1. Last Structure Example: Perovskites
Last Structure Example: Perovskites
Superconductors
Ferroelectrics (BaTiO3)
Colossal Magnetoresistance (LaSrMnO3)
Multiferroics (BiFeO3)
High εr Insulators (SrTiO3)
Low εr Insulators (LaAlO3)
Conductors (Sr2RuO4)
Thermoelectrics (doped SrTiO3)
Ferromagnets (SrRuO3)
A-site (Ca)
Oxygen
B-site (Ti)
CaTiO3 eg
t2g
In addition to being among the most abundant minerals on earth, complex oxides give some of the most varied and interesting properties. These include their use as dielectric and superconducting materials. Yet, only recently has the research in the field of complex oxides flourished because they were long thought to be …well complex. This complexity comes from the strong coupling with charge, spin, and lattice dynamics, which often results in very full phase diagrams.
Though the coupling may lead to complex behaviors, the structure of these materials can be quite simple, such as the perovskite form shown here, where the A and B sites are typically different cations and X is an anion that bonds to both. This octahedral arrangement (imagine connecting the oxygens) gives rise to a crystal field potential, hinders the free rotation of the electrons and quenches the orbital angular momentum by introducing the crystal field splitting of the d orbitals. Even among only perovskite structures, we can see these varied behaviors.
Formula unit – ABO3
A atoms at the corners
B atoms (smaller) at the body-center
O atoms at the face centers
In groups, define the crystal structure.

2. Is this cube a primitive lattice?
PEROVSKITES
A-site (Ca)
Oxygen
B-site (Ti)
CaTiO3
Lattice: Simple Cubic (idealized structure)
1 CaTiO3 per unit cell
Cell Basis/Motif: (see A cell diagram) Ti at (0, 0, 0); Ca at (1/2, 1/2, 1/2); 3 O at (1/2, 0, 0), (0, 1/2, 0), (0, 0, 1/2) could label differently
How many nearest neighbors for A and B cations?
Ca 12-coordinate by O, Ti 6-coordinate by O, O distorted octahedral
Yes primitive—only one lattice point per unit cell. Don’t confuse lattice points with atoms, of which there are many.
Anion doesn’t have to be oxygen, even though that is common. For example, fluorides are also of interest. Current theory suggests that the properties in fluorides might be even better than oxides.
Is this cube a primitive lattice?

3. Learning Objectives for Diffraction
After our diffraction class you should be able to:
Explain why diffraction occurs
Utilize Bragg’s law to determine angles of diffraction
Briefly discuss some different diffraction techniques and what can be learned from diffraction
(Future) Determine the lattice type and lattice parameters of a material given an XRD pattern
Test: Sept 12, bring calculator and handwritten sheet
Particularly useful to put this lecture in context with what we’ve learned so far to give the big picture (BP)
Perhaps I should reverse Laue/Powder order and go over some data in more detail? Could drop Laue if run out of time. BP

4. Applications of XRD (X-ray diffraction)
XRD is a nondestructive and cheap technique. Some of the uses of x-ray diffraction are:
Determination of the structure and periodicity of crystalline materials
Determination of the orientation of single crystals
Differentiation between crystalline and amorphous materials, or if secondary phases
Measurement of (epitaxial) strain
Determination of electron distribution within the atoms, and throughout the unit cell (harder)
Measurement of layer thickness (related XRR)
Before we get into what diffraction is, let’s discuss why you should care.
Secondary phases, compare to CaTiO3. Discuss valence. Can also form TiO2 and CaO. Also sometimes get different structures for the same atoms.
Thickness, either fringes if really thin and smooth or can do at low angle (x-ray reflectivity)

