinn d Bragg’s Law:
When Bragg’s Law is satisfied “reflected” beams are in phaseWhen Bragg s Law is satisfied, reflected beams are in phase
and interfere constructively. Specular “reflections” can
occur only at these angles. 229

DIFFRACTION ORDERS
1st order:
2 sind  12 sind 
2nd order:
22 2 sind  2s 
By convention we set the diffraction order = 1 for XRDBy convention, we set the diffraction order = 1 for XRD.
For instance, when n=2 (as above), we just halve the d-spacing to make n=1.
2 2 sind  2( / 2)sind 22 2 sind  22( / 2)sind 
e.g. the 2nd order reflection of d100 occurs at same θ as 1st order reflection of d200

XRD TECHNIQUES AND APPLICATIONS
• powder diffraction
• single-crystal diffractions ng crysta ffract on
• thin film techniques
• small-angle diffraction
Uses:
• phase identification
• crystal structure determination
• crystal size
• residual stress/strain
Uses:
y
• radial distribution functions
• thin film quality
• crystallographic texture
• percent crystalline/amorphous
• defect studies
• in situ analysis (phase transitions,
thermal expansion coefficients, etc)
• superlattice structure• percent crystalline/amorphous • superlattice structure

POWDER X-RAY DIFFRACTION
• uses monochromatic radiation scans angle• uses monochromatic radiation, scans angle
• sample is powder → all orientations simultaneously presented to beam
• some crystals will always be oriented at the various Bragg angles
• this results in cones of diffracted radiationthis results in cones of diffracted radiation
• cones will be spotty in coarse samples (those w/ few crystallites)
no restrictionn n
on rotational orientation
relative to beam
crystallitecrystallite
2 sind 
232
2 sinhkl hkld 

DEBYE-SCHERRER METHOD
diff i i f h
2 sinhkl hkld 
…or we can use a diffractometer to intercept sections of the cones
234

BASIC DIFFRACTOMETER SETUP
235

DIFFRACTOMETERS
General Area Detector Diffraction System (GADDS)

THIN FILM SCANS
4-axis goniometer
237

THETA-2THETA GEOMETRY
• X ray tube stationary• X-ray tube stationary
• sample moves by angle theta, detector by 2theta
238

THETA-THETA GEOMETRY
• sample horizontal (good for loose samples)
• tube and detector move simultaneously through theta
239

POWDER DIFFRACTOGRAMS
In powder XRD, a finely powdered sample is probed with monochromatic X-rays of a
known wavelength in order to evaluate the d-spacings according to Bragg’s Law.
Cu Kα radiation: λ = 1 54 ÅCu Kα radiation: λ 1.54 Å
peak positions depend on:
• d spacings of {hkl}
Minimum d?
• d-spacings of {hkl}
• “systematic absences”
increasing θ, decreasing d
Minimum d?
min / 2d  240

ACTUAL EXAMPLE: PYRITE THIN FILM
FeS2 – cubic (a = 5 43 Å)FeS2 cubic (a = 5.43 Å)
Random crystal orientations Cu Kα = 1.54 Å
ity
“powder pattern”
111
200
210
211
220
311
Intensi
111 211
220
311
2 Th t2 Theta
2θ = 28.3° → d = 1.54/[2sin(14.15)]
= 3.13 Å = d111
reference pattern from ICDD
(250,000+ datasets)
On casual inspection, peaks give us d-spacings, unit cell size, crystal
symmetry, preferred orientation, crystal size, and impurity phases (none!)
111

Layered Cuprate Thin film growth oriented along c axis
EXAMPLE 2: La2CuO2
Layered Cuprate Thin film, growth oriented along c axis
2 theta d
002 sinld  
(hkl)12 2 Å
(00l)
2 theta d
7.2 12.1
(hkl)
(001)
c = 12.2 Å
Epitaxial film is textured.
14.4 6.1
( )
(002)
Epitaxial film is textured.
(It has crystallographic
orientation).
Many reflections are “missing”
22 4.0 (003)
Many reflections are missing
243

POWDER DIFFRACTION
Peak positions determined by size and shape of unit cell
(d-spacings and systematic absences)
Peak intensities determined by the atomic number and
position of the various atoms within the unit cellp f m
Peak widths determined by instrument parameters,Peak widths determined by instrument parameters,
temperature, and crystal size, strain, and inhomogeneities
we will return to this later…
244

GENERATION OF X-RAYS
X-rays beams are usually generated by colliding high-energy electrons with metals.y y g y g g gy
2p3/2 → 1s
X-ray emission
spectrum
+ HEAT
Siegbahn notation

Generating Bremsstrahlung
Generating Characteristic X-rays
246

GENERATION OF X-RAYS
Side window Coolidge X ray tubeSide-window Coolidge X-ray tube
X-ray energy is determined by anode material, accelerating voltage,
and monochromators:
Co Kα1 : 1.79 Å
C K 1 54 Å ( 8 k V)
 /hchE 
and monochromators:
Cu Kα1 : 1.54 Å (~8 keV)
Mo Kα1 : 0.71 Å
1/2
( )C Z 
 Moseley’s Law:
247

ROTATING ANODES
• 100X higher powers possible by spinning the anode
at > 6000 rpm to prevent melting it → brighter source
248

SYNCHROTRON LIGHT SOURCES
G V l t l t
• brightest X-ray sources
GeV electron accelerators
brightest X ray sources
• high collimation
• tunable energy
• pulsed operation
SOLEIL
p p
Bremsstrahlung (“braking radiation”)
249

MONOCHROMATIC X-RAYS
Filters (old way)Filters (old way)
A foil of the next lightest element
(Ni in the case of Cu anode) can(Ni in the case of Cu anode) can
often be used to absorb the
unwanted higher-energy radiation to
give a clean Kα beamgive a clean Kα beam
Monochromators
Use diffraction from a curvedUse diffraction from a curved
crystal (or multilayer) to select
X-rays of a specific wavelength
250

