# 361: Crystallography and Diffraction

## 1 Catalog Description

Elementary crystallography. Basic diffraction theory; reciprocal space. Applications to structure analysis, preferred orientation. Point and 2D Detector techniques. Lectures, laboratory. Prerequisites: GEN ENG 205 4; PHYSICS 135 2,3. Mathematics including Calculus 1-3 and Linear Algebra will be required.

## 3 361: Crystallography and Diffraction

At the conclusion of the course students will be able to:
1. Identify different types of crystal structures and symmetry elements such as space groups, point symmetry and glide planes that occur in metals, ceramics, and polymers.
2. Perform standard x-ray diffraction measurements on metals, ceramics and polymers and quantitatively determine lattice constants, grain size, texture, and strain in bulk crystals and epitaxial films.
3. Describe the basic particle and wave physical processes underlying x-ray emission, elastic and inelastic scattering, absorption, and interference of coherent waves.
4. Identify symmetry elements in (point, translation) in 2D and 3D patterns and crystals.
5. Describe basic principles underlying synchrotron x-ray sources, x-ray fluorescence spectroscopy, and electron and neutron diffraction.
6. Use reciprocal space graphical constructions and vector algebra to interpret diffraction from 3D and 2D single crystals, and random and textured polycrystalline samples.

## 4 Symmetry of Crystals

### 4.1 Types of Symmetry

Symmetry elements can be of the following types:
(or a combination of these):

#### 4.1.1 Mirror Planes and Chirality

Probably the most familar form of symmetry from daily life is reflection symmetry. We refer to some object as being symmetric with respect to reflection if putting a mirror crossing through the object would result in no apparrent change to the object (we call this lack of change self coincidence). The lines (or planes in 3D) on which these hypothetical mirrors could be placed are appropriately called mirror planes. Observe Figure 4.1, noticing that the image could be replicated by holding half of it up to an appropriately placed mirror. In a sense, symmetry thus allows us to define structures or patterns in terms of unique, fundamental units. Mirror planes will typically be denoted in figures by bolded lines. It is also worth noting that the two shapes in Figure 4.1 have an opposite handedness, or chirality. The points of the two objects proceed outward in either a clockwise or counterclockwise manner; as a result, the object on the right can be identified as distinct from the one on the left. Thus, we see that a reflection changes the chirality of an object. We encounter this phenomenon in daily life; it is possible to tell a person's right hand from their left because of the arrangement of their fingers, but an image of a left hand reflected from a mirror will have the same arrangement as that of a right hand (similar conclusions follow for the reflection of the right hand).
Figure 4.1: A figure containing a mirror plane. Note that the objects on either side of the plane have opposite chirality

#### 4.1.2 Proper Rotation Axes

A rotational axis refers to some line which can be utilized as the center of some specific rotation that will leave the system (such as a pattern or crystal) seemingly unchanged (invariant). Rotational axes can occur in both three dimensions, or two dimensions, with the “line” being represented as a point in the latter case, indicating lines perpendicular to the plane of view. The plane object pictured in Figure 4.2 demonstrates rotational symmetry, as a 120${}^{○}$ turn about its center brings it into self coincidence. In addition, the object has three mirror planes, and the object's reflection across these planes also brings it into self coincidence.
Figure 4.2: A figure containing 3-fold rotational symmetry, as well as 3 mirror planes
Figure 4.3 shows the types of proper rotation axes that we're concerned with, as well as the symbols which will be used to represent each type. Referring back to Figure 4.2, the object pictured has 3-fold rotational symmetry about an axis directed out of the page, as a $36{0}^{○}/3$ or $12{0}^{○}$rotation allows for self coincidence. In other words, an object has $n$-fold symmetry about an axis if a $36{0}^{○}/n$ rotation about that axis brings it into self coincidence. All objects exhibit 1-fold symmetry, or are brought into self coincidence by a full $36{0}^{○}$ turn. This case is trivial (hence why there is no associated symbol) but important nonetheless.
Figure 4.3: Symbols for proper and improper rotational symmetries.
You may notice that other types of rotational axes (i.e. 5, 7 or greater) are not provided; This is due to the fact that such rotational axes are incompatable with translation symmetry (which will be discussed in Section 4.3). In brief, this simply refers to the fact that objects with, for example, 5-fold symmetry cannot be translated in a way that fills all of space. For instance, Figure 4.4 demonstrates that pentagons cannot be used to fill space without gaps (this also known as tiling).

#### 4.1.3 Inversion Centers

An object is said to contain an inversion center if there is some point (the center) in which identical elements would be encountered when preceding forward or backwards (with the point as the origin) along any line passing through the point. An example of inversion center symmetry is pictured in Figure 4.5. When a line is drawn through the center of this plane object, corresponding points of the object are equidistant from the center line. It will come into coincidence with itself if all points on the object are inverted through the inversion center which is also on this center line. You will note that this object also exhibits rotational symmetry, as a 180${}^{○}$turn about its center brings it into self coincidence. The nature of the inversion operation depends on dimensionality of the system. In 2D, inversion is identical to 2-fold rotation, and preserves the chirality of objects, while in 3D, inversion is distinct from other symmetries and inverts the chirality of objects. Open circles are commonly used to denote an inversion center in figures.
Figure 4.5: A figure with an inversion center and 2-fold rotational symmetry

#### 4.1.4 Improper Rotation Axes

Figure 4.3 shows the symbols utilized to denote rotoinversion axes in images. Note that the 1-fold rotoinversion operation ($\stackrel{‾}{1}$) is denoted by an open circle, as it is exactly equivalent to the inversion operation (given that 1-fold rotation is equivalent to not changing the structure at all). In order to help visualize these operations, observe Figure 4.6, which is adapted from Vainhstein's Modern Crystallography Vol I. The arrangements depicted consist of perfectly asymmetrical, three dimensional, chiral objects related by both proper and improper rotational axes. Note the nature of the 3D inversion operation demonstrated by the $\stackrel{‾}{1}$ example, and that the $\stackrel{‾}{2}$ operation is equivalent to a mirror plane perpendicular to the axis of rotoinversion.
Figure 4.6: Proper Rotation Axes (2D & 3D), Improper Rotoinversion Axes (3D only): From Vainshtein, “Modern Crystallography”, Vol. I Springer

### 4.2 Projections of Symmetry Elements and Point Groups

There are simpler ways of representing symmetry operations, namely through a system of stereographic projections. The top half of Figure4.7 depicts a view of a body with its symmetry axis pointing out of the page and the circle representing its perimeter. The symbol in the center indicates the 2-fold rotation symmetry operation, and the two solid circles depict the $18{0}^{○}$ rotation of the body. The darkened circle in the $\stackrel{‾}{2}$ symmetry operation projection represents a mirror plane.
Figure 4.7: Stereographic representation of symmetry operations

#### 4.2.1 Arrangement of symmetry equivalent objects as an effect of a rotoinversion axis

Figure 4.8: Three bar symmetry operation
1. Start at ${s}_{1}$
2. Rotate 120º
3. Invert to arrive at ${s}_{2}$
4. Rotate 120º
5. Invert to arrive at ${s}_{3}$,
6. Continue until you return to ${s}_{1}$
7. $x$'s and $\circ$'s mark all symmetry-equilivelent positions (see Fig. 4.6).
Figure 4.9: $\stackrel{‾}{3}$ symmetry stereographic projection. $x$'s represent a geometric entity above the plane, while $○$'s represent the same geometric entity (with opposite chirality) below the plane. ${s}_{n}$ (with integer $n$) represent the steps applied after each $\stackrel{‾}{3}$ operation (starting at ${s}_{1}$) until the entity returns to ${s}_{1}$. For 3D representation, see Fig. 4.6.
Certain symmetry elements can be combined to form new patterns, such as the product of a 2-fold and inversion center shown in 4.10. This produces a 2-fold rotation axis which is perpendicular to a mirror plane as indicated by the darkened circle around the object.
Just like an algebraic operation, if A is one symmetry operation and B is another, they may be grouped through some operator to produce a third symmetry element, or C.
$A\cdot B=C$
$2\cdot \stackrel{‾}{1}=\frac{2}{m}\ne \stackrel{‾}{2}$
Figure 4.10: Stereographic projection of a figure displaying a 2-fold rotation axis, inversion symmetry, and the compound $\stackrel{‾}{2}$ operation. Note that the $\stackrel{‾}{2}$ operation is equivalent to a mirror plane perpendicular to the axis of rotation, and that these three operations together lead to a projection of $\frac{2}{m}$ symmetry
For the $2mm$ symmetry operation shown in 4.11, the mirror plane lies parallel to the rotation axis. There is also a third mirror plane that arises from this combination, as shown by the second line drawn through the center of the dashed circle.
$2\cdot m=2mm$
Figure 4.11: Stereographic projection of $2mm$ symmetry
Figure4.12 illustrates an object with symmetry that combines two rotation axes, or a 3-fold rotation axis perpendicular to a 2-fold rotation axis. Note that it does not matter whether the 3-fold or 2-fold operation is performed first, but the 3 2 symmetry is not equivalent to the 2 3 symmetry.
$3\cdot 2=3\phantom{\rule{6px}{0ex}}2$
Figure 4.12: 3-fold perpendicular to a 2-fold rotation axis