5. While there are many forms of diffraction, we start with
x-ray diffraction.
X-ray Diffraction is an optical technique.
There are many optical techniques.
Most optical techniques operate in the
Continuum limit:
Where the wavelength is bigger than the spacing between atoms. Otherwise diffraction effects dominate.
Let’s simplify our complicated crystal to just a linear chain of atoms. Our example of a chain of oscillators is nice because it is easy to visualize such a system, namely, a chain of masses connected by springs. But the ideas of our example are far more
useful than might appear from this one simple mechanical model. Indeed, many materials
to a first approximation. In a sense we will explore later, even the electromagnetic field
small) perturbations behaving just as if they were a bunch of coupled oscillators — at least
(including solids, liquids and gases) have some aspects of their physical response to (usually
behaves this way! This “harmonic oscillator” response to perturbations leads — in a
with various boundary conditions. Because the harmonic approximation is often a good
caught a glimpse of this when we examined the normal modes for a chain of oscillators
continuum model — to the appearance of wave phenomena in the traditional sense. We
first approximation to the behavior of systems near stable equilibrium, you can see why
media (earth, water, air, etc.) which are modeled as continuous rather than discrete.
wave phenomena are so ubiquitous. Wave phenomena are typically associated with propagation
function of time and space) and we want to ignore most of the details of the microscopic
macroscopic properties of some material (e.g., the behavior of a plucked guitar string as a
The basic physical idea is reasonably simple. Often times we are interested in certain
are much larger than the length scales associated with the microscopic structure (e.g., the
length scales associated with the macroscopic behavior of the material (e.g., wavelengths)
make-up of the material since they should be irrelevant for the most part. So long as the
inter-particle spacing) we can approximate the behavior of the material by taking a limit
become arbitrarily large (“approach infinity”).
in which the inter-particle spacing approaches zero while letting the number of oscillators
Why might you want to think about optical properties? There are at least two reasons. One is that you can make use of known optical materials to design and build devices to manipulate light: mirrors, lenses, filters, polarizers, and a host of others. The second is that you can measure the optical response of a new material and obtain a wealth of information about the low energy excitations in the solid.
With the use of your own eyes, you could see that solids have a wide range of optical properties. Silver is a lustrous metal, with a high reflectance over the whole visible range. Silicon is a crystalline semiconductor and the basis of modern electronics. With the surface oxide freshly etched off, silicon is also rather reflective, although not as good a mirror as silver. ∗ Salt (sodium chloride) is a transparent ionic insulator, is necessary for life, and makes up about 3.5% (by weight) of seawater. A crystal of salt is transparent over the entire visible spectrum; because the refractive index is about 1.5, the reflectance is about 4%.
In contrast, the reflectance of silicon would be better than in the visible, reaching up
poor reflector, with at most 20% reflectance and trailing off to zero at the shortest wavelengths.
If you had ultraviolet eyes, you would see these materials differently. Silver would be a
to 75%. Sodium chloride would be opaque over much of the spectrum, with a reflectance a
appear opaque at the shortest infrared wavelengths but would then become transparent, so
visible or uv-sensitive individuals. Silver would have a reflectance above 99%. Silicon would
bit higher than in the visible. Those with infrared eyes would also see things differently from
over much of the infrared
that you could see through even meter-thick crystals.† Sodium chloride remains transparent
But the continuum model does not work when the wavelength is on the order of the spacing between the atoms. Then you have diffraction.
What is this spacing approximately?

6. DIFFRACTION
Diffraction is a wave phenomenon in which the apparent bending and spreading of waves when they meet an obstruction is measured.
Diffraction occurs with electromagnetic waves, such as light and radio waves, and also in sound waves and water waves.
X-ray diffraction is optimally sensitive to the periodic nature of the solid’s atomic structure.
These shown water waves have wavelengths a little smaller than the geographical formations and thus we see diffraction.

7. When X-rays interact with atoms, you get scattering
Scattering is the emission of X-rays of the same frequency/energy as the incident X-rays in all directions (but with much lower intensity)
Similar to the double slit experiment, this scattering will sometimes be constructive
Second order process due to phase ambiguity, doesn’t matter if initial photon is absorbed before emitted photon.
See from Fermi’s golden rule (Ch.4 in the book).

8. Will look at this again shortly
Incident beam
Will look at this again shortly
Second order
Note diameter of smallest circle is one wavelength. Third circular is two wavelengths. The light must be incident over some time to create the wavefronts shown.
Careful study of the pattern of theses beams can tell you a lot about the structure (see applications of XRD slide).
Zeroth Order

9. Physical Model for X-ray Scattering
Consider a plane wave scattering on an atom.
Atom
The book lists ko as just k, but I find that confusing, so I typically add the subscript
Reason for adding subscript: in most texts just k would stand for the final k value and here it’s opposite so I like to add the o. That way it’s absolutely clear which I mean.
For now you can just realize from Modern physics and quantum mechanics that the wavevector k is related to the energy by E=hbar^2 k^2 /(2m) for a particles or in the case of a photon E=hbar*velocity*k (You can figure this out from starting with E=hbar omega and using v=frequency times wavelength and so on)
Why isn’t r squared? We are looking at psi, not intensity.
This is a different approach from the book, but I think its helpful for visualizing the geometry involved. It will also help to understand what the structure factor is (in next class).