DETECTION OF X-RAYS
Detection principles
• gas ionizationg
• scintillation
• creation of e-h pairs
• Point detectors• Point detectors
• Strip detectors
• Area detectors
251

DETECTION OF X-RAYS
S i till ti tPoint detectors Scintillation countersPoint detectors
Gas proportional counters
252

X-RAY DETECTORS
Area detectors
• film
• imaging plate
• CCD
Charge-coupled devices
CCD
• multiwire
253

X-RAY DETECTORS
Imaging platesImaging plates
“photostimulable phosphor”
254

The Reciprocal Lattice and the
Laue Description of DiffractionLaue Description of Diffraction
Reading: A/M 5-6
G/S 3
255

PLANE WAVES
A wave whose surfaces of constant phase are infinite parallel planes ofA wave whose surfaces of constant phase are infinite parallel planes of
equal spacing normal to the direction of propagation.
ik r |k|=2π/λ
( ) i
r Ae  k r
ψ: wave amplitude at point r
k
|k|=2π/λ
ψ: wave amplitude at point r
A: max amplitude of wave
k: wave vector
r: space vector from arbitrary origin
k
r space vector from arbitrary origin
Amplitude is constant in any plane normal to k because k•r is a constant
for such planes:
kk
k•r1 = kr1
k k √2( 45) k
k
r
wavefront
k
r 45°
k•r2 = kr1√2(cos45) = kr1
r2
origin
r1
45
k•r is indeed constant on wavefronts

THE RECIPROCAL LATTICE
The reciprocal lattice of a Bravais lattice is the set of all vectors K such thatp
for all real lattice position vectors R.
1i
e K R
Reciprocal lattice: The set of all wave vectors K that yield plane waves
with the periodicity of a given Bravais lattice.
R = n1a1 + n2a2 + n3a3
Direct lattice position vectors:
Reciprocal lattice vectors: K = hb1 + kb2 + lb3
 
2

2 3
1
1 2 3
a a
b
a a a


where the primitive vectors
of the reciprocal lattice are:
 
2

3 1
2
1 2 3
a a
b
a a a

 and {ni} and {h,k,l}
are integers
 
2

1 2
3
1 2 3
a a
b
a a a


are integers
257

1i
e K R
is satisfied when K•R = 2πn, with n an integer
To verify that the {bi} are primitive vectors of the reciprocal
lattice, let’s first show that bi•aj = 2πδij
 a a a
 
 
 
2 2 2  

    
 
1 2 32 3
1 1 1
1 2 3 1 2 3
a a aa a
b a a
a a a a a a

 
 
2 0   

3 1
2 1 1
1 2 3
a a
b a a
a a a


(since cross product of two
vectors is perpendicular to both)
 
2 0   

1 2
3 1 1
1 2 3
a a
b a a
a a a


Indeed, bi•aj = 2πδij
 1 2 3
so, K•R = (hb1 + kb2 + lb3)•(n1a1 + n2a2 + n3a3)
= 2π(hn1 + kn2 + ln3) = 2π × integer
K is indeed a reciprocal lattice vector

WHAT IS A RECIPROCAL LATTICE VECTOR?
The reciprocal lattice is defined at the lattice generated from the set of allp g
vectors K that satisfy
for all direct lattice position vectors R
1i
e K R
for all direct lattice position vectors R.
What is K?
a wave vector of a plane wave that has the periodicity of the direct latticea wave vector of a plane wave that has the periodicity of the direct lattice
The direct lattice is periodic (invariant under translation by R)
(
( ) ( )i i
e e   K r+R) K r
r + R r 
1i
e K R
Reciprocal lattice vectors = wave vectors of plane waves that are unity
at all direct lattice sites 259

THE RECIPROCAL LATTICE
th i l l tti i d fi d i t f B i l tti• the reciprocal lattice is defined in terms of a Bravais lattice
• the reciprocal lattice is itself one of the 14 Bravais lattices
• the reciprocal of the reciprocal lattice is the original direct lattice
i l bi di l ie.g., simple cubic direct lattice
â1a x â2a y â3a z
 
2
3
2
ˆ ˆ2 2
a 
   2 3
1
a a
b x x

  3
a a1 2 3a a a
2
ˆ

b y
2
ˆ

b z → simple cubic reciprocal lattice
a
2b y
a
3b z p p
with lattice constant 2π/a
→ b1 parallel to a1, etc. 260

Crystals with orthogonal axes (cubic, tetragonal, orthorhombic)
b b b are parallel to a a a respectivelyb1, b2, b3 are parallel to a1, a2, a3, respectively.
ˆaa x ˆba y ˆca z
b i l l tti 2
a1a x b2a y c3a z
b3 reciprocal lattice 2
ˆ
a

1b x
a3
a2
b2
direct lattice
2
ˆ
b

2b y
b1 a1
a2
2
ˆ
c

3b z
c
261

RECIPROCAL LATTICE OF FCC IS BCC
FCC primitive vectors:
Note: not orthogonal
2
 
2
3
ˆ ˆ ˆ( )
4 14 ˆ ˆ ˆ2 2 ( )
2
(2)
a
a a

 

   

2 3
1
1 2 3
y z - x
a a
b y z - x
a a a

  (2)
8
1 2 3
4 1
ˆ ˆ ˆ( + )
2

2b z x - y
4 1
ˆ ˆ ˆ( + )
2

3b x y - z
2a
2 ( )
2a
3 y
→ BCC reciprocal lattice with lattice constant 4π/a 262

RECIPROCAL LATTICE OF BCC IS FCC
BCC primitive vectors (not orthogonal):
2
 
2
3
ˆ ˆ(2 2 )
4 14 ˆ ˆ2 2 ( )
2
(4)
a
a a

 