#### 4.2.2 Quartz Crystal

Note the point symmetry of a quartz crystal, as pictured in the figure below.
Figure 4.13: Quartz crystal
The perfect shape (or habit) is brought into self coincidence by (or is invariant to) the following symmetry operations:
${g}_{0}=e,=1$
${g}_{1}=3,\phantom{\rule{6px}{0ex}}{g}_{2}={3}^{2}$
${g}_{3}={2}_{X},\phantom{\rule{6px}{0ex}}{g}_{4}={2}_{Y},\phantom{\rule{6px}{0ex}}{g}_{5}={2}_{U}$
Note that successive symmetry operations also bring the object into self coincidence.
${g}_{1}\cdot {g}_{1}={g}_{1}^{2}={g}_{2},\phantom{\rule{6px}{0ex}}{g}_{1}\cdot {g}_{4}={g}_{3}$
Note that ${g}_{1}$and ${g}_{2}$are the inverse of each other and that ${g}_{3}$, ${g}_{4}$and ${g}_{5}$are their own inverse operation. For instance, if you perform the ${g}_{1}$operation followed by the ${g}_{2}$operation, the object is at its starting position.
${g}_{1}\cdot {g}_{2}={g}_{0}\to {g}_{1}={g}_{2}^{-1}$ ${g}_{3}\cdot {g}_{3}={g}_{3}^{2}={g}_{0}\to {g}_{3}={g}_{3}^{-1}$
This is the 3 2 or D${}_{3}$ point symmetry group.

#### 4.2.3 Group Theory Axioms

1. A group G consists of an operator “multiplication” and a set of elements ${g}_{i}\in G$, such that the group is closed under “multiplication.” ${g}_{i}\cdot {g}_{j}={g}_{k}\in G$
2. Multiplication is associative: ${g}_{i}\cdot \left({g}_{j}\cdot {g}_{k}\right)=\left({g}_{i}\cdot {g}_{j}\right)\cdot {g}_{k}$
3. There exists an identity element $e\in G$ such that $e\cdot {g}_{i}={g}_{i}$
4. For any ${g}_{i}$ there is an inverse ${g}_{i}^{-1}\in G$ such that ${g}_{i}\cdot {g}_{i}^{-1}=e$
The completion of the following table is left as an exercise for the reader.
 ${g}_{0}$ ${g}_{1}$ ${g}_{2}$ ${g}_{3}$ ${g}_{4}$ ${g}_{5}$ ${g}_{0}$ ${g}_{0}$ ${g}_{1}$ ${g}_{2}$ ${g}_{0}$ ${g}_{2}$ ${g}_{2}$ ${g}_{3}$ ${g}_{4}$ ${g}_{0}$ ${g}_{5}$
Table 1: Multiplication of symmetry elements within the 3 2 point group
The following demonstrates the multiplication for the 3 2 point symmetry group:
${g}_{0}=e,{g}_{1}=3,{g}_{2}={3}^{2},{g}_{3}={2}_{x},{g}_{4}={2}_{Y},{g}_{5}={2}_{U}$
Note that the Cyclic group ${C}_{3}=\left\{0,1,2\right\}$ under modulo-3 addition is directly analogous to the 3-fold rotation axes point group.
 Crystal system Minimal symmetry† Crystal classes Triclinic 1(or $\stackrel{‾}{1}$) $1,\stackrel{‾}{1}$ Monoclinic One 2 (or $\stackrel{‾}{2}$) $2,m,\frac{2}{m}$ Orthorhombic Three 2's (or $\stackrel{‾}{2}$) $222,2mm,mmm$ Tetragonal One 4 (or $\stackrel{‾}{4}$) $4,\stackrel{‾}{4},\frac{4}{m},422,4mm,\stackrel{‾}{4}2m,\frac{4}{m}\frac{2}{m}\frac{2}{m}$ Cubic(isometric) Four 3's (or $\stackrel{‾}{3}$) $23,432,\frac{2}{m}\stackrel{‾}{3},\stackrel{‾}{4}3m,\frac{4}{m}\stackrel{‾}{3}\frac{2}{m}$ Hexagonal One 3 or 6 (or $\stackrel{‾}{3}$ or $\stackrel{‾}{6}$) $3,\stackrel{‾}{3},32,3m,\stackrel{‾}{3}\frac{2}{m},6,\stackrel{‾}{6},622,6mm,\stackrel{‾}{6}m2,\frac{6}{m},\frac{6}{m}\frac{2}{m}\frac{2}{m}$
† The symmetry elements shown in this column are the least symmetry that a crystal can have and still belong to the corresponding crystal system. In this sense, this column shows the diagnostic symmetry elements of each system.
Table 2: Azaroff Table 1-2; Crystal systems and crystal classes
Figure 4.15: Rotation symmetry axes of a cube
Figure 4.16: Crystal habits (Schartz & Cohen Fig. 1-6)

#### 4.2.4 The 5 Point Groups belonging to a Cubic Crystal System

Note that a cubic crystal does not necessarily have 4-fold axes. (See Vainshtein “Modern Cryst.” Vol 1 Fig 2.45-2.47)
-
Figure 4.18: Crystal habits
Note the symmetry of the octahedron in Figure 4.19 with vertices at midpoints of square faces of cube.
A regular octahedron has $m\stackrel{‾}{3}m$ point group symmetry, also known as $\frac{4}{m}\stackrel{‾}{3}\frac{2}{m}$. Figure 4.20 shows the projection of the octahedron on the [111] plane, with a $\stackrel{‾}{3}$ improper rotation axis.
Figure 4.19: Octahedron

Figure 4.21: Stereographic projection of octahedron symmetry

### 4.3 Translational Symmetry

The object in Figure 4.22 demonstrates translation center symmetry or a repetitiveness in space. This figure may be interpreted as a flat representation of a diatomic molecule, which is being translated in equal steps both to the right and to the left.
Figure 4.22: Translational symmetryof a schematic diatomic molecule in space by an operator $T$.

## 5 Crystal Lattices

### 5.1 Indexing within a crystal lattice

The unit cell of a crystal may be described by the following three parameters:
Crystal axes: a, b, c
Length: a, b, c
Angles: $\alpha ,\beta ,\gamma$
Crystal axes, illustrated in Figure 5.1, share an origin at one corner of the unit cell. The angle between these axes, or $\alpha ,$$\beta$ and $\gamma$, are based on the symmetry of the crystal. For instance, only the cubic crystal system has $\alpha =\beta =\gamma =90{}^{○}.$ The c-axis usually points to the highest symmetry element. Finally, the lengths of the axes are determined by the natural repeat lengths of the unit cell and based on the crystal habit.
Figure 5.1: Crystal axes
1. Determine the plane's intercepts with the crystal axes
2. Take reciprocals of the intercepts
3. Remove common factors
4. Put in ( ), no commas
All parallel planes have the same (hkl) Miller Indices. If a plane crosses through the origin, as shown in the left axes in Figure 5.2, it can be translated to determine its intercepts. By convention, a bar over the number is used to indicate a negative index.
Figure 5.2: Determining Miller indices
a, b ,c Intercepts: $\infty ,\frac{-1}{4}$,$\frac{1}{5}$
Reciprocals: 0, -4, 5
Miller Indices: (0 $\stackrel{‾}{4}$5)
Miller indices: If you need an additional refresher on Miller indices, the Wikipedia page (https://en.wikipedia.org/wiki/Miller_index) is very helpful. Here we include a reminder of the notation used to indicate planes and directions.
• Specific planes: We use round brackets - $\left(hk\ell \right)$
• Class of planes: To indicate all planes that are crystallographically identical, we use curly brackets - $\left\{hk\ell \right\}$
• Specific direction: We use square brackets - $\left[uvw\right]$
• Class of directions: To indicate all directions that are crystallographically identical, we use angle brackets - $⟨hk\ell ⟩$
Determine the hkl indices of the plane shown in the figure below:
Figure 5.3: Plane in cubic unit cell
Note: Faces of cube are of the same ((hkl)) form ((100))