10. Diffraction In a Crystal of Many Atoms
Generic incoming radiation amplitude is:
To calculate amplitude of scattered waves at detector position, sum over contributions of all scattering centers Pi with scattering amplitude (form factor) f: ko
Detector Pi ri R’
R’-ri R
source
Lots doesn’t depend on the sum over positions
Drop the time dependence for now to make it simpler; it’s just a phase that will disappear if you multiple by complex conjagate.
Trace out triangle for R’-ri on powerpoint and/or put on board.
Could plug into top equation on the board to show that you get the value after the arrow.
Generic for any point in the sample. (ignoring the time dependence, will disappear when we multiply by the complex conjugate)
The intensity that is measured (can’t measure amplitude) is
Scattering vector
Kittel calls G, but K is another common notation.

11. Diffraction Theory k’ G=k’-ko koPi ri ko ko R
R’-ri R’
source
Generic for any point in the sample. (ignoring the time dependence, will disappear when we square it)
Detector
The intensity that is measured (can’t measure amplitude) is
Scattering vector
Kittel calls G, but K is another common notation.

12. The Bottom Line
If you do the math you can prove that the peaks only occur when (a1, a2, a3 = lattice vectors):
n1, n2, n3 integers
This requires a bunch of math to show that could take a good portion of the class today if I wanted to. However, that’s not really the point of this class. Better just to trust me here and learn how to us it.
Compare these relations to the properties of reciprocal lattice vectors:

13. The Laue Condition
Replacing n1 n2 n3 with the familiar h k l, we see that these three conditions are equivalently expressed as: G
(Max von Laue, 1911)
So, the condition for nonzero intensity is that the scattering vector G is a translation vector of the reciprocal lattice.

14. Show vector subtraction on the board
From Laue to Bragg
The magnitude of the scattering vector G depends on the angle between the incident wave vector and the scattered wave vector:
Show vector subtraction on the board
=/2
Notice this angle is 2!
Elastic scattering requires:
A very common kind of measurement is the theta-2theta scan. Discuss. Easy to get confused about the angle.
Draw vector k-ko on the board to illustrate correct direction of K.
Elastic scattering requires no energy lose, so k’s magnitude is the same
Use right triangles to show 2ksin theta
You can use a similar argument to show how d and lambda are related from the diagram on the left. a=dsin theta (D=lambda/2 sin theta) Define sin of theta to show this.
So from the wave vector triangle and the Laue condition we see:
Leaving Bragg’s law:
If the Bragg condition is not met, the incoming wave just moves through the lattice and emerges on the other side of the crystal (neglecting absorption)

15. How does this limit ?
where, d is the spacing of the planes and n is the order of diffraction.
Bragg reflection can only occur for wavelength
This is why we cannot use visible light. No diffraction occurs when the above condition is not satisfied.
What jumps out at you as a necessary condition for this to be true?

16. Bragg Equation:
The diffracted beams (reflections) from any set of lattice planes can only occur at particular angles predicted by the Bragg law.
Bragg-Brentano diffractometer (θin=θout)
When atoms are specific phases apart. Otherwise they destructively interfere over the many layers of the crystal. Also shows why higher order diffraction occurs (just off by a multiple of the wavelength).
Note: While peaks should not technically be labeled with parentheses, it is very common to see them labelled that way. Likely even my examples will sometimes show them that way.
Above are 1st, 2nd, 3rd and 4th order “reflections” from the (111) face of NaCl. Orders of reflections are given as 111, 222, 333, 444, etc. (without parentheses!)

17. Why might you use this technique?
A single crystal specimen in a Bragg-Brentano diffractometer (θin=θout) would produce only one family of peaks in the diffraction pattern.
Why might you use this technique? 2q
Good for determining lattice parameters. It’s also simple. It’s not too hard to figure out what all of your peaks are.
The (110) planes would diffract at 29.3 °2q; however, the detector is not at that position (the perpendicular to those planes does not bisect the incident and diffracted beams). Only background is observed.
At 20.6° (2q), Bragg’s law fulfilled for the (100) planes, producing a diffraction peak.
The (200) planes are parallel to the (100) planes. Therefore, they also diffract for this crystal. Since d200 is ½ d100, they appear at 42° (2q).