   

2 3
1
1 2 3
y z
a a
b y z
a a a

  (4)
8
1 2 3
4 1
ˆ ˆ( )
2a

2b z + x
4 1
ˆ ˆ( )
2a

3b x + y
2a 2a
→ FCC reciprocal lattice with lattice constant 4π/a 263

RECIPROCAL LATTICES
• simple orthorhombic → simple orthorhombic
• FCC → BCC
BCC FCC• BCC → FCC
• simple hexagonal → simple hexagonal (rotated)
264

= b1
r. l.
= b1
= b2
= b
β ≠ 90°
= b3
d. l.
265

α,β,γ ≠ 90°,β,γ ≠
266

Note that these formulas
are missing a factor of 2π
267

FIRST BRILLOUIN ZONES
The Wigner-Seitz cell of the reciprocal lattice is called the first Brillouin zone
(FBZ).
Wigner-Seitz cell: primitive cell with lattice point at its center
d.l. FCC
r.l. BCC
d.l. BCC
r.l. FCC. .
1st Brillouin zone:
r.l. FCC
1st Brillouin zone:
enclosed region is W-S cell
for 2D hexagonal lattice
truncated octahedron
rhombic dodecahedron
268

LL

X (2a 0 0)
LL
X (2a 0 0)
269

FIRST BRILLOUIN ZONES
Greek letters: points within the FBZp
Roman letters: points on the FBZ surface
270

Electronic band structure is calculated within the 1st Brilluoin zone
271

Th
K and LATTICE PLANES
Theorem:
For any family of lattice planes separated by distance d, there are
reciprocal lattice vectors perpendicular to the planes, the shortest of
hi h h l th f 2 /dwhich has a length of 2π/d.
Conversely, any reciprocal lattice vector K has a family of real-space
planes normal to it separated by dplanes normal to it, separated by d.
ˆ2 / dK = n
here, g = K
hk in 2D
hkl in 3D
272

O i t ti f l i d t i d b it l t
MILLER INDICES OF LATTICE PLANES
Orientation of a plane is determined by its normal vector
It is natural to pick the shortest perpendicular reciprocal
lattice vector to represent the normal
Miller indices: coordinates of this reciprocal lattice vector
i.e., A plane with Miller indices hkl is normal to
the reciprocal lattice vector K hb + kb + lbthe reciprocal lattice vector K = hb1 + kb2 + lb3
→ Definition #2: directions in k-space
(Definition #1: inverse intercepts in the real lattice)
273

Proof that K = hb1 + kb2 + lb3 is normal to (hkl)
If K = hb1 + kb2 + lb is normal to the plane at leftIf K = hb1 + kb2 + lb3 is normal to the plane at left,
its dot product with any in-plane vector is zero.
Consider vector AB that lies in the plane.
k
2a
(hkl)
1a
By vector addition,
h l
  AB
31 aa
The dot product,
(hkl)
hAB
l
3
a
( )h k l
h l
 
   
 
AB K = 
31
1 2 3
aa
b b b
2 2 0 t2 2 0  =
So the reciprocal vector formed by using the Miller indices of a plane as its
components forms a vector in space that is normal to the Miller plane.
etc.
Furthermore, the length of the shortest vector K is equal to 2π/dhkl.
In the figure above, the spacing between the planes is the projection of :on
h
K
K
1
a
h K
2 2
hkl
h
d
h h
 
  
K
K K K

1a
0
2
hkl
K
d

 K→
274

REMINDER on ELASTIC SCATTERING
scattering vector
 p' p qMomentum conservation:
scattering vector
p p' pelastic scattering:
Elastic scattering geometry
q 2 sin
p 
q 2 sin
2



von LAUE DESCRIPTION OF DIFFRACTION
reciprocal space description equivalent to Bragg description but• reciprocal space description, equivalent to Bragg description but
more powerful for crystallography & solid state physics
Equivalence to Bragg picture:
2
2 sind n n
k

  
Equivalence to Bragg picture:
K
k
0
2
2 sink n nK K
d

   0
d
p ksince scattering is elastic and ,
2 sin 2 sin
2
p
q k K

  

q K von Laue: “Constructive interference occurs when
scattering vector is a reciprocal lattice vector.”

DERIVATION of von LAUE CONDITION
Consider two scatterers:
ˆ ˆ( )d d ' ' Path difference between the rays: ˆ ˆcos cos ( )d d ' = '  d n -n
Condition for constructive interference: ˆ ˆ( ) =' nd n -n
Multiply through by 2π/λ: ( - ) = 2' nd k k
For the Bravais lattice of scatterers: ( - ) = 2' nR k k
Multiply by i and raise to e:
( - ) 2
= 1i ' i n
e e 
k k R
Diff ti h th h i
So, - ='k k K
Diffraction occurs when the change in
wave vector, k’-k, is a vector of the
reciprocal lattice. 277

'
K = k k Laue condition
Alternativelyk-space Bragg plane
= / 2Kk K
Alternatively,k-space Bragg plane
Reciprocal lattice vectors are perpendicular to direct lattice planes
Bragg: diffraction when path length difference = nλ
Laue: diffraction when scattering vector = recip. vector
equivalently, when tip of wave vector lies on a k-space Bragg plane

EWALD (“e-val”) SPHERE
A geometrical construction that provides the relationship between the orientation ofA geometrical construction that provides the relationship between the orientation of
a crystal and the direction of the beams diffracted by it .
A sphere of radius k centered on the base of the incident wave vector
k drawn to the origin O (hkl = 000) of the reciprocal lattice.
Projected Ewald sphere (Ewald circle)j p
origin of reciprocal space
radius = k
k
k’
θ
θ O
'
K = k k
Laue condition:
k’ θ O
K
K k k
real space
origin of diffraction
direction of
diffracted beam
reciprocal lattice
(-2,-1)
279

