UCSD NANO106 - 12 - X-ray diffraction

X-rays diffraction
Shyue Ping Ong
Department of NanoEngineering
University of California, San Diego

Intensities
¡Electrons, atoms and unit cells scatter X-rays
¡Expected knowledge
¡ Know how to compute intensities given the atomic
parameters and a unit cell
¡ Understand the concept of systematic extinctions, and
interpret the International Tables of Crystallography on
reflection and extinction conditions.
¡ Able to discuss in qualitative terms how measured
intensities differ from calculated intensities
NANO 106 - Crystallography ofMaterials by Shyue Ping Ong - Lecture 12
2

Scattering of X-rays by electrons
¡ Consider electron at origin, with X-ray coming in
from –x direction.
¡ Scattered radiation at P can be computed form
the Thomson equation
I = I0
K
r2
sin2
α
where α is the angle between the scattering direction and
the direction of acceleration of the electron, and K is given by
K =
µ0
4π
!
"
#
$
%
&
2
e4
m2
!
"
#
$
%
& = 7.94×10−30
m2
3

Scattering of X-rays by electrons,
contd.
¡ Incident X-ray causes electron to oscillate in a
direction parallel to electric field E.
¡ Given that E is perpendicular to propagation of
wave, the oscillation will in general be in the y-z
plane (known as the polarization direction).
¡ For a typical X-ray, the polarization will be random
and the average component in the y and z-
directions will be equal.
¡ Let’s consider each of these components
separately.
I0y = I0z =
1
2
I
4

Scattering of X-rays by electrons,
contd.
¡ For y-component,
¡ For z-component,
¡ Total scattered intensity
¡ Typically, the K/r2 will be omitted as most diffraction experiments rescale the
intensities so that the highest peak have a value of 100.
α =
π
2
Iy (P) = I0y
K
r2
α =
π
2
− 2θ
Iz (P) = I0z
K
r2
sin2
(
π
2
− 2θ)
= I0z
K
r2
cos2
(2θ)
Iz (P) = I0
K
r2
(1+cos2
2θ)
Strongest for θ= 0 and θ= π.
Weakest for θ= π/2.
5

Compton Scattering
¡ EM photon transfers part of its momentum to electron, and
changes its wavelength in the process. This process is
also known as inelastic scattering.
¡ Radiation scattered in this manner is known as Compton-
modified radiation. The phase of the radiation is changed
in a random way and no longer carries phase-sensitive
information (i.e. no diffraction information).
¡ Thomson scattering on the other hand have a fixed phase
shift of π (half a wavelength) and is known as coherent
scattering.
6

Scattering of X-rays by an atom
¡ All electrons in an atom (with atomic number Z) will
be scattered by X-ray beam.
¡ For forward scattering, no path difference and total
scattering is equal to Z times that of a single electron.
¡ In all other directions, there will be a path length
difference and total scattered intensity < forward
scattered beam.
¡ Diffraction process is governed by
s =
sinθ
λ
7

Atomic scattering factor
¡ Ratio of amplitude scattered by
atom to amplitude scattered in
the same direction by a single
electron
¡ Values are ai and bi are
tabulated. Note that values are
for angstroms.
f (s) = Z − 41.78214s2
aie−bis2
i=1
N
∑
s =
sinθ
λ
8

Example: W
¡ Consider contribution of single W atom to Cu Kα radiation
from (222) plane of bcc crystal with lattice parameter a =
3.1653 A.
Blackboard
9

Scattering of X-rays by a unit cell
¡ Extending the concept further, the scattering of X-rays by
a unit cell can be computed by taking into account the
relative positions of atoms.
¡ As seen previously, destructive interference can occur in
an atom because of the relative positions of electrons in
an atom. The same can happen in a unit cell.
10

Scattering of X-rays by a unit cell,
contd.
¡ Rays 1 and 2 ((100) plane) are diffracted if they satisfy the Bragg equation, i.e., path
difference with length λ.
¡ If atom exists at half the distance between the planes ((200) plane), path length
difference is λ/2 and destructive interference occurs. No diffraction despite satisfying
Bragg condition for (100) plane!
¡ If instead the Bragg condition for the (200) plane is satisfied (different θ), a diffraction
beam will be seen.
¡ Scattering by atom does not depend on position within a plane. Only variable that
matters is the distance between the atom and the (100) plane.
11

Scattering of X-rays by a unit cell,
contd.
¡ Distance of atom to a particular plane is given by the dot product with
the normal to the plane, i.e., the reciprocal lattice vector!
¡ This can be translated to a phase difference by multiplying by 2π
¡ Combining the results from the earlier slide on scattering by an atom
with the relative phase of an atom at position r, we have
ghkl ⋅r
φ = 2πghkl ⋅r = 2π(hx + ky+lz)
f (s)eiφ
= f
sinθ
λ
!
"
#
$
%
&ei2πghkl⋅r
12