18. THE EWALD SPHERE (Will show a few ways)
Consider an arbitrary sphere
passing through the reciprocal lattice,
with the crystal arranged in the center of the sphere.
We specify two conditions:
the sphere radius is 2 / - the inverse wavelength of X-ray radiation
the origin of the reciprocal lattice lies on the surface of the sphere
X-rays are ON
diffracted ray
Sometimes you’ll see this as just one over lambda (dropping the 2 pi), when reciprocal lattice also drops 2 pi
At this point, the Bragg condition is said to be satisfied
I personally don’t like the Ewald sphere, but a lot of researchers think it’s the best way to understand diffraction points, so I feel I should at least show it.
So far it looks easy enough, but it gets more confusing because you have to rotate the sphere. 2 O
2/
The diffraction spot will be observed when a reciprocal lattice point crosses the Ewald sphere

19. 1. Longitudinal or θ-2θ scan
Sample moves as θ, Detector follows as 2θ k0 k’

20. 1. Longitudinal or θ-2θ scan
Sample moves on θ, Detector follows on 2θ G k0 k’
Reciprocal lattice rotates by θ during scan

21. 1. Longitudinal or θ-2θ scan
Sample moves on θ, Detector follows on 2θ G k0 k’ 2q

22. 1. Longitudinal or θ-2θ scan
Sample moves on θ, Detector follows on 2θ G k0 k’ 2q

23. 1. Longitudinal or θ-2θ scan
Sample moves on θ, Detector follows on 2θ G k0 2q k’

24. 1. Longitudinal or θ-2θ scan
Sample moves on θ, Detector follows on 2θ G k0 2q k’

25. 1. Longitudinal or θ-2θ scan
Sample moves on θ, Detector follows on 2θ k’ G k0 2q
Provides information about relative arrangements, angles, and spacings between crystal planes.

26. Higher order diffraction peaks
Even if overlaps, will not have peak if the structure factor is nonzero (later).

27. 3 COMMON X-RAY DIFFRACTION METHODS
Laue
Rotating Crystal
Powder
Orientation
Single Crystal
Polychromatic Beam
Fixed Angle
Lattice constant
Single Crystal
Monochromatic Beam
Variable Angle
Lattice Parameters
Polycrystal/Powder
Monochromatic Beam
Fixed Angle

28. Back-reflection vs. Transmission Laue Methods
In the back-reflection method, the film is placed between the x-ray source and the crystal. The beams which are diffracted backward are recorded.
Which is this?
X-rays have wide wavelength range (called white beam).
Single
Crystal
X-Ray
Film
Single
Crystal
Film
X-Ray
The diffraction spots generally lay on:
an ellipse
a hyperbola

29. LAUE METHOD
The diffracted beams form arrays of spots, that lie on curves on the film.
The symmetry of the pattern reflects the symmetry of the crystal when viewed along the direction of the incident beam.
Each set of planes in the crystal picks out and diffracts a particular wavelength from the white radiation that satisfies the Bragg law for the values of d and θ involved.
Laue on one of our thin films.

30. Crystal structure determination by Laue method?
Although the Laue method can be used, several wavelengths can reflect in different orders from the same set of planes, making structure determination difficult (use when structure known for orientation or strain).
Rotating crystal method overcomes this problem. How?
You would interpret from the intensity of the spots, but if multiple spots are at the same location, that’s a problem.
Materials scientists typically already know the crystal structure for a material they’ve worked on a lot and then this method can be used to understand orientation or strain tensors.

31. ROTATING CRYSTAL METHOD
A single crystal is mounted with a rotation axis perpendicular to a monochromatic x-ray beam.
A cylindrical film is placed around it and the crystal is rotated. Or a detector is rotated.
Sets of lattice planes will at some point make the correct Bragg angle, and at that point a diffracted beam will be formed.  

32. Rotating Crystal Method
Reflected beams are located on imaginary cones. 
By recording the diffraction patterns (both angles and intensities), one can determine the shape and size of unit cell as well as arrangement of atoms inside the cell.
But around what axis should you rotate?
Requires various orientations measured and you may not know where those orientations are, so hard if that is the case.
You might get lucky if you guess or you could…
Film

33. THE POWDER METHOD Least crystal information needed ahead of time
If a powder is used, instead of a single crystal, then there is no need to rotate the sample, because there will always be some crystals at an orientation for which diffraction is permitted.
A monochromatic X-ray beam is incident on a powdered or polycrystalline sample.
Common method if you don’t know much about your material (or if you already have it in powder form).

34. The Powder Method
A sample of some hundreds of crystals (i.e. a powdered sample) show that the diffracted beams form continuous cones.
A circle of film is used to record the diffraction pattern as shown.
Each cone intersects the film giving diffraction arcs.
If a monochromatic x-ray beam is directed at a single crystal, then only one or two diffracted beams may result.
If the sample consists of some tens of randomly orientated single crystals, the diffracted beams are seen to lie on the surface of several cones.
The cones may point both forwards and backwards.

35. Powder diffraction film
When the film is removed from the camera, flattened and processed, it shows the diffraction lines and the holes for the incident and transmitted beams. Likewise, detectors can show the same thing.