Diffraction occurs only when a reciprocal lattice point
lies on the surface of the Ewald sphere.
Laue condition:
p
'
K = k k
kθ
K
k
k’
θ
θ
k’
θ
θ O
K
k’
(-2,-1)
2 / dK =
In this case, hkl = -2,-1,0 so diffraction occurs from the (210) planes
d h diff d b ff l k’
--
2102 / dK
and the diffracted beam moves off along k’.
280

In general, a sphere in k-space with the origin on its
surface will have no other reciprocal lattice points onp p
its surface:
O
No Bragg peaks for a general incident X-ray!
In order to record diffraction patterns, we must:
• use polychromatic radiation (vary the sphere size) → Laue method
• rotate the crystal (and thus the reciprocal lattice) → rot. cryst. methody ( p ) y
• use a powder sample (equivalent to rotating reciprocal
space about all angles) → powder method
282

LAUE METHOD
• fixed crystal & angle, variable wavelengthy g , g
284

The Laue method is mainly used to determine the orientation of
large single crystals.g g y
When the zone axis lies along the symmetry axis of the crystal, the
pattern of Bragg spots will have the same symmetry. 285

ROTATING CRYSTAL METHOD
• fixed wavelengthg
• aligned crystal is rotated about one axis to rotate reciprocal lattice
• produces spots on layer lines
k’
k
k
http://escher.epfl.ch/x-ray/diff.mpeg
286

POWDER (DEBYE-SCHERRER) METHOD
• fixed wavelengthg
• fixed powder sample
• equivalent to rotating the reciprocal lattice through all possible
angles about the origin
every point in Each shell with radius K < 2kevery point in
reciprocal space
traces out a shell
of radius K
intersects the Ewald sphere to
form a circle.
All h diff d b fAll the diffracted beams from a
powder lie on the surface of cones
288

PEAK INTENSITIES
Peak intensities depend on (in large part):
1) intensity scattered by individual atoms (form factors)
2) the resultant wave from atoms in unit cell (structure factor)2) the resultant wave from atoms in unit cell (structure factor)
In many cases, the intensity from certain planes (hkl) is zero
• symmetry of crystal causes complete cancellation of beam
“systematic absences”
• happenstance
Possible reasons:
• happenstance
Other factors that affect intensity: • scattering angle
• multiplicities
• temperature factor
• absorption factorabsorption factor
• preferred orientation
289

MONOATOMIC BASES
up to now we have considered diffraction only from Bravais lattices
( - ) 2
= 1i ' i n
e e 
k k R
up to now w ha cons r ffract on on y from ra a s att c s
with single atom bases (i.e., atoms only at the lattice points R).
We found the diffraction condition: 1e ef ff
=1i
e K Rwhich is the same as:
( ) i
F f e 
  K R
K R K
The scattering amplitude FK is the sum over the lattice sites:
( )F f eK R
R
K
where fR(K) is the “atomic form factor” for a given atom (disc. later).
The scattered intensity is proportional to the absolute square of the
amplitude: 2amplitude: 2
0I I FK K
…this is what is actually measured in an experiment.

Crystals with multiple atoms in each primitive cell must be further
n-ATOM BASES
rysta s w th mu t p atoms n ach pr m t c must furth r
analyzed into a set of scatterers at positions d1, d2 … dn within each
primitive cell.
( )i

K R+d
( )  A R R dj
The positions of the atoms are:
( )
( ) ji
j
j
F f e

 
K R+d
K
R
Kmaking the scattering amplitude:
i
 
K d
( ) jii
j
j
e f e

  
K dK R
R
K
i
L e 
  K R
RR
( ) ji
j
j
f e

  
K d
K K
“Lattice sum”
j
“Structure factor” of the basis
*If the structure factor = 0, there is no diffraction peak.

The structure factor gives the amplitude of a scattered wave arising
STRUCTURE FACTOR
( ) ji
jf e

  
K d
K K
h structur factor g s th amp tu of a scatt r wa ar s ng
from the atoms with a single primitive cell.
( )j
j
fK
For crystals composed of only one type of atom, it’s common to split
th t t f t i t t tthe structure factor into two parts:
( )f S K KK
ji
S e

 
K d
K
“atomic form factor”
j
K
“geometric structure factor”
S = 0 gives a systematic absence (i.e., absence of expected diff. peak).
292

The amplitude of the rays scattered at positions d d
STRUCTURE FACTORS
1iK d iK d
The amplitude of the rays scattered at positions d1, …, dn
are in the ratios:
1i
e K d ni
e K d
…
Th d b h ll h fThe net ray scattered by the entire cell is the sum of
the individual rays:
1
j
n
i
j
S e


 
K
K
dGeometric
structure
factor
-Adds up scattered
waves from unit cell
2
1jfactor
2
( )hklI S K
-In particular, no
peak when SK = 0
293

SIMPLE CUBIC
For simple cubic: one atom basis (0,0,0)
d1 = 0a1 + 0a2 + 0a3d1 0a1 0a2 0a3
0iK 0
1i
S e 
 K
K
294

MONO-ATOMIC BCC
For monoatomic BCC:
we can think of this as SC with two point basis (0,0,0), (½,½,½)
2 a   
2 ( )
0 2
1
j
a
i x y zi i
j
S e e e
  
   

  
KK K
K
d
2
ˆ ˆ ˆ( )h k l
a

  K x y zFor SC,
lkh 
)1(1
( )
1 i h k l
e   
 
lkh 
 )1(1
S 2 when h + k + l evenS = 2, when h + k + l even
S = 0, when h + k + l odd, systematic absences
295

e.g. consider the powder pattern of BCC molybdenum
Powder card shows only even hkl sums b/c Mo is BCCy
Why?
- Diffraction from other (hkl) results in destructive interference:
(100) ( 00)(100) (200)
d100
d200
Beam cancels b/c body center atoms
scatter exactly 180° out of phase
Strong reflection b/c all atoms lie on
200 planes and scatter in phase 296