Structure Factor
¡ Scattering by complete unit cell is simply obtained by summing the scattering
for all atoms in a unit cell. We can define a structure factor
¡ Intensity of the diffracted beam is then given by
¡ Geometric interpretation of structure factor (Argand diagram)
Fhkl = fj
sinθhkl
λ
!
"
#
$
%
&ei2πghkl⋅r
j=1
N
∑ = fj
sinθhkl
λ
!
"
#
$
%
&e
i2π (hxj +kyj +lzj )
j=1
N
∑
Ihkl = Fhkl
2
= FhklFhkl
*
f (s)eiφ Structure factor is complex sum of
scattering from each atom. If sum goes
back to origin, there is no diffracted beam
even if Bragg condition is satisfied (known
as extinction).
13

Effect of lattice centering on F
¡ We will now consider the effect of the different kinds of
lattice centerings on the structure factor, F.
¡ Recall that for 3D Bravais lattices, there are four different
centerings:
¡ Primitive – no centering
¡ C-centering (A and B-centering are similar)
¡ I-centering (or body-centering)
¡ F-centering (or face-centering)
14

Structure factor for primitive lattice
¡ For a primitive lattice, there are no other symmetrically
equivalent atoms in the unit cell.
¡ If we only have a single atom in the unit cell at r = (0, 0, 0)
[note we have dropped the functional dependence of f for
brevity]
¡ Note that f does depend on the lattice plane (hkl) through
the dependence on θ.
¡ No extinctions and all planes give rise to diffracted beam.
Fhkl = f
I = f 2
15

Structure factor for C-centered lattice
¡ Identical atom at r + C (where C is the centering vector)
¡ If we assume an atom at r = (0, 0, 0)
¡ Systematic absences or systematic extinctions independent of shape
or dimensions of cell.
C
Blackboard
16

Structure factor for I-centered lattice
¡I-centered
17
Fhkl = f ⋅ 1+ eiπ (h+k+l)
( )
= f ⋅ 1+ (−1)h+k+l
( )
I = f 2
⋅ 1+ (−1)h+k+l
( )
2
Extinction when h + k + l = 2n + 1

Structure factor for F-centered lattices
¡ F-centered lattices have equivalent atoms at r + A, r + B,
and r + C
¡ Extinctions occur whenever a mix of odd and even
numbers occur in (hkl). Hence, only planes with all odd or
all even indices can give rise to a diffracted beam in face-
centered crystals!
18
Fhkl = f ⋅ 1+ eiπ (h+k)
+ eiπ (h+l)
+ eiπ (k+l)
( )
= f ⋅ 1+ (−1)h+k
+ (−1)h+l
+ (−1)k+l
( )
I = f 2
⋅ 1+ (−1)h+k
+ (−1)h+l
+ (−1)k+l
( )
2

Symmetry implications on the structure
factor
¡ Inversion symmetry (atoms at r and –r).
¡ Screw axis – e.g., 41 – (x, y, z) -> (-x, y, z + ¼) -> (-x, -y, z + ½) -> (x, -y, z +
¾)
Fhkl = fj (ei2πghkl⋅r
+e−i2πghkl⋅r
)
j=1
N
∑
= 2 fj cos(2πghkl ⋅r)
j=1
N
∑
Real number (no complex
component)
Fhkl = fj e
i2π (hxj +kyj +lzj )
+ e
i2π (−hxj +kyj +l(zj +1/4))
+ e
i2π (−hxj −kyj +l(zj +1/2))
+ e
i2π (hxj −kyj +l(zj +3/4))"
#
$
%
j=1
N/4
∑
For planes given by (00l), we have
F00l = fje
i2πlzj
1+ e
i
πl
2
+ eiπl
+ e
i
3πl
2
"
#
'
$
%
(
j=1
N/4
∑
= 1+il
+i2l
+i3l"# $% fje
i2πlzj
j=1
N/4
∑
Always zero unless l = 4n!
19

Symmetry implications on the structure
factor, contd.
¡ Glide planes, e.g., n-glide. Atoms at (x, y, z) and (x + ½, y
+ ½ , -z)
Extinction when h + k = 2n+1
20
Fhkl = fj e
i2π hxj +kyj +lzj( ) + e
i2π h(xj +0.5)+k(yj +0.5)−lzj( )
( )j=1
N/2
∑
For planes given by (hk0), we have
Fhk0 = fje
i2π hxj +kyj( ) 1+ ei2π 0.5h+0.5k( )
( )j=1
N/2
∑
= cos
π
2
(h + k)
⎛
⎝⎜
⎞
⎠⎟ 2e
i
π
2
hxj +kyj( )
fje
i2π hxj +kyj( )
j=1
N/2
∑