#### 5.1.1 Hexagonal Prism

Indices of hexagonal prism faces are not of the same (hkl) form. Instead, a new index $i$ is introduced to identify planes, in which $i=-\left(h+k\right)$ or ${a}_{1}+{a}_{2}=-{a}_{3}$. Thus, there are only three independent indices. The four-index ($hkil$) Bravais-Miller system more easily defines planes within a hexagonally structured crystal. Figure 5.4 shows indexing within a hexagonal structure using the three Miller indices, whereas Figure5.5 uses the ${a}_{1},{a}_{2},{a}_{3},c$ axes in the Bravais-Miller scheme.
Figure 5.4: Using Miller indices (hkl)
By using the Bravais-Miller scheme, the symmetry equivalent faces are all of the same form (1100).
Figure 5.6: 3 indices

#### 5.1.2 Form

Example: Octahedron
Faces are of the {111} plane form
What Crystal System?
Figure 5.7: Octahedron formed by eight congruent isosceles triangles (From Azaroff Fig. 1-24)

#### 5.1.3 Direction Indices

Direction indices define a direction with respect to the unit cell axes - a, b and c.
$\left[uvw\right]=ua+vb+wc$
5.8 demonstrates direction indices with respect to unit cell axes.
Figure 5.8: Direction indices

#### 5.1.4 Zone

For instance:
• Let ${\sigma }_{hkl}$ be the normal vector for the plane (hkl).
• Then ${\sigma }_{hkl}\cdot \left[uvw\right]=0$ if hkl plane belongs to [uvw] zone
• For cubic unit cell ${\sigma }_{hkl}=c\left[hkl\right]$ $\to$$hu+kv+lw=0$
• [010] is the zone axis for (001), (100), (102), (10$\stackrel{‾}{2}$) planes

### 5.2 Lattices

The entire infinite array of objects can be described by giving the structure within the unit cell and by giving $\stackrel{\to }{{t}_{1}},\stackrel{\to }{{t}_{2}},\stackrel{\to }{{t}_{3}}$.
 1-D ${t}_{1}$ $\left(a\right)$ 2-D ${t}_{1},{t}_{2}$ $\left(a,b\right)$ 3-D ${t}_{1},{t}_{2},{t}_{3}$ $\left(a,b,c\right)$
Table 3: Translation symmetry operations
Figure 5.10: Array of 7's
2-D $\to$Plane Lattice
3-D $\to$Space Lattice
There are five possible plane lattices, or ways to arrange points in two dimensions such that each has identical surroundings. The five plane lattices, shown in Figure 5.11, include the oblique p-lattice, rectangular p-lattice, rectangular c-lattice, square p-lattice, and hexagonal p-lattice.
Figure 5.11: Five plane lattices (Hammond Fig. 2.4)
Figure 5.12: The Rectangular-C lattice is described by a non-primitive unit cell with 2 lattice points per unit cell; one at the corners (shared by 4 surrounding unit cells) and one at the center. The primitive unit cell for this lattice (diamond) is also shown with $a=b={a}_{1}$ and $\gamma \ne 90°$.
 Azaroff International Tables Parallelogram Oblique p Rectangular Rectangular P Diamond Rectangular C Triangular Hexagonal P Square Square P
Table 4: Two-dimensional lattices
 Axial System Min. Sym. Oblique 1 $a\ne b$ $\gamma \ne 90{}^{○}$ Rectangular m $a\ne b$ $\gamma =90{}^{○}$ Hexagonal 3 $a=b$ $\gamma =120{}^{○}$ Square 4 $a=b$ $\gamma =90{}^{○}$
Table 5: Four axial systems for 2D lattices along with their minimal symmetry requirements and lattice vector conditions.
Note: Only 1,2,3,4,6 are consistent with translation symmetry.

#### 5.2.1 Unit cell

Figure 5.13: Unit cell edges

#### 5.2.2 Plane Groups

$A\cdot t=B$
Figure 5.14: 3-fold and Hexagonal translation operators are combined to create the $p3$ plane group.
There are 17 such plane groups from which any two dimensional periodic pattern may be classified. The shorthand names of these plane groups, as shown in Figure5.15 below, are generated by the following scheme: the “1” represents no symmetry, the integers indicate the type of rotational symmetry, the $"m$” indicates the presence of a mirror plane, and the “g” stands for glide symmetry. Glide symmetry is a hybrid symmetry, resulting from the combination of a mirror reflection followed by a translation along the mirror line direction. The first image in Figure 5.16 demonstrates a familiar mirror plane symmetry, with R's reflected across the solid lines, while the second image shows reflection with one R translated half a lattice spacing down. The dashed lines in the second image indicate a special type of symmetry known as a reflection glide or a glide line of symmetry. The glide lines give this pattern a translational form of symmetry along the lattice rather than a simple reflection. In crystal structures, this shift in glide symmetry would only be observable in an electron microscope which allowing for sight of distances of the order of 0.5–2 Å (50–200 pm). Lastly, note that 3-fold symmetry axes do not exist in oblique, rectangular or square lattices.
Figure 5.15: 17 Plane Groups

#### 5.2.3 Space Lattices or Bravais Lattices

 Crystal System Lattice Types Triclinic P $a\ne b\ne c$ $\alpha \ne \beta \ne \gamma$ Monoclinic P,C $a\ne b\ne c$ $\alpha =\gamma =90{}^{○}\ne \beta$ Orthorhombic P,I,F,C $a\ne b\ne c$ $\alpha =\beta =\gamma =90{}^{○}$ Tetragonal P,I $a=b\ne c$ $\alpha =\beta =\gamma =90{}^{○}$ Trigonal P,R $a=b=c$ $\alpha =\beta =\gamma \ne 90{}^{○}$ Hexagonal P $a=b\ne c$ $\alpha =\beta =90{}^{○};\gamma =120{}^{○}$ Cubic P,I,F $a=b=c$ $\alpha =\beta =\gamma =90{}^{○}$
P: Primitive (1/cell) I:Body Centered (2/cell) C: Base Centered (2/cell)
F: Face Centered (4/cell) Rhombohedral pattern in Hex. Cell (3/cell)
Table 6: 7 Crystal Systems

#### 5.2.4 Positions Within the Unit Cell

Translation to an identical point is achieved by the following: $t={n}_{1}a+{n}_{2}b+{n}_{3}c$ , where n is an integer.
Figure 5.17: Possible locations for origin of unit cell for a square lattice
 Body Centered I 000;$\frac{1}{2}\frac{1}{2}\frac{1}{2}$ C-Centered C 000;$\frac{1}{2}\frac{1}{2}0$ Face Centered 000+fct F 000; $\frac{1}{2}\frac{1}{2}0$;$\frac{1}{2}0\frac{1}{2}$; 0$\frac{1}{2}\frac{1}{2}$
Table 7: Lattice Point locations for non-primitive unit cells

#### 5.2.5 Hexagonal Close Packed Structure HCP

• For Zn, Mg, or Be
Note that he HCP structure has two atoms per hexagonal-Punit cell ( or two atoms per lattice point). The 000 and $\frac{1}{3}\frac{2}{3}\frac{1}{2}$ points do not have equivalent surroundings and therefore do not qualify as lattice points.
Figure 5.18: c-axis projection of HCP

#### 5.2.6 Rhombohedral (Hexagonal)

For a Rhombohedral Primitive Cell:
• 1 lattice point per unit cell at 000
For a Hexagonal R Non-Primitive Unit Cell:
• 3 lattice points per unit cell at 000, $\frac{2}{3}\frac{1}{3}\frac{1}{3}$, $\frac{1}{3}\frac{2}{3}\frac{2}{3}$
Figure 5.19: Hexagonal R Non-Primitive Unit Cell
Counting Lattice Points: 1 corner point per unit cell at 000, since 8 corners shared with 8 cells.
Figure 5.20: 14 Bravais Lattices

## 6 Stereographic Projections

Recall the regular octahedron illustrated in Figure 6.1 exhibiting various types of symmetry.
Figure 6.1: Octahedron - Cubic Crystal System
How did we position these Symmetry Rotation Axes and Mirror Planes in this 2D projection?
Figure 6.2: Octahedron - Stereographic Projection
Stereographic projections are a two-dimensional form of representing three-dimensional crystal structure. Namely, these are graphic representation of the orientation of planes. Stereographic projections allow for an easy visualization of crystallographic features. Looking at Figure 6.3, the normal to the (hkl) plane of the unit cell at the center of the sphere, (${\sigma }_{hkl}$), intersects the reference sphere at the hkl pole.
Consider that all hkl planes pass through the center of the reference sphere.
Figure 6.3: Reference Sphere normal vectors, and poles $P1$ and $P2$ for two planes at center of sphere seperated by angle alpha.
A plane can be represented by a trace made by the plane through the reference sphere.
Traces for planes perpendicular to diameter of reference sphere, but not passing through center are Small Circles or Latitudes (except those at the equator).
Figure 6.4: Reference Sphere with Traces
Rather than measuring angles between planes on the surface of the sphere, we employ an equi-angular stereographic projection.
Note: geographers use equal area projection.
Angle $\alpha$ between two planes in Fig. 6.4 can be measured as being
=Angle between their traces along Great Circles (Angle-True)
=Angle between their Normal's
=Angle along Great Circle connecting their Poles
The plane of the crystal being described is located at the center of the reference sphere, labeled “C” in Figure 6.5. The following terms describe useful parameters to define in the stereographic projection method:

### 6.2 Point of projection

Other end of chosen diameter. “B”
Note the projections of the Great Circles.