MONO-ATOMIC FCC
For monoatomic FCC:
SC with four point basis (0,0,0), (½,½,0), (0,½,½), (½,0,½)
4 ( ) ( ) ( )
0 2 2 2
1
j
a a a
i x y i y z i x zi i
j
S e e e e e
     
      
    
K K KK K
K
d
1j
2
ˆ ˆ ˆ( )h k l
a

  K x y zFor SC,
( ) ( ) ( )
1 i h k i k l i h l
S e e e    
   K
a
S = 4 when h + k, k + l, h + l all even (h, k, l all even or all odd)
K
S = 0 otherwise 297

POLYATOMIC STRUCTURES
Atoms of different Z in the unit cell have different scattering
powers, so we explicitly include the form factors:
Total
structure ( ) ji
jf e

  
K d
K K
factor
{fj }: atomic form factors
j
j
K
{fj }: atomic form factors
 # of electrons
298

CsCl STRUCTURE
Cesium Chloride is primitive cubic
Cs (0,0,0)
Cl (1/2,1/2,1/2)
( ) ji
j
j
f e

  
K d
K K
( )i h k l
Cs Clf f e   
  K
Cs+ and Cl- are not isoelectronic
→ systematic absences unlikely
but what about CsI?
299

(hkl) CsCl CsI
Cs+ and I- are isoelectronic, so CsI looks like BCC lattice:
(hkl) CsCl CsI
(100) 
(110)  (110)  
(111) 
(200)  (200)  
(210) 
(211)  (211)  
(220)  
(221) (221) 
(300) 
(310)  (310)  
(311)  300

Diatomic FCC Lattices
Sodium Chloride (NaCl)
Na: (0,0,0)(0,1/2,1/2)(1/2,0,1/2)(1/2,1/2,0)
Add (1/2,1/2,1/2)
Cl: (1/2,1/2,1/2) (1/2,1,1)(1,1/2,1)(1,1,1/2)
301

( ) ( ) ( ) ( )
[ ][1 ]i h k l i h k i h l i l k
Na Clf f e e e e       
     K
S 4(f + f ) h h k l ll
[ ][ ]Na Clf fK
S = 4(fNa + fCl) when h, k, l, all even
S = 4(fNa - fCl) when h, k, l all odd
S = 0 otherwise
302

Once again, there are more systematic absences for
isoelectronic ions (e.g., K and Cl)
(110) always absent in RS
(hkl) NaCl KCl
(100)
(110) always absent in RS
(110)
(111) 
 (200)  
(210)
(211)
(220)   (111) sometimes absent
(221)
(300)
(310)
(311)  303

For FCC, we expect the intensity of the (111) reflection
t i ith i i ∆Z b t ti d ito increase with increasing ∆Z between cation and ion
I KCl KF KB KII111 : KCl < KF < KBr < KI
L l t d t ti i t fLess complete destructive interference
304

DIAMOND STRUCTURE
Diamond: FCC lattice with two-atom basis (0 0 0 ) (¼ ¼ ¼)Diamond: FCC lattice with two atom basis (0,0,0,), (¼,¼,¼)
12 ( )iK
  
2 ( )
0 4
1
[ ]j
iK x y ziK d iK
diamond FCC FCC
j
S S e S e e
   

  
( /2)( )
[1 ]i h k l
FCCS e   
 
Only for all even or all odd hkl is S ≠ 0 For these unmixed values
S = 2, h + k + l twice an even number
S = 1 +/- i h + k + l odd
S 0 h k l t i dd b
Only for all even or all odd hkl is S ≠ 0. For these unmixed values,
Additional condition:
S = 0, h + k + l twice an odd number
IFCC : all nonvanishing spots have equal intensity
Idiamond : spots allowed by FCC have intensities of 4, 2, 2, or 0
305

(hkl) Al Si
FCC diamond
( ) S
(100)
(110)( )
(111)  
(200) ( ) 
(210)
(211)( )
(220)  
(221)( )
(300)
(310) What about( )
(311)  
zinc blende?
306

SUMMARY OF SYSTEMATIC ABSENCES
t l t t diti f k tcrystal structure condition for peak to occur
SC any h,k,l
BCC h + k + lBCC h + k + l = even
FCC h,k,l all even or all odd
N Cl h k l llNaCl h,k,l all even,
or all odd if fA ≠ fB
di d h k l ll d t i #diamond h,k,l all even and twice an even #,
or all odd
HCP any h k l except when h + 2k = 3nHCP any h,k,l except when h + 2k 3n
and l is odd
( ) ji
f

 
K d
K( ) ji
j
j
f e  
K d
K K
307

SIMPLE ANALYSIS OF SIMPLE PATTERNS
What will we see in XRD patterns of SC BCC FCC?
222
lkh 
What will we see in XRD patterns of SC, BCC, FCC?
lkh 
SC: 1,2,3,4,5,6,8,9,10,11,12,…
BCC: 2,4,6,8,10,12,...
FCC: 3,4,8,11,12,16,24,…
Observable diffraction
, , , , , , ,
Observable diffraction
peaks for monoatomic
crystals
SC FCC BCC
We can take ratios of (h2 + k2 + l2) to determine structure.
308

 nd sin2 2
2
1
sin
hkld
 
222
a
dhkl For cubic crystals:
hkld
222
lkh
hkl

y
2 2 2 2
i ( )h k l2 2 2 2
sin ( )h k l   
2 2 2 22 2 2 2
th peak th peak
2 2 2 2
1 k 1 k
sin ( )
sin ( )
n nh k l
h k l