36. K

37. Useful for Phase Identification
The diffraction pattern for every phase is as unique as your fingerprint
Phases with the same element composition can have drastically different diffraction patterns.
Use the position and relative intensity of a series of peaks to match experimental data to the reference patterns in the database

38. Databases such as the Powder Diffraction File (PDF) contain dI lists for thousands of crystalline phases.
The PDF contains over 200,000 diffraction patterns.
Modern computer programs can help you determine what phases are present in your sample by quickly comparing your diffraction data to all of the patterns in the database.

39. Quantitative Phase Analysis
With high quality data, you can determine how much of each phase is present
The ratio of peak intensities varies linearly as a function of weight fractions for any two phases in a mixture
RIR method is fast and gives semi-quantitative results
Whole pattern fitting/Rietveld refinement is a more accurate but more complicated analysis
Reference Intensity Ratio Method
RIR method: need to know the constant of proportionality

40. Applications of Powder Diffractometry
phase analysis (comparison to known patterns)
unit cell determination (dhkl′s depend on lattice parameters)
particle size estimation (line width)
crystal structure determination (line intensities and profiles)

41. Extra slides
There is a lot of useful information on diffraction. Following are some related slides that I have used or considered using in the past.
A whole course could be taught focusing on diffraction so I can’t cover everything here.

42. XRD: “Rocking” Curve Scan
Sample normal
“Rock” Sample
Vary ORIENTATION of G relative to sample normal while maintaining its magnitude. How? “Rock” sample over a very small angular range.
Resulting data of Intensity vs. Omega (w, sample angle) shows detailed structure of diffraction peak being investigated. Can inform about quality of sample.
Draw a flat line versus a line with small tilts in either direction

43. XRD: Rocking Curve Example
GaN Thin Film
How do you know if this is good?
Compare to literature to see how good (some materials naturally easier than others)
(002) Reflection
16000
Intensity (Counts/s)
8000
Generally limited by quality of substrate
Could give a whole course on diffraction, but lots more to cover
I’m not talking politics, I just think having Obama cropped in the picture is funny.
16.995
17.195
17.395
17.595
17.795
Omega (deg)
Rocking curve of single crystal GaN around (002) diffraction peak showing its detailed structure.

44. X-ray reflectivity (XRR) measurement
A glancing, but varying, incident angle, combined with a matching detector angle collects the X rays reflected from the samples surface
Calculation of the density, composition, thickness and interface roughness for each particular layer
r t [Å] s [Å]
Edge of TER
Kiessig oscillations (fringes) Mo Mo Mo W Si
The surface must be smooth (mirror-like)

45. XRD: Reciprocal-Space Map
GaN(002)
AlN 
/2
Vary Orientation and Magnitude of k (G).
Diffraction-Space map of GaN film on AlN buffer shows peaks of each film.

46. The X-ray Shutter is the most important safety device on a diffractometer
X-rays exit the tube through X-ray transparent Be windows.
X-Ray safety shutters contain the beam so that you may work in the diffractometer without being exposed to the X-rays.
Being aware of the status of the shutters is the most important factor in working safely with X rays.

47. Non-xray Diffraction Methods (more in later chapters)
Any particle will scatter and create diffraction pattern
Beams are selected by experimentalists depending on sensitivity
X-rays not sensitive to low Z elements, but neutrons are
Electrons sensitive to surface structure if energy is low
Atoms (e.g., helium) sensitive to surface only
For inelastic scattering, momentum conservation is important
X-Ray
Neutron
Electron
Replace top figure (not so informative, probably more confusing)
λ = 1A°
E ~ 104 eV
interact with electron
Penetrating
λ = 1A°
E ~ 0.08 eV
interact with nuclei
Highly Penetrating
λ = 2A°
E ~ 150 eV
interact with electron
Less Penetrating

48. Group: Consider Neutron Diffraction
Qualitatively discuss the atomic scattering factor (e.g., as a function of scattering angle) for neutron diffraction (compared to x-ray) by a crystalline solid.
For x-rays, we saw that f is related to Z and has a strong angular component. For neutrons?
The same equation applies, but since the neutron scatters off a tiny nucleus, scattering is more point-like, and f is ~ independent of .
3.9

49. Preferred Orientation (texture)
Diffracting crystallites
Preferred Orientation (texture)
Preferred orientation of crystallites can create a systematic variation in diffraction peak intensities
can 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)
(220)
(222)
(400) 40 50 60 70 80 90
100
Two-Theta (deg)
x10 3
2.0
4.0
6.0
8.0
10.0
Intensity(Counts)
> Gold - Au