#### 6.2.1 Basic Circle

N' E' S' W' = Projection of Great Circle N E S W from plane N E S W with normal A B.
Steps for representing the angle for a plane C on a projection:
1. Plane C $\to$Normal CP
2. Normal CP $\to$Pole P
3. Pole P $\to$Stereographic Projection P'

#### 6.2.2 Wulff Net

Pole P1 at 20${}^{○}$N, 30${}^{○}$E
Pole P2 at 30${}^{○}$S, 40${}^{○}$E
Rotate until both poles like on the same great circle $\to$ 50$°$ is the angle between the planesFigure 6.6: Wulff Net

#### 6.2.3 Standard Projections in a Cubic System

Figure 6.7 below illustrates $\left(001\right)$ and $\left(011\right)$ stereographic projections of cubic system poles for low indexed $\left(hkl\right)$ planes..
Figure 6.7: Standard projections of cubic crystals, from C&S textbook
Figure 6.8: Standard (001) projection of cubic crystal, from C&S textbook. The lines in this projection represent zones whose $hkl$ plane normals are perpendicular to the listed $\left[uvw\right]$ zone axis. Visualize this by looking at the $\left[111\right]$ zone and $111$ pole.
As shown in figure above, the [uvw] zone has (hkl) planes whose normals ${\sigma }_{hkl}$are perpendicular to $\left[uvw\right].$These normals lie in a plane whose intersection with the reference sphere forms a trace.
Note: See C&S Table 2-4 for listing of cubic inter-planar angles.

### 6.3 Representing Atomic Planes with Vectors

When representing a set of periodic planes within a crystal with a vector, its direction is perpendicular to the (hkl) plane. An example for a cubic system is shown in Figure 6.9. The magnitude of this vector is the interplanar spacing, or d-spacing, d-spacing (${d}_{hkl}\right)$, given by the following expression:
$\begin{array}{cc}{d}_{hkl}=\frac{a}{\sqrt{{h}^{2}+{k}^{2}+{l}^{2}}}& \left(6.1\right)\end{array}$
Examples for (110) and (220) planes are:
$\begin{array}{cc}\begin{array}{c}{d}_{110}=\frac{a}{\sqrt{2}}\\ {d}_{220}=\frac{a}{2\sqrt{2}}\end{array}& \left(6.2\right)\end{array}$
Define vector ,${\stackrel{\to }{d}}_{hkl}$to represent cubic (hkl) planes:
${\stackrel{\to }{d}}_{hkl}\perp \left(hkl\right)$ planes

Direction: $\left[hkl\right]=h{\stackrel{‾}{a}}_{1}+k{\stackrel{‾}{a}}_{2}+l{\stackrel{‾}{a}}_{3}$
Figure 6.10: Defining${\stackrel{‾}{d}}_{hkl}$
Unfortunately, this scheme only works for cubic.
Try orthorhombic, $\alpha =\beta =\gamma =90{}^{○}$
Figure 6.11: Orthorhombic Example

### 6.4 Concept of Reciprocal Lattice

Any (hkl) set of diffraction planes will be represented by a point whose coordinate is hkl, such that:
1. The vector ${\stackrel{\to }{r}}_{hkl}^{*}$ which starts at the origin $000$ and ends at hkl is normal to the $\left(hkl\right)$ planes.
2. The distance from the origin to $hkl$ $|{\stackrel{\to }{r}}_{hkl}^{*}|=1/{d}_{hkl}$where ${d}_{hkl}$is the inter-planar spacing
Consider a monoclinic crystal with $a=4Å,b=3Å,c=6Å,$and $\gamma =70{}^{○}$
Show reciprocal lattice points 000,100, and 010 in relation to direct space axes a and b.
Figure 6.12: Reciprocal lattice points for a monoclinic crystal. The $\stackrel{\to }{c}$ axis is perpendicular to the plane of the paper (screen). Note the placement of the reciprocal space origin $000$ is arbitrary. The direction of ${\stackrel{\to }{r}}_{100}^{*}$ that extends from 000 to 100 must be perpendicular to b and c, since by definition it is perpendicular to the (100) planes. While the length of ${\stackrel{\to }{r}}_{100}^{*}$ is arbitrary the ratio of the lengths is not arbitrary. In this case of $\frac{{\stackrel{\to }{r}}_{100}^{*}}{{\stackrel{\to }{r}}_{010}^{*}}=\frac{b}{a}$ .
Construct a hk0 reciprocal lattice corresponding to the (hkl) planes belonging to the [001] zone.
Figure 6.13: hk0 grid
Note: ${r}_{210}^{*}\perp \left(210\right)$plane
Consider a primitive lattice in direct space, whose unit cell is defined by
$\stackrel{‾}{a,}\stackrel{‾}{b},\stackrel{‾}{c}$$\phantom{\rule{6px}{0ex}}\left(a,b,c,\alpha ,\beta ,\gamma \right)$.
Figure 6.14: Unit cell defined by $\stackrel{‾}{a,}\stackrel{‾}{b},\stackrel{‾}{c}$ axes
Figure 6.15: Reciprocal axes
The following equations define the reciprocal axes:
${\stackrel{‾}{a}}^{*}=\frac{\stackrel{‾}{b}×\stackrel{‾}{c}}{\stackrel{‾}{a}\cdot \left(\stackrel{‾}{b}×\stackrel{‾}{c}\right)}$
${\stackrel{‾}{b}}^{*}=\frac{\stackrel{‾}{c}×\stackrel{‾}{a}}{\stackrel{‾}{b}\cdot \left(\stackrel{‾}{c}×\stackrel{‾}{a\right)}}$
${\stackrel{‾}{c}}^{*}=\frac{\stackrel{‾}{a}×\stackrel{‾}{b}}{\stackrel{‾}{c}\cdot \left(\stackrel{‾}{a}×\stackrel{‾}{b\right)}}$
Note:
• ${\stackrel{‾}{a}}^{*}\perp \phantom{\rule{6px}{0ex}}bc$ plane ${\stackrel{‾}{b}}^{*}\perp \phantom{\rule{6px}{0ex}}ac$plane ${\stackrel{‾}{c}}^{*}\perp \phantom{\rule{6px}{0ex}}ab$ plane
• $\stackrel{‾}{a}\cdot \left(\stackrel{‾}{b}×\stackrel{‾}{c}\right)=\stackrel{‾}{b}\cdot \left(\stackrel{‾}{c}×\stackrel{‾}{a}\right)=\stackrel{‾}{c}\cdot \left(\stackrel{‾}{a}×\stackrel{‾}{b}\right)=V$ =Volume of unit cell
• ${\stackrel{‾}{a}}^{*}\cdot \phantom{\rule{6px}{0ex}}\stackrel{‾}{a}=1$, ${\stackrel{‾}{b}}^{*}\cdot \stackrel{‾}{\phantom{\rule{6px}{0ex}}b}=1$, ${\stackrel{‾}{c}}^{*}\cdot \phantom{\rule{6px}{0ex}}\stackrel{‾}{c}=1$ Reciprocal Relationship
Special Case:
For $\alpha =\beta =90{}^{○}$(i.e. all but triclinic or rhombohedral)
Figure 6.16: Special Case ($\alpha =\beta =90{}^{○}\right)$
${\stackrel{‾}{c}}^{*}$$\stackrel{‾}{c}$
${\gamma }^{*}=180{}^{○}-\gamma$
Figure 6.17: Special Case
$\gamma =90{}^{○}-{\gamma }^{*}+90{}^{○}-{\gamma }^{*}+{\gamma }^{*}$
$\gamma =180{}^{○}-{\gamma }^{*}$
For $\alpha =\beta =\gamma =90{}^{○}\to$Cartesian axial system
(Cubic, tetragonal, orthorhombic)
${\stackrel{‾}{a}}^{*}$$\stackrel{‾}{a},$ ${\stackrel{‾}{b}}^{*}$$\stackrel{‾}{b}$, ${\stackrel{‾}{c}}^{*}$$\stackrel{‾}{c}$
And...
$|\stackrel{‾}{b}×\stackrel{‾}{c}|=bc$
$|\stackrel{‾}{a}\cdot \stackrel{‾}{b}×\stackrel{‾}{c}|=abc=V$
$|{\stackrel{‾}{a}}^{*}|=|\frac{\stackrel{‾}{b}×\stackrel{‾}{c}}{\stackrel{‾}{a}\cdot \stackrel{‾}{b}×\stackrel{‾}{c}}|=\frac{bc}{abc}=\frac{1}{a}$
${a}^{*}=\frac{1}{a}$, ${b}^{*}=\frac{1}{b}$, ${c}^{*}=\frac{1}{c}$
...hence “reciprocal.”