 

 1st peak 1st peaksin ( )h k l  
309

α-Fe is cubic. Is it FCC or BCC?
BCC!
2 2
2
2 2
sin sin 33
2
sin sin 22


 


110
211
α e s cub c s t CC o CC
BCC!
1sin sin 22
200
211
What about Al?
2 2
sin sin 22 5 111 FCC!
2
2 2
1
sin sin 22.5
1.33
sin sin 19


 
200 311200
220
311
222
400
331 420
310

Ex: An element, BCC or FCC, shows diffraction
peaks at 2: 40 58 73 86 8 100 4 and 114 7peaks at 2: 40, 58, 73, 86.8,100.4 and 114.7.
Determine:(a) Crystal structure?(b) Lattice constant?
(c) What is the element?(c) What is the element?
2theta theta (hkl)
40 20 0 117 1 (110)
2
sin 222
lkh 
normalized
40 20 0.117 1 (110)
58 29 0.235 2 (200)
73 36 5 0 3538 3 (211)73 36.5 0.3538 3 (211)
86.8 43.4 0.4721 4 (220)
100 4 50 2 0 5903 5 (310)100.4 50.2 0.5903 5 (310)
114.7 57.35 0.7090 6 (222)
BCC, a =3.18 Å  W
311

SCATTERING OF X-RAYS BY ATOMS
Atoms scatter X-rays because the oscillating electric field of an X-ray sets
each electron in an atom into vibration. Each vibrating electron acts as a
secondary point source of coherent X-rays (in elastic scattering).
The X-ray scattered from an atom is the resultant wave from all its electrons
Particle picture:Particle picture:
h diff f f d/b k d i• zero phase difference for forward/backward scattering
→ scattering factor (form factor, f ) proportional to atomic number, Z
• increasingly destructive interference with larger scattering angle
• for a given angle, intensity decreases with decreasing X-ray wavelength
Thomson equation:
for a given angle, intensity decreases with decreasing X ray wavelength
21
(1 cos 2 )I  q
• max scattering intensity at 2θ = 0 & 180°
• gradual decrease to 50% as 2θ approaches 90°
( )
2

ATOMIC FORM FACTORS
Form factor f = scattering amplitude of a wave by an isolated atomForm factor f = scattering amplitude of a wave by an isolated atom
• Z (# electrons)
• scattering angle
X l th
For X-rays, f depends on:
scattering vector q
• X-ray wavelength


scattering vector q
General elastic formula:
Cl-
0
( ) ( ) i
j jf e d 
 
q r
q r r
4 sin 
K+
Cl
Cl
4 sin
q
 

with,
For θ = 0 (forward scattering),
O
0
(0) ( )jf d # electrons

  r r =
O
θ = 37°
consequences: • powder patterns show weak lines at large 2θ.
• light atoms scatter weakly and are difficult to see.

ELECTRON DENSITY MAPS
The electron density as a function of position x,y,z is
the inverse Fourier transform of the structure factors:
1 2 ( )1
( ) i hx ky lz
hkl hkl
xyz e
V

   
 
The electron density map
d ib th t t fdescribes the contents of
the unit cells averaged
over the whole crystal (not
the contents of a single unitg
cell)
314

PEAK WIDTHS
Peak shape is a Voigt function (mixture of Gaussian and Lorentzian)a shap s a Vo gt funct on (m tur of Gauss an an Lor ntz an)
• Gaussian component arises from natural linewidth and strain
• Lorentzian component arises from coherent domain size
Pure
Gaussian
Pure
Lorentzian
Gaussian
Peak width (broadening) is determined by several factors:( g) m y f
• natural linewidth of X-ray emission
• instrumental effects (polychromatic λ, focusing, detector)
• specimen effects• specimen effects
1) crystallite size
2) crystallite strain 315

Instrument and Sample Contributions to the Peak Profile
must be Deconvolutedmust be Deconvoluted
• In order to analyze crystallite size, we must deconvolute:
– Instrumental Broadening FW(I)
• also referred to as the Instrumental Profile, Instrumental
FWHM Curve, Instrumental Peak Profile,
– Specimen Broadening FW(S)
• also referred to as the Sample Profile, Specimen Profile
• We must then separate the different contributions to specimen
broadening
C t llit i d i t i b d i f diff ti k– Crystallite size and microstrain broadening of diffraction peaks
317

SIZE BROADENING
Small crystallites (< 200 nm) show broadened diffraction linesSma crysta t s ( nm) show roa n ffract on n s
Nanocrystal X-ray
Diffraction
318

Which of these diffraction patterns comes from a
nanocrystalline material?nanocrystalline material?
(a.u.)Intensity
66 67 68 69 70 71 72 73 74
2 ( )2(deg.)
These diffraction patterns were produced from the same sample!
• Two different diffractometers, with different optical configurations, were used
• The apparent peak broadening is due solely to the instrumentation
320

at Bragg angle 
The finite size of real crystals results
in incomplete destructive interference
over some range of angles
B 1
2
at Bragg angle,
phase lag between two planes =
perfectly in phase, constructive
B

g g
2
B 1
At some angle
Ph l b t t l  1
2
3
4
Phase lag between two planes:
At (j+1)th plane:
Phase lag:
 
j-1
j
j+1
Phase lag:
• Rays from planes 1 and j+1 cancel
2

  j
T = (2j-1)d
2j-1
2j
Rays from planes 1 and j 1 cancel
• Ditto for 2 & j+2, … j & 2j
• Net diffraction over 2j planes = 0
Crystal with 2j planes
T t l thi k TTotal thickness T The angular range θB to θ1 is the range where
diffracted intensity falls from a maximum to
zero (half of Bragg peak profile)! 321

If we have more than 2j planes:

