UCSD NANO106 - 12 - X-ray diffraction

$: UCSD NANO106 - 12 - X-ray diffraction$
X-rays diffractionShyue Ping OngDepartment of NanoEngineeringUniversity 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 of Materials 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 = I0Kr2 sin2α

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.


I0 y = I0z =12I

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 =π2

Iy (P) = I0 yKr2

α =π2− 2θ

Iz (P) = I0zKr2sin2(π

2− 2θ )

= I0zKr2cos2(2θ )

Iz (P) = I0Kr2(1+ cos2 2θ ) Strongest for θ= 0 and θ= π.

Weakest for θ= π/2.

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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.


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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 = sinθλ

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−bis

2

i=1

N

∑

s = sinθλ

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.


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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 sinθλ

!

"#

$

%&ei2πghkl⋅r

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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 = f jsinθhklλ

!

"#

$

%&ei2πghkl⋅r

j=1

N

∑ = f jsinθhklλ

!

"#

$

%&ei2π (hx j+kyj+lz j )

j=1

N

∑

Ihkl = Fhkl2= 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)


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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 = fI = 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

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Structure factor for I-centered lattice

¡ I-centered


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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!


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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 = f j (ei2πghkl⋅r + e−i2πghkl⋅r )

j=1

N

∑

= 2 f j cos(2πghkl ⋅ r)j=1

N

∑Real number (no complex component)

Fhkl = f j ei2π (hx j+kyj+lz j ) + ei2π (−hx j+kyj+l (z j+1/4)) + ei2π (−hx j−kyj+l (z j+1/2)) + ei2π (hx j−kyj+l (z j+3/4))"

#$%

j=1

N /4

∑

For planes given by (00l), we have

F00l = f jei2πlz j 1+ e

iπl2 + eiπl + e

i3πl2

"

#'

$

%(

j=1

N /4

∑

= 1+ il + i2l + i3l"# $% f jei2πlz j

j=1

N /4

∑

Always zero unless l = 4n!

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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

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Fhkl = f j ei2π hx j+kyj+lz j( ) + ei2π h(x j+0.5)+k (yj+0.5)−lz j( )( )

j=1

N /2

∑For planes given by (hk0), we have

Fhk0 = f jei2π hx j+kyj( ) 1+ ei2π 0.5h+0.5k( )( )

j=1

N /2

∑

= cos π2

(h + k)⎛⎝⎜

⎞⎠⎟ 2e

iπ2hx j+kyj( )

f jei2π hx j+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* = FhklFhkl

* = Ihkl

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International tables and systematic absences


International tables page for Cmm2IMPORTANT: 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!

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International tables and systematic absences


International tables page for Fm-3mEarlier, 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!

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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.


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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

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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

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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


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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


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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.


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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


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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

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Lorentz Polarization Geometric Factor 1

¡Diffractometer measures a fixed area of overall conical region of peaks.

¡ Intensity per unit length is related to


1sin2θ

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¡ 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 ± Δθ2

90−θ ± Δθ2

ΔNN

=rΔθ2πrsin(90−θ )

4πr2=Δθ cosθ2

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¡Planes do not diffract only at exact Bragg angle.

¡Can be shown that integrated intensity is proportional to


1sin2θ

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Overall correction¡ The overall Lorentz polarization

factor is given as:

¡ Total measured intensity is therefore given by:


Lp(θ ) =1+ cos2 2θsin2θ cosθ

Ihkl = Fhkl2 phklLp(θ )A(θ )e

−2B(T )s2

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¡α-CsCl

Example of XRD computation


Calculated

Experimental

http://nbviewer.ipython.org/github/materialsvirtuallab/nano106/blob/master/lectures/lecture_13_xrd/XRD%20CsCl.ipynb

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Example of XRD computation¡β-CsCl


Calculated Experimental

http://nbviewer.ipython.org/github/materialsvirtuallab/nano106/blob/master/lectures/lecture_13_xrd/XRD%20CsCl.ipynb

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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


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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

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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 = 414fNie

−BNis2

+34fCue

−BCus2"

#$

%

&'

Ihkl = phkl Fhkl2 23LP (θhkl,Kα1 )+

13LP (θhkl,Kα 2 )

!

"#

$

%&

Relative intensities of two Kα lines

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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.


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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

"

#$$

%$$

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Computed intensities vs experimental intensities


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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.


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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


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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


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Computed lattice parameters assuming different cubic lattices


From expt pattern

Lowest standard deviation!

Increasing dhkl

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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


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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.


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UCSD NANO106 - 12 - X-ray diffraction

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