### 6.5 Reciprocal Lattice Vector

The reciprocal lattice vector extends in reciprocal space from origin 000 to the lattice point $hkl$ and is perpendicular to the corresponding $\left(hkl\right)$ planes. The set of all such $hkl$ points corresponds to the reciprocal lattice. These vectors have units of 1/length, so a larger vector indicates smaller spacing between planes.
${r}_{hkl}^{*}=h{\stackrel{\to }{a}}^{*}+k{\stackrel{\to }{b}}^{*}+l{\stackrel{\to }{c}}^{*}$
Note: Hammond - ${r}_{hkl}^{*}={d}_{hkl}^{*}$ Cullity - ${r}_{hkl}^{*}={H}_{hkl}$
The following examples show applications of reciprocal lattice vectors in determining inter-planar relationships.
1. Find inter-planar angle $\phi$ between planes$\left({h}_{1}{k}_{1}{l}_{1}\right)$and $\left({h}_{2}{k}_{2}{l}_{2}\right)$
i.e.find angle between reciprocal lattice vectors ${\stackrel{\to }{r}}_{1}^{*}$& ${\stackrel{\to }{r}}_{2}^{*}.$
$cos\left(\phi \right)=\frac{{\stackrel{‾}{r}}_{1}^{*}\cdot {\stackrel{‾}{r}}_{2}^{*}}{{r}_{1}^{*}{r}_{2}^{*}}$ , where ${r}_{n}^{*}=|{\stackrel{\to }{r}}_{n}^{*}|=\sqrt{{\stackrel{‾}{r}}_{n}^{*}\cdot {\stackrel{‾}{r}}_{n}^{*}}$
${\stackrel{‾}{r}}_{1}^{*}\cdot {\stackrel{‾}{r}}_{2}^{*}=\left({h}_{1}\stackrel{‾}{a}*+{k}_{1}\stackrel{‾}{b}*+{l}_{1}\stackrel{‾}{c}*\right)\cdot \left({h}_{2}\stackrel{‾}{a}*+{k}_{2}\stackrel{‾}{b}*+{l}_{2}\stackrel{‾}{c}*\right)$
$={h}_{1}{h}_{2}a{*}^{2}+{k}_{1}{k}_{2}b{*}^{2}+{l}_{1}{l}_{2}c{*}^{2}+\left({h}_{1}{k}_{2}+{k}_{1}{h}_{2}\right)\stackrel{‾}{a}*\cdot \phantom{\rule{6px}{0ex}}\stackrel{‾}{b}*+\left({h}_{1}{l}_{2}+{l}_{1}{h}_{2}\right)\stackrel{‾}{a}*\cdot \phantom{\rule{6px}{0ex}}\stackrel{‾}{c}*+\left({k}_{1}{l}_{2}+{l}_{1}{k}_{2}\right)\stackrel{‾}{b}*\cdot \phantom{\rule{6px}{0ex}}\stackrel{‾}{c}*$
Note: $\stackrel{‾}{a}*\cdot \phantom{\rule{6px}{0ex}}\stackrel{‾}{b}*=a*b*cos\left(\gamma *\right),$etc.
Example:
Find angle between (111) and (110) in cube?
$cos\left(\phi \right)=\frac{{\stackrel{‾}{r}}_{111}^{*}\cdot \phantom{\rule{6px}{0ex}}{\stackrel{‾}{r}}_{110}^{*}}{{r}_{111}^{*}{r}_{110}^{*}}=\frac{\left(1+1+0\right){a}^{*2}}{\sqrt{3}{a}^{*}\sqrt{2}{a}^{*}}=\frac{2}{\sqrt{6}}⇒\phi =35.26{}^{○}$
Figure 6.18: Angle between (111) and (110) in cube
2. Find the zone axis for two intersecting planes
(${h}_{1}{k}_{1}{l}_{1}\right)$and $\left({h}_{2}{l}_{2}{k}_{2}\right)$
Miller indices: If you need a refresher on Miller indices, the Wikipedia page (https://en.wikipedia.org/wiki/Miller_index) is very helpful. Here we include a reminder of the notation used to indicate planes and directions.
• Specific planes: We use round brackets - $\left(hk\ell \right)$
• Class of planes: To indicate all planes that are crystallographically identical, we use curly brackets - $\left\{hk\ell \right\}$
• Specific direction: We use square brackets - $\left[hk\ell \right]$
• Class of directions: To indicate all directions that are crystallographically identical, we use angle brackets - $⟨hk\ell ⟩$
$\stackrel{‾}{z}=\left[uvw\right]=u\stackrel{‾}{a}+v\stackrel{‾}{b}+w\stackrel{‾}{c}$
Since ${\stackrel{‾}{r}}_{hkl}*\perp \left(hkl\right)$plane, ${\stackrel{‾}{r}}_{hkl}*\perp \stackrel{‾}{z}$ for any (hkl) belonging to zone $\stackrel{‾}{z}$
$\stackrel{‾}{z}\cdot \phantom{\rule{6px}{0ex}}\stackrel{‾}{{r}_{1}}*=u{h}_{1}+v{k}_{1}+w{l}_{1}=0$
$\stackrel{‾}{z}\cdot \phantom{\rule{6px}{0ex}}\stackrel{‾}{{r}_{2}}*=u{h}_{2}+v{k}_{2}+w{l}_{2}=0$
Solve simultaneous equations for u, v, w
$u={k}_{1}{l}_{2}-{l}_{1}{k}_{2}$
$v={l}_{1}{h}_{2}-{h}_{1}{l}_{2}$
$w={h}_{1}{k}_{2}-{k}_{1}{h}_{2}$
2 equations and 3 unknowns $\to$uvw determined to within a scale factor
Example:
What is the zone axis (or edge formed by) for $\left(\stackrel{‾}{1}10$) and (010)?
$u=\left(1\right)\left(0\right)-\left(0\right)\left(1\right)=0$
$v=\left(0\right)\left(0\right)-\left(-1\right)\left(0\right)=0$
$w=\left(-1\right)\left(1\right)-\left(1\right)\left(0\right)=-1$
$\stackrel{‾}{z}=\left[uvw\right]=\left[00\stackrel{‾}{1}\right]$or $\left[001\right]=\stackrel{‾}{c}-axis$
Equivalent to doing vector cross product:
${\stackrel{‾}{r}}_{\stackrel{‾}{1}10}^{*}×\phantom{\rule{6px}{0ex}}{\stackrel{‾}{r}}_{010}^{*}$
Figure 6.19: Zone axis for (
3. Find d-spacing ${d}_{hkl}$
Use $|{\stackrel{‾}{r}}_{hkl}^{*}|={r}_{hkl}^{*}=\frac{1}{{d}_{hkl}}$
${d}_{hkl}=$$\perp$distance between 2 ‖ planes with intercept differences = $\frac{a}{h},\frac{b}{k},\frac{c}{l}$
${r}_{hkl}^{*}=\sqrt{{\stackrel{‾}{r}}_{hkl}^{*}\cdot \phantom{\rule{6px}{0ex}}{\stackrel{‾}{r}}_{hkl}^{*}}$
${r}_{hkl}^{*2}={h}^{2}a{*}^{2}+{k}^{2}b{*}^{2}+{l}^{2}c+2hk\stackrel{‾}{a}*\cdot \phantom{\rule{6px}{0ex}}\stackrel{‾}{b}*+2hl\stackrel{‾}{a}*\cdot \phantom{\rule{6px}{0ex}}\stackrel{‾}{c}*+2kl\stackrel{‾}{b}*\cdot \phantom{\rule{6px}{0ex}}\stackrel{‾}{c}*$
For $\alpha =\beta =\gamma =90{}^{○}$ , $2hk\stackrel{‾}{a}*\cdot \phantom{\rule{6px}{0ex}}\stackrel{‾}{b}*+2hl\stackrel{‾}{a}*\cdot \phantom{\rule{6px}{0ex}}\stackrel{‾}{c}*+2kl\stackrel{‾}{b}*\cdot \phantom{\rule{6px}{0ex}}\stackrel{‾}{c}*=0$
Figure 6.20: ${d}_{100}$ spacing
Note: $\stackrel{‾}{a}*\cdot \phantom{\rule{6px}{0ex}}\stackrel{‾}{b}*=a*b*cos\left(\gamma *\right)$
For Orthorhombic - ${d}_{hkl}=\frac{1}{{r}_{hkl}*}=\frac{1}{\left(\frac{{h}^{2}}{{a}^{2}}+\frac{{k}^{2}}{{b}^{2}}+\frac{{l}^{2}}{{c}^{2}}{\right)}^{1/2}}$
For Cubic - ${d}_{hkl}=\frac{a}{\left({h}^{2}+{k}^{2}+{l}^{2}\right)}$
Monoclinic
($\alpha =\beta =90{}^{○}\right)$ $a=4Å,\phantom{\rule{6px}{0ex}}b=3Å,\phantom{\rule{6px}{0ex}}c=6Å,\phantom{\rule{6px}{0ex}}\gamma =70{}^{○}$
${\stackrel{\to }{r}}_{100}^{*}={a}^{*}=\frac{1}{{d}_{100}}\ne \frac{1}{4\stackrel{˚}{A}}$   ${d}_{100}=a{}^{*-1}$  ${a}^{*}=?\ne \frac{1}{a}$ since $\gamma \ne 90{}^{○}$
Use ${\stackrel{‾}{a}}^{*}\cdot a=1\to a\cdot acos\left(\frac{\pi }{2}-\gamma \right)=1$
${a}^{*}=\frac{1}{asin\left(\gamma \right)}=\frac{0.250}{0.94}=0.266{\stackrel{˚}{A}}^{-1}$  $\therefore {d}_{100}=asin\left(\gamma \right)=3.759\stackrel{˚}{A}\ne 4\stackrel{˚}{A}$
Likewise, ${b}^{*}=\frac{1}{bsin\left(\gamma \right)}=\frac{0.333}{0.94}=0.355{\stackrel{˚}{A}}^{-1}$   $\therefore {d}_{010}=bsin\left(\gamma \right)=2.819\stackrel{˚}{A}$.
Figure 6.21: Monoclinic ${d}_{100}$spacing. Green lines are edge on view of (100) planes. Red lines are for (010) planes
Note: For monoclinic, $\frac{a*}{b*}=\frac{b}{a}$$c*=\frac{1}{c}=\frac{1}{6}{\stackrel{˚}{A}}^{-1}$ since $\stackrel{‾}{c}\parallel \stackrel{‾}{c}*$
Figure 6.22: Direct vs Reciprocal Space for Ortho-P. Both lattices are Ortho-P with inverted aspect ratio. Namely, $\frac{{a}^{*}}{{b}^{*}}=\frac{b}{a}$
Figure 6.23: Direct vs Reciprocal Space for Monoclinic-P
4. Cell axis transformations (primitive $\to$non-primitive)
e.g. Rhombohedral $\left(\alpha =\beta =\gamma =60{}^{○}\right)$ $\to$FCC