If we have fewer than 2j planes:

B 1
2
B 1
2
1 1
2
3
4
j-1
j
2
3
4
j-1
jj
j+1
2j+1
j
j+1
2j-32j 1
2j+2
2j 3
2j-2
still zero intensity at θ1 nonzero intensity at θ1
Rays from planes j-1 & j not canceledRays from new planes are canceled
still zero intensity at θ1 nonzero intensity at θ1
Thinner crystals result in broader peaks! 322

Same arguments apply to
B 2
So we see diffracted X-rays over all scattering angles between 2θ1
and 2θ2.
– If we assume a triangular shape for the peak, the full width at
half maximum of the peak will be B = (2θ1 – 2θ2)/2 = θ1 – θ2
323

Let’s derive the relation between crystal thickness T and peak width B:
2 sind  2 sind  
2 i (2 1)T j 
Considering the path length differences between X-rays scattered from the front
and back planes of a crystal with 2j+1 planes and total thickness T:
1
2
2 sin (2 1)
2 sin (2 1)
T j
T j
 
 
 
 
1 2 1 2    
1 2(sin sin )T    
1 2 1 2
2 (cos( )sin( ))T
   

 

If we subtract them:
Using trig identity:
1 2 1 2
1 2sin sin 2cos sin
2 2
   
 

 
2 (cos( )sin( ))
2 2
T Using trig identity:
Since and1 2 


 1 2 1 2
sin( )
    

1 2
2 (cos )( ))T
 
 

 T

1 2
2( )B
 

Since and ,
2
B sin( )
2 2

But so2 (cos )( ))
2
BT  
cos B
T
B 
2( )
2
B . But, , so
324

SCHERRER FORMULA
A more rigorous treatment includes a unitless shape factor:
K
T


g p
Scherrer Formula (1918)
T = crystallite thickness
λ (X-ray wavelength, Å)
cos B
T
B 
m ( )
K (shape factor) ~ 0.9
B, θB in radians
2 2 2
B B B 
Accurate size analysis requires correction for instrument broadening:
M RB B B
BM: Measured FWHM (in radians)
B C di FWHM f b lk f (l i i 200 )BR: Corresponding FWHM of bulk reference (large grain size, > 200 nm)
Readily applied for crystal size of 2-50 nm.
325

K
SCHERRER CONSTANT
0 94
cos B
K
T
B



0.94
cos B
T
B



• The constant of proportionality, K (the Scherrer constant)
depends on the how the width is determined, the shape of the
B B
crystal, and the size distribution
– the most common values for K are:
• 0.94 for FWHM of spherical crystals with cubic symmetry
0 89 f i t l b dth f h i l t l / bi t• 0.89 for integral breadth of spherical crystals w/ cubic symmetry
• 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• 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) 102-113.
326

Suppose =1.5 Å, d=1.0 Å, and =49°. Then for a crystal 1
mm in diameter, the width B, due to the small crystal
effect alone, would be about 2x10-7 radian (10-5 degree),effect alone, would be about 2x10 radian (10 degree),
too small to be observable. Such a crystal would contain
some 107 parallel lattice planes of the spacing assumed
aboveabove.
However, if the crystal were only 50 Å thick, it would
contain only 51 planes, and the diffraction curve would be
very broad, namely about 43x10-2 radian (2.46°), which is
easily measurable.easily measurable.
“Incomplete destructive interference
327
p
at angles slightly off the Bragg angles”

What do we mean by crystallite size?
“A t lli d i th t tt h tl ”“A crystalline domain that scatters coherently”
• A particle may be made up of several different crystallinep y p y
grains
• Crystallite size often matches grain size, but there are
tiexceptions
328

Though the shape of crystallites is usually irregular we can often
CRYSTALLITE SHAPE
• Though the shape of crystallites is usually irregular, we can often
approximate them as:
– sphere, cube, tetrahedra, or octahedra
– parallelepipeds such as needles or platesparallelepipeds such as needles or plates
– prisms or cylinders
• Most applications of Scherrer analysis assume spherical crystallitepp y p y
shapes
• If we know the average crystallite shape from another analysis, we
can select the proper value for the Scherrer constant K
• Anisotropic crystal shapes can be identified by anisotropic peak
b d ibroadening
– if the dimensions of a crystallite are 2x * 2y * 200z, then (h00) and (0k0)
peaks will be more broadened then (00l) peaks.
329
e.g., a nanowire

STRAIN EFFECTS
L
L


Strain:
330

Non-Uniform Lattice Distortions
• Rather than a single d-spacing,
th t ll hi l hthe crystallographic plane has a
distribution of d-spacings
• This produces a broaderp
observed diffraction peak
• Such distortions can be
introduced by:introduced by:
– surface tension of
nanocrystals
h l f t l h
nsity(a.u.)
– morphology of crystal shape,
such as nanotubes
– interstitial impurities
Inte
p
26.5 27.0 27.5 28.0 28.5 29.0 29.5 30.0
2(deg.)
332

EPITAXY - “above in an ordered fashion”
when one crystal grows on another with a well-defined 3D crystallographicy g y g p
relationship
Homoepitaxy: epitaxy between identical crystals (e.g., Si on Si)
Heteroepitaxy: the two crystals are different (e g ZnO on Al O )Heteroepitaxy: the two crystals are different (e.g., ZnO on Al2O3)
requirements = lattice symmetry & lattice constant matching
335

Molecular picture – Si growth on Si (100)
336
Dan Connelly

Rock salt PbS “nanotrees”
338
Jin group – U. Wisc. branches grow epitaxially –
each tree is a single crystal!