Friedel’s Law
¡ X-ray diffraction data will always display a center of
symmetry, even if crystal structure does not.
¡ Hence, point group symmetry of diffraction data must
belong to one of the 11 Laue classes!
Fhkl = Fhkl
*
Fhkl
= Fhkl
*
Ihkl = FhklFhkl
*
= Fhkl
Fhkl
*
= Ihkl
21

International tables and systematic
absences
International tables page for Cmm2
IMPORTANT: The International Tables
report the conditions for reflection, which is
the opposite of the extinction conditions! DO
NOT GET THE TWO MIXED UP!
Earlier, we derived that for a C-centered lattice, the extinction
condition for a general (hkl) is that h + k = 2n + 1, i.e., odd. So
the reflection condition is that h + k = 2n, i.e., even!
22

International tables and systematic
absences
International tables page for Fm-3m
Earlier, we derived that for a F-centered lattice, extinctions occur whenever a mix of odd
and even numbers occur in (hkl). So the reflection condition is that we must have all odd
or all even indices, i.e., the sum of any two indices must be even!
23

Practical calculations of the Structure
Factor
¡We will now illustrate how you can compute the
structure factor for a given crystal structure, and
also discuss how missing reflections provide
information about the arrangement of atoms in
the structure.
24

Example: CsCl
¡ Pm-3m unit cell:
¡ Cs (0, 0, 0)
¡ Cl (0.5, 0.5, 0.5)
¡ Two sets of reflections:
¡ Fundamental reflections: Proportional to sum of atomic scattering factors
¡ Superlattice reflections: Proportional to difference of atomic scattering factors
¡ If Cs and Cl are completely randomly located:
¡ Fundamental reflection = (fCs+fCl)/2
¡ Superlattice refection = 0 ç Indication of order of structure
Blackboard
25

Example: NaCl
¡ Two interpenetrating fcc lattices of Na and Cl, with Cl lattice displaced
by (½, ½, ½)
¡ Recall that h, k, l must be either all odd or all even for reflections in
fcc. Hence
Blackboard
Blackboard
26

Measured diffraction intensities
¡ XRD experiments are carried out for a certain period of
time with a detector with a certain aperture.
¡ Measured XRD patterns are time averages of the
scattered intensity and only a small fraction of total
scattered intensity is measured.
¡ To go from computed intensities to measured intensities,
we need to consider the effects of
¡ Temperature
¡ Absorption
¡ Multiplicity
¡ Geometry
27

Temperature
¡ At finite temperatures, atoms in a crystal vibration around their lattice sites.
Amplitude of vibration is determined by the temperature.
¡ Atomic vibrations result in more diffuse electron clouds, which in turn affect
the atomic scattering factor.
¡ To account for temperature effects, the atomic scattering factor must be
multiplied by an exponential attenuation or damping factor, the Debye–Waller
factor.
¡ B(T) is unknown for most crystal systems.
¡ Larger for elements in the left-most columns of the periodic table
¡ Group 1 ~ 0.1 nm2
¡ Group 2 ~0.01-0.03 nm2
28

Absorption factor
¡ X-rays travelling through a media is partially absorbed.
¡ Intensities must be scaled by absorption factor, A.
¡ For standard powder diffractometer, however, it can be
shown that this factor is a constant independent of
diffraction angle. Hence, this may be ignored since we are
usually interested only in relative intensities.
29

Multiplicity factor
¡ Diffracted intensity for a particular plane (and angle 2θ) may have
contributions from several planes (due to symmetry).
¡ E.g., Intensity for (200) in Cu has contributions from (020) and (002) planes
and their negatives.
¡ Total intensity scattered from a plane (hkl) must be multiplied by its
multiplicity phkl
30

Lorentz Polarization Factor
¡ Unpolarized X-ray is scattered in different directions by
single electron.
¡ Trigonometric factor given by:
¡ In addition, three geometric factors need to be taken into
account.
P(θ) =
1+cos2
2θ
2
31

Lorentz Polarization Geometric Factor 1
¡Diffractometer measures a fixed area of overall
conical region of peaks.
¡Intensity per unit length is related to
1
sin2θ
32

¡ Powder sample has randomly oriented grains
¡ Detector detects intensity over
¡ Normal of planes lie within
¡ Fraction of grains = No. of grains with normals within band
/ total number of grains
¡ Hence factor of cosθ
2θ ±
Δθ
2
90 −θ ±
Δθ
2
ΔN
N
=
rΔθ2πrsin(90 −θ)
4πr2
=
Δθ cosθ
2
33

¡Planes do not diffract only at exact Bragg angle.
¡Can be shown that integrated intensity is
proportional to
1
sin2θ
34