Figure 6.24: Cell axis transformation
Let old rhombohedral unit cell axes be ${\stackrel{‾}{a}}_{0},{\stackrel{‾}{b}}_{0},{\stackrel{‾}{c}}_{0}$
The new FCC unit cell axes ${\stackrel{‾}{a}}_{n},{\stackrel{‾}{b}}_{n},{\stackrel{‾}{c}}_{n}$are related to the old by:
${\stackrel{‾}{a}}_{n}={u}_{1}{\stackrel{‾}{a}}_{0}+{v}_{1}{\stackrel{‾}{b}}_{0}+{w}_{1}{\stackrel{‾}{c}}_{0}$
${\stackrel{‾}{b}}_{n}={u}_{2}{\stackrel{‾}{a}}_{0}+{v}_{2}{\stackrel{‾}{b}}_{0}+{w}_{2}{\stackrel{‾}{c}}_{0}$
${\stackrel{‾}{c}}_{n}={u}_{3}{\stackrel{‾}{a}}_{0}+{v}_{3}{\stackrel{‾}{b}}_{0}+{w}_{2}{\stackrel{‾}{c}}_{0}$
or:
$\left(\begin{array}{c}{\stackrel{‾}{a}}_{n}\\ {\stackrel{‾}{b}}_{n}\\ {\stackrel{‾}{c}}_{n}\end{array}\right)=\left(\begin{array}{ccc}{u}_{1}& {v}_{1}& {w}_{1}\\ {u}_{2}& {v}_{2}& {w}_{2}\\ {u}_{3}& {v}_{3}& {w}_{3}\end{array}\right)\left(\begin{array}{c}{\stackrel{‾}{a}}_{0}\\ {\stackrel{‾}{b}}_{0}\\ {\stackrel{‾}{c}}_{0}\end{array}\right)$
Matrix of transformation
For rhombohedral to FCC example:
${\stackrel{‾}{a}}_{n}={\stackrel{‾}{a}}_{0}+{\stackrel{‾}{b}}_{0}-{\stackrel{‾}{c}}_{0}$
${\stackrel{‾}{b}}_{n}=-{\stackrel{‾}{a}}_{0}+{\stackrel{‾}{b}}_{0}+{\stackrel{‾}{c}}_{0}$
${\stackrel{‾}{c}}_{n}={\stackrel{‾}{a}}_{0}-{\stackrel{‾}{b}}_{0}+{\stackrel{‾}{c}}_{0}$
Transformation Matrix: $\left(\begin{array}{ccc}1& 1& -1\\ -1& 1& 1\\ 1& -1& 1\end{array}\right)$
Modulus of Transformation:
Determinant $|\begin{array}{ccc}1& 1& -1\\ -1& 1& 1\\ 1& -1& 1\end{array}|=4$ $\to {V}_{n}=4{V}_{0}$ ( The FCC unit cell has four times the volume of the rhombohedral unit cell)
For transforming $\left({h}_{0}{k}_{0}{l}_{0}\right)$$←\to$$\left({h}_{n}{k}_{n}{l}_{n}\right)$
Since the inter-planar spacing and direction are the same, the planes have the same reciprocal lattice vector, ${\stackrel{‾}{r}}_{{h}_{0}{k}_{0}{l}_{0}}^{*}={\stackrel{‾}{r}}_{{h}_{n}{k}_{n}{l}_{n}}^{*}.$
$\left({h}_{0}{\stackrel{‾}{a}}_{0}^{*}+{k}_{0}{\stackrel{‾}{b}}_{0}^{*}+{l}_{0}{\stackrel{‾}{c}}_{0}^{*}=\left({h}_{n}{\stackrel{‾}{a}}_{n}^{*}+{k}_{0}{\stackrel{‾}{b}}_{n}^{*}+{l}_{n}{\stackrel{‾}{c}}_{n}^{*}\right)$
take dot product with ${\stackrel{‾}{a}}_{n}={u}_{1}{\stackrel{‾}{a}}_{0}+{v}_{1}{\stackrel{‾}{b}}_{0}+{w}_{1}{\stackrel{‾}{c}}_{0}$
$\left($${u}_{1}{\stackrel{‾}{a}}_{0}+{v}_{1}{\stackrel{‾}{b}}_{0}+{w}_{1}{\stackrel{‾}{c}}_{0}\right)\cdot \phantom{\rule{6px}{0ex}}\left({h}_{n}{\stackrel{‾}{a}}_{n}^{*}+{k}_{0}{\stackrel{‾}{b}}_{n}^{*}+{l}_{n}{\stackrel{‾}{c}}_{n}^{*}\right)={h}_{n}$
$\begin{array}{c}{h}_{n}={u}_{1}{h}_{0}+{v}_{1}{k}_{0}+{w}_{1}{l}_{0}\\ {k}_{n}={u}_{2}{h}_{0}+{v}_{2}{k}_{0}+{w}_{2}{l}_{0}\\ {l}_{n}={u}_{3}{h}_{0}+{v}_{3}{k}_{0}+{w}_{3}{l}_{0}\end{array}\to \left(\begin{array}{c}{h}_{n}\\ {k}_{n}\\ {l}_{n}\end{array}\right)=\left(\begin{array}{ccc}{u}_{1}& {v}_{1}& {w}_{1}\\ {u}_{2}& {v}_{2}& {w}_{2}\\ {u}_{3}& {v}_{3}& {w}_{3}\end{array}\right)\left(\begin{array}{c}{h}_{0}\\ {k}_{0}\\ {l}_{0}\end{array}\right)$
Same matrix of transformation used for axes and indices.
Old Rhombohedral $\to$New FCC example
$\left(\begin{array}{c}{h}_{n}\\ {k}_{n}\\ {l}_{n}\end{array}\right)=\left(\begin{array}{ccc}1& 1& -1\\ -1& 1& 1\\ 1& -1& 1\end{array}\right)\left(\begin{array}{c}{h}_{0}\\ {k}_{0}\\ {l}_{0}\end{array}\right)$
$\begin{array}{c}{h}_{n}={h}_{0}+{k}_{0}-{l}_{0}\\ {k}_{n}=-{h}_{0}+{k}_{0}+{l}_{0}\\ {l}_{n}={h}_{0}-{k}_{0}+{l}_{0}\end{array}$
Note that ${h}_{n}+{k}_{n}=2{k}_{0}$, or an even integer and ${k}_{n}+{l}_{n}=2{l}_{0}$, also an even integer.
Therefore,${h}_{n},{k}_{n},{l}_{n}$must all be even or all odd “unmixed.”
Figure 6.25: Rhombohedral indexed (110) plane in an FCC lattice
Transformation of the (110) plane in the rhombohedral structure to its (hkl) in the face-centered cubic structure.
$\left(110{\right)}_{R}=\left(???{\right)}_{FCC}$ $\left(\begin{array}{c}{h}_{n}\\ {k}_{n}\\ {l}_{n}\end{array}\right)=\left(\begin{array}{ccc}1& 1& -1\\ -1& 1& 1\\ 1& -1& 1\end{array}\right)\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)=\left(200\right)$
In Figure 6.26 below, a non-primitive orthorhombic-c unit cell is illustrated, with the c axis pointing out of the page. The transformation of its lattice in real space to reciprocal space is shown in Figure 6.27. Notice that the a and b axes only reach halfway to the first lattice point in their respective directions. The reason for this is due to the organization of lattice planes in the a and b directions of the real space. For instance, going vertically in the a direction of Figure 6.26, the first set of planes appears halfway between the lattice spacing. The planes that follow in this a direction appear at the same spacing. Thus, this set of $\left(200\right)$ planes within the Orthorhombic-c unit cell, can represent the fundamental period of planes in the a direction. The same analysis can be done in the b direction, yielding the $\left(020\right)$ fundamental period of planes. From this, the first point in the reciprocal space appears at 200 in the ${\stackrel{\to }{a}}^{*}$direction and 020 in the ${\stackrel{\to }{b}}^{*}$ direction. Note that in the c direction, which is not shown, the first plane is found at $\left(001\right)$, due to the base-centered structure.
Figure 6.26: Orthorhombic-C
Note that the original, real lattice and the reciprocal lattice have a similar shape, with an inverted aspect ratio in the reciprocal lattice. Furthermore, they are both base-centered and have a similar primitive unit cell, which is monoclinic. The missing base-centered reciprocal lattice points (e.g. 010, 100, 210, etc.) are due to the use of a non-primitive unit cell in real space for the transformation. Had the primitive cell been used, all allowed reciprocal lattice points, corresponding to allowed $\left(hkl\right)$ Bragg diffraction planes, would be present. Anytime a nonprimative unit cell is used to describe the lattice, there will be missing $hkl$ reciprocal lattice points.
A two dimensional projection of the FCC unit cell on the a and b axes is illustrated in Figure 6.28 below. The solid circles represent the lattice points in the plane of the paper, and the circles with the $\frac{1}{2}$ label represent lattice points half a lattice spacing above the plane. The dashed lines indicate the set of (110) planes within this unit cell. The transformation from the FCC to its reciprocal lattice generates a BCC structure as shown in Figure 6.29, and vice versa. Note that the (100) planes are missing, since the fundamental period of planes is (200) in the a direction. This reasoning can be applied to other missing reciprocal space points, using the fundamental period of planes in the real FCC lattice.
Figure 6.28: FCC unit cell lattice points