A polycrystalline sample should contain thousands of
crystallites. Therefore, all possible diffraction peaks should
b b dbe observed.
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. 339

A single crystal specimen in a Bragg-Brentano
diffractometer would produce only one family of peaks indiffractometer would produce only one family of peaks in
the diffraction pattern.
2
At 20.6 °2, Bragg’s law
fulfilled for the (100) planes,
The (110) planes would diffract at 29.3
°2; however, they are not properly
aligned to produce a diffraction peak
The (200) planes are parallel to the (100)
planes. Therefore, they also diffract for this
crystal Since d is ½ d they appear atproducing a diffraction peak. aligned to produce a diffraction peak
(the perpendicular to those planes does
not bisect the incident and diffracted
beams). Only background is observed.
crystal. Since d200 is ½ d100, they appear at
42 °2.
340

Wurtzite ZnO nanowire arrays
Poor
uniaxial
texture
Uniaxial
t t
c
texture
Biaxial
texture
General route to vertical ZnO
nanowire arrays using textured
ZnO seeds. texture
(growth on Al2O3)
341
ZnO seeds.
Greene, L. E., Law, M., Tan,
D. H., Montano, M., Goldberger,
J., Somorjai, G., Yang, P. Nano
Letters 5, 1231-1236 (2005).

ROCKING CURVE EXAMPLES
Thickness, composition, and strain state of epitaxial single crystal films344

ROCKING CURVE EXAMPLE
Thi k i i d i f i i l i l l fil
(1° = 3600 arcsec)
Thickness, composition, and strain state of epitaxial single crystal films
345

PHI SCANS
flsit on a reflection,
then spin in-plane
346

k-SPACE GEOMETRY
for rotation around [001]
of cubic crystal:
two possibilities:
monitor {011}: expect 4 peaks separated by 90° rotation.
two possibilities:
347
monitor {111}: expect 4 peaks separated by 90° rotation.
(ignoring possible systematic absences)

PHI SCAN EXAMPLE
1 um GaN (wurtzite) on Silicon(111) (1011)
(002)
2-theta scan proves
uni-axial texture phi scan provesp p
bi-axial texture (epitaxy)
In plane alignment: GaN[1120]//Si[110] 348

Epitaxial YBa2Cu3O7 on Biaxially Textured Nickel (001):
An Approach to Superconducting Tapes with High Critical Current Density
Science, Vol 274, Issue 5288, 755-757 , 1 November 1996
349

phi
Epitaxial YBa Cu O on Biaxially
omega
p
Epitaxial YBa2Cu3O7 on Biaxially
Textured Nickel (001):
An Approach to Superconducting
Tapes with High Critical Current Density
Science, Vol 274, Issue 5288,
755-757 , 1 November 1996
350

TEXTURE MEASUREMENT (POLE FIGURES)
ψ
351

P f d i t ti f t llit t
POLE FIGURES
• Preferred orientation of crystallites can create a
systematic variation in diffraction peak intensities
– can qualitatively analyze using a 1D diffraction patterncan qualitatively analyze using a 1D diffraction pattern
– a pole figure maps the intensity of a single peak as a
function of tilt and rotation of the sample
• this can be used to quantify the texture
(111)
(311)
(200)6 0
8.0
10.0
Counts)
00-004-0784> Gold - Au
(200)
(220)
(222)
(400)
2.0
4.0
6.0
Intensity(C
(400)
40 50 60 70 80 90 100
Two-Theta (deg)
x10
3
352

POLE FIGURE EXAMPLE – PHI ONLY
Example: c-axis aligned superconducting thin films.

(b)

(a)

Biaxial Texture (105 planes) Random in-plane alignmentp g
353

SMALL ANGLE X-RAY SCATTERING
354

SAXS: diffraction from planes with > 1 nm d spacingSAXS: diffraction from planes with > 1 nm d-spacing
355

Small Angle
X-ray Diffractiony
Direct Visualization of Individual
Cylindrical and Spherical
Supramolecular DendrimersSupramolecular Dendrimers
Science 17 October 1997; 278: 449-452
356

HCP
Triblock Copolymer Syntheses of Mesoporous Silica with Periodic 50 to 300 Angstrom
Pores Science, Vol 279, Issue 5350, 548-552 , 23 January 1998 357

UCI XRD
Rigaku SmartLab XRD
 0D 1D 2D detectors 0D, 1D, 2D detectors
 In-plane & Out-of-plane
 Thin-film XRDThin film XRD
 High resolution XRD
 SAXS
 μ-XRD
 Capillary transmission
 1500°C heating stage
 1100°C dome stage

Why ED patterns have so many spots
Typically in X-ray on neutron diffraction only one reciprocal latticeTypically, in X ray on neutron diffraction only one reciprocal lattice
point is on the surface of the Ewald sphere at one time
In electron diffraction the Ewald sphere is not highly curved b/c of the
h t l th l t th t d Thi l t fl t E ldvery short wavelength electrons that are used. This almost flat Ewald
sphere intersects with many reciprocal lattice points at once
- In real crystals reciprocal lattice points are not infinitely small and in a
l i th E ld h i t i fi it l thi
λX-ray = hc/E = 0.154 nm @ 8 keV
λ = h/[2m eV(1 + eV/2m c2)]1/2 = 0 0037 nm @ 100 keV
real microscope the Ewald sphere is not infinitely thin
λe- = h/[2m0eV(1 + eV/2m0c2)]1/2 = 0.0037 nm @ 100 keV
361

DIFFRACTION FROM DISORDERED SOLIDS
amorphous solids
362

DIFFRACTION: WHAT YOU SHOULD KNOW
Index planes
Calculate crystal density
Calculate d-spacings
Derive/use Bragg’s Law
Index diffraction peaksIndex diffraction peaks
Determine lattice constants
Reciprocal lattice
Ewald sphere constructionEwald sphere construction
Calculate structural factors, predicting X-ray diffraction pattern
(systematic absences)
Use of Scherrer relation
363

Crystal structure analysis

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