Overall correction
¡ The overall Lorentz polarization
factor is given as:
¡ Total measured intensity is
therefore given by:
Lp (θ) =
1+cos2
2θ
sin2
θ cosθ
Ihkl = Fhkl
2
phklLp (θ)A(θ)e−2B(T )s2
35

¡α-CsCl
Example of XRD computation
Calculated
Experimental
http://nbviewer.ipython.org/github/materialsvirtuallab/nano106/blob/master/lectures/lecture_13_xrd/XR
D%20CsCl.ipynb
36

Example of XRD computation
¡β-CsCl
Calculated Experimental
http://nbviewer.ipython.org/github/materialsvirtuallab/nano106/blob/master/lectures/lecture_13_xrd/XR
D%20CsCl.ipynb
37

Crystal structures and powder
diffraction patterns
¡We will use two examples – Ni and NaCl
¡For each structure, we will perform:
¡ Crystal structure è Diffraction pattern
¡ Diffraction pattern è Crystal structure
38

Ni-coin pattern from known crystal
structure
¡ CuKα radiation
¡ US “Nickel” coin – 75 at % Cu, 25 at % Ni
¡ Both Cu and Ni are fcc and fully soluble in one other, i.e., alloy of Cu
and Ni
¡ Vegard’s law: approximate empirical rule which holds that a linear
relation exists, at constant temperature, between the crystal lattice
parameter of an alloy and the concentrations of the constituent
elements
¡ Hence, lattice parameter of 0.75Cu-0.25Ni:
a = 0.75aCu + 0.25aNi = 0.75(0.36078nm)+ 0.25(0.35868nm) = 0.35868nm
39

Ni-coin pattern from known crystal
structure
¡Structure factor for solid solution of Ni and Cu
¡Debye–Waller factors at T =290 K:
¡ BNi =0.0035 nm2 and BCu =0.0054 nm2
Fhkl = 4
1
4
fNie−BNis2
+
3
4
fCue−BCus2"
#
$
%
&
'
Ihkl = phkl Fhkl
2 2
3
LP (θhkl,Kα1
)+
1
3
LP (θhkl,Kα 2
)
!
"
#
$
%
&
Relative intensities of two Kα lines
40

Computed vs experimental intensities
of nickel coin
¡ Agreement of intensities is not good because of the assumption that
all crystal orientations are present with equal probability is not valid
due to the processing of the coin.
41

NaCl diffraction pattern from known
crystal
¡ Earlier, we derived the intensities (pre-correction) as
¡ Debye-Waller factors:
¡ BNa = 0.0172 nm2
¡ BCl = 0.0141 nm2
Ihkl =
0 if h,k,l not all odd or all even
16( fNa + fCl )22
if h,k,l all even
16( fNa − fCl )22
if h,k,l all odd
"
#
$$
%
$
$
42

Computed intensities vs experimental
intensities
43

Rietveld refinement
¡Previous examples show that good accuracy can be
obtained with relatively simple approaches, but more
sophisticated approaches exist
¡Rietveld refinement is a commonly used method that
uses a least squares approach to refine a theoretical
line profile until it matches the measured profile.
¡Able to deal reliably with strongly overlapping
reflections.
¡Accounts for the background intensity, the peak
shape, the atom positions and occupations in the unit
cell, the cell parameters, thermal Debye–Waller
factors, etc.
44

Crystal structure from diffraction
pattern
¡Far more difficult task than computation of
diffraction patterns
¡Start from the measured diffraction pattern and a
few other pieces of information, e.g., density
45

Example: Ni-coin structure from
diffraction pattern
¡ Density
¡ Ni: 8.912 g/cm3
¡ Cu: 8.933 g/cm3
¡ Measured alloy ~ 8.923 g/cm3
¡ Rule of thumb:
¡ No. of peaks in powder pattern is inversely
proportional to complexity of structure, including
symmetry
¡ Start from assumption of cubic lattice -> three
Bravais lattices (cP, cI and cF)
¡ Calculate Miller indices and lattice parameters
for three centering operations
46

Computed lattice parameters assuming
different cubic lattices
From expt pattern
Lowest
standard
deviation!
Increasing
dhkl
47

Caveats
¡Ni coin example seems simple, but in reality,
structure determination is difficult.
¡ Started from cubic assumption -> only 1 lattice
parameter (rather than 6)
¡Infinite number of choices for unit cell
¡Ideally, use the most symmetric unit cell, but even
if another unit cell had been chosen, should still
be able to assign Miller indices to all peaks
48

Practical structure determination
¡ Nowadays unit cell determination is essentially a non-
linear least-squares problem that is readily solved by
means of standard numerical algorithms
¡ Determination of unit cell parameters is first step in
structure determination
¡ Once the lattice parameters are known, then the positions
of all diffracted beams are known, and one can focus on
the intensities of the individual peaks, i.e., on the
determination of the atom positions.
49

UCSD NANO106 - 12 - X-ray diffraction

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