## 7 Representative Crystal Structures

### 7.1 Crystal Structure Examples

The combination of the previously discussed point and translation symmetry elements with the fourteen Bravais lattices give rise to 230 possible three-dimensional space groups. For these three-dimensional crystal structures, an identical atom or a fixed arrangement of atoms (ions, molecules) is found at each Bravais lattice point.
The following examples show a one atom per lattice point structure. Another example which is not shown below is the simple cubic structure.
Ex.: BCC Metals $\alpha -Fe,$Cr, Mo, V, etc.
Ex.: FCC Metals $\gamma -$Fe, Cu, Pb, Ni, etc.

### 7.2 The hexagonal close packed structure, unlike the examples in Figures 7.1 and 7.2, has two atoms per lattice point. These two atoms are considered to be the motif, or repeating object within the HCP lattice.

Ex.: HCP (hexagonal close-packed)
Figure 7.3 shows four different representations of a hexagonal close-packed structure with two atoms per unit cell - one atom at 000, and one atom at $\frac{2}{3}\frac{1}{3}\frac{1}{2}$or $\frac{1}{3}\frac{2}{3}\frac{1}{2}$. The atoms in Figure 7.3(b) correspond to the nine atoms illustrated in Figure 7.3(a). The association of two atoms with a single lattice point is represent by the dashed lines connecting the pairs. Furthermore, (c) and (d) show more of a hexagonal representation of the HCP atomic structure with the stacking of atomic planes in a periodic arrangement. Some elements with this structure include zinc, magnesium, and beryllium.
Figure 7.3: HCP structure
The FCC crystal structure is as close-packed as the HCP structure. Though its relation to the HCP structure is not immediately apparent, Figure 7.4 demonstrates the stacking of FCC (111) planes in a hexagonal pattern, similar to the (0002) HCP planes stacking. One distinction between the two is that the HCP layer stacking occurs in an ABAB pattern in the $\stackrel{‾}{c}$ or [001] direction, and the layer stacking for FCC occurs in an ABCABC pattern in the [111] directions. This is illustrated in Figures 7.5 and 7.7 below.
Figure 7.4: Face-centered cubic planes
Figure 7.6: Hexagonal close-packed planes
The following are other examples of crystal structures with two atoms per lattice point. The first example, in Figure 7.8, is the diamond crystal structure that is seen in other atoms besides carbon, such as silicon and germanium. In the examples following Figure 7.8, compounds consisting of two unlike atoms follow unique characteristics of arrangement. For instance, if a compound's structure consists of atoms A and B, the structure of A atoms must possess the same symmetry elements of the crystal as a whole. Namely, a symmetry operation characteristic of the entire crystal, when performed on a given A atom, must bring it into coincidence with another A atom. Furthermore, body-centering, face-centering, and base-centering translations, if present, should begin and end on like atoms. An example of this is detailed above figure 7.11.
Diamond Structure
ex. C, Si, Ge
4C at 000 + face-center translations (fct)
4C at $\frac{1}{4}\frac{1}{4}\frac{1}{4}+fct$
Figure 7.8: Diamond
Zinc Blend
The face-centered cubic zinc blend structure can be likened to the diamond structure in Figure 7.8 above, but the atoms at 000+fct (S) are now different from those at the second set of positions - $\frac{1}{4}\frac{1}{4}\frac{1}{4}+fct$ (Zn).
Ex. GaAs, ZnS
4S at 000+fct
4Zn at $\frac{1}{4}\frac{1}{4}\frac{1}{4}+fct$
Figure 7.9: Zinc Blend
The CsCl structure illustrated in Figure 7.10 below is common to compounds such as CsBr, NiAl, ordered $\beta -brass,$ and ordered CuPd. Though it appears to be a body-centered cubic structure at first glance, the body centering translation of $\frac{1}{2}\frac{1}{2}\frac{1}{2}$ connects two unlike atoms, rather than the required self-coincidence of a single atomic element. Therefore, the structure is simple cubic, and one may think of the Cs atom at 000 and the $C{l}^{-}$atom at $\frac{1}{2}\frac{1}{2}\frac{1}{2}$ as being related to the single lattice point at $000.$
In the NaCl or rock salt crystal structure of Figure 7.11 below, the sodium atoms are face-centered. When face centering translations are applied to the chlorine atom at $\left(\frac{1}{2}\frac{1}{2}\frac{1}{2}\right)$ , all chlorine atom positions are mapped. In addition, a $9{0}^{○}$rotation about the four fold rotation axis at [010] brings the chlorine atoms into self coincidence.
Figure 7.10: CsCl structure is Cubic-P

### 7.3 Voids in FCC

The FCC structure has two types of voids, namely octahedral and tetrahedral voids, which naturally arise from the arrangement of atoms in its crystal lattice.
Octahedral voids are located at $\frac{1}{2}\frac{1}{2}\frac{1}{2}$+fct in the FCC structure. There is one located at the center of the unit cell, and one at the center of each of the unit cell edges. Since there are twelve edges and each edge is shared with four other unit cells, this yields three edge octahedral voids per unit cell in addition to the one at the center. Thus, there are a total of four octahedral voids per unit cell. The space labeled “A” in Figure 7.12 is an octahedral hole in the structure, which is surrounded by eight sulfur atoms. In Figure 7.11, note that $C{l}^{-}$is NaCl octahedrally surrounded by $N{a}^{+}$.
The Zn atom at $\frac{3}{4}\frac{3}{4}\frac{1}{4},$ marked “B” in Figure7.12, is surrounded by four S atoms at the corners of a tetrahedron. This is known as a tetrahedral void in the structure, as the Zn atom is surrounded by a tetrahedral structure of sulfur atoms. In the zinc blend structure as well as other FCC structured crystals, there are a total of eight octahedral voids per unit cell. For the zinc blend structure, however,only half of these tetrahedral sites are occupied by Zn atoms, which is illustrated more clearly in Figure 7.9.
Figure 7.12: Portion of the zinc blend structure

### 7.4 Atom Sizes and Coordination

Ex.: CsCl
Structure - Cubic P (not BCC)
$C{s}^{+}$at 000 $\to$CN=8
$C{l}^{-}$at $\frac{1}{2}\frac{1}{2}\frac{1}{2}$ $\to$ CN=8
Figure 7.13: CsCl
Using the hard sphere model for atoms, the placement of atoms in the unit cell of a given structure, as well as known atomic radii, the lattice constant of a crystal structure may be determined.
The following procedure demonstrates the evaluation lattice constant “$a$” in CsCl using known ionic radii:
${r}_{C{s}^{+}}=1.69\stackrel{˚}{A}$${r}_{C{l}^{-}}=1.81\stackrel{˚}{A}$
Body diagonal of cube = $\sqrt{3}a=2{r}_{C{s}^{+}}+2{r}_{C{l}^{-}}$
$a=\frac{2}{\sqrt{3}}\left({r}_{C{s}^{+}}+{r}_{C{l}^{-}}\right)=4.04\stackrel{˚}{A}$
Note that the nearest neighbors experience an attractive force due to their opposite charges and that the second nearest neighbors experience a repulsive force due to their like charges. Therefore, if the two ionic sizes become more disparate than CsCl, then the packing changes to an NaCl structure.
NaCl structure:
4 $N{a}^{+}$at 000 + fct   CN=6 (octahedral surroundings)
4 $C{l}^{-}$at $\frac{1}{2}\frac{1}{2}\frac{1}{2}$+ fct  CN=6 (octahedral surroundings)
fct $\equiv$face-center translations
$\left\{\left(000\right),\left(\frac{1}{2}\frac{1}{2}0\right),\left(\frac{1}{2}0\frac{1}{2}\right),\left(0\frac{1}{2}\frac{1}{2}\right)\right\}$
${r}_{N{a}^{+}}=0.95\stackrel{˚}{A}$${r}_{C{l}^{-}}=1.81\stackrel{˚}{A}$
If the NaCl structure were like “CsCl” structure,
then $a=\frac{2}{\sqrt{3}}\left({r}_{N{a}^{+}}+{r}_{C{l}^{-}}\right)=3.19\stackrel{˚}{A}$
Therefore, the NaCl CN=8 structure would not bond since the $C{l}^{-}$-$C{l}^{-}$distance $a=3.19\stackrel{˚}{A}<2{r}_{C{l}^{-}}=3.62\stackrel{˚}{A}$
In other words, for the CN=8 CsCl structure, the distance of the required lattice parameter would be greater than the available distance of two radii of the larger ions that make up the lattice parameter “a”. Thus, there is a critical or minimum radius ratio for which cation–anion contact is established in the CsCl structure.
$a>2{r}_{>},$ where ${r}_{>}=$radius of larger ion & ${r}_{<}=$radius of smaller ion
$\therefore Limit$ $a=2{r}_{>}\to \frac{2}{\sqrt{3}}\left({r}_{<}+{r}_{>}\right)=2{r}_{>}$
Radius Ratio $\equiv \frac{{r}_{<}}{{r}_{>}}=\sqrt{3}-1=0.732$
This radius ratio rule is different for different crystal structures, as shown in Table 8 below. Deviation from the radius ratio rule indicates a shift from ionic to covalent bonding.
Ex. Zinc blend structure: ${r}_{Z{n}^{2+}}=0.6\stackrel{˚}{A}$${r}_{{S}^{2-}}=1.80\stackrel{˚}{A}$ $\to$ $\frac{{r}_{<}}{{r}_{>}}<0.33$
Figure 7.14: Close-packed structure
What about HCP, FCC close-packed structures? For the HCP structures, the coordination number is 12. An ionic crystal cannot have this coordination number because there would be an equal number of anions and cations surrounding each ion.

## 8 Introduction to Diffraction

The experimental technique of X-ray diffraction in crystals, since early applications at the start of the 20th century, has been an important method for mapping crystal structures. The following section introduces the basic, key concepts of X-ray diffraction such as the nature of X-rays and Bragg's Law.

### 8.1 X-ray

An X-ray is an electromagnetic wave similar to light, but of a much shorter wavelength. As illustrated in Figure 8.1 below, a beam of X-rays of a single wavelength traveling in the same direction (i.e. $\stackrel{\to }{k}$) has an electric field $\left(\stackrel{‾}{E}$) perpendicular to its propagation direction and a magnetic field $\left(\stackrel{‾}{H}$) perpendicular to its electric field. A plane-polarized wave is one in which the electric field is confined to the a plane, such as the x-y plane, as the wave travels along the x direction, labeled the Poynting vector in Figure 8.1. The magnetic field is not considered in the following applications.
Figure 8.1: X-ray
The following equation models the movement of the electric field as a transverse plane wave traveling in the x-direction, with an amplitude of ${\stackrel{‾}{E}}_{0}$ in the and a time varying $\stackrel{‾}{E}$ in the y direction:
$E={E}_{0}cos\left[\frac{2\pi }{\gamma }\left(x-vt\right)\right]$ $\gamma$ = wavelength,$v$=velocity traveling in x-direction
Let ${\stackrel{‾}{E}}_{0}={E}_{0}\stackrel{ˆ}{j}$
After a time, t, has elapsed the wave has been phase shifted to the right by $x\text{'}=vt$.

### 8.2 Interference

Interference occurs when two or more coherent, or same wavelength, waves are superimposed. If two interfering waves are in phase, then constructive interference occurs, as illustrated in Figure 8.2. Waves out of phase, as shown in Figure 8.3, may produce a resultant wave wit
Figure 8.2: Constructive interference
(1800) Young's two-slit experiment
Young's two slit experiment, as illustrated in Figure 8.4 below, is a classic example of the interference of light waves and resulting diffraction patterns.
Figure 8.4: Young's two-slit experiment
In Young's experiment, a plane light wave of wavelength $\gamma$ traveling to the right with an “in-phase” wave front parallel to the grating produces two coherent circular waves emanating from the 2 slits at A and B. The two waves will add up in phase at point P on a screen, if the path length difference, $s=BD$ , is a multiple of wavelength $\gamma$.
$s=dsin\theta =n\gamma$ $\to$ interference maxima at y
To produce interference fringe pattern, of dark destructive interference and bright constructive interference spots, $d\approx \gamma$
If $d<\gamma ,\phantom{\rule{6px}{0ex}}sin\theta =\frac{n\gamma }{d}>1$$\to$ impossible
If $d>>\gamma ,\phantom{\rule{6px}{0ex}}sin\theta =\frac{\gamma }{d}$ $\to$ too small to separate n=1 max. from direct beam

### 8.3 X-ray Diffraction History

In 1912, Max von Laue, a young Assistant Professor in Munich, was inspired by PhD thesis work of Ewald who was working in Sommerfeld Lab at the University of Munich. Ewald's thesis concerned solving for a mathematical physics relationship of the scattering of waves from a periodic array of dipole oscillators. Laue realized that Ewald's discovery directly related to scattering of waves from a crystal's periodic structure, and collaborated with Friedrich and Rongten's (discoverer of X-rays) student Knipping to explore this idea. Their experiment demonstrated an X-ray diffraction pattern from a crystal, and proved that X-rays were waves with wavelength of the order of a d-spacing within a crystal.
Later in 1912, W.L. Bragg, excited by the Laue discovery, worked with his father to develop a simpler picture to explain X-ray diffraction.

### 8.4 How does X-ray diffraction work?

First, consider reflection from a plane of atoms, as represented by a horizontal line in Figure 8.5. Note that the atoms do not have to evenly spaced within the plane.
Figure 8.5: Reflection from a plane of atoms
A plane-wave is represented by rays 1 & 2 in-phase at AC in Figure 8.5, and also reflecting in phase . The path length difference of rays 1-1' and rays 2-2' are equal when $AD=BC$, and by trigonometry, when the incident angle is equal to the outgoing angle.
Since path lengths will be equal for all rays under the specular reflection condition, the outgoing angle will equal the angle of incidence. This is known as the Law of Reflection.
$AD=CB$ $\to {\theta }_{i}={\theta }_{o}$
Law of Reflection
Scattering from two planes
Second, consider the reflection from a set of equally spaced atom planes as illustrated in 8.6.
Figure 8.6: Reflection from a set of equally spaced atom planes
In this case, scattering occurs in both planes. Here, we would like to know when the scattering from both planes will result in constructive interference. If the path difference, $\delta$=BCD of rays 1 and 2 is equal to the length equal to an integer multiple of the wavelength of the X-rays ($n\gamma$ ,where n is an integer) then the scattered rays 1' & 2' are in-phase at yy.' If the angle of incidence were tilted down slightly, then the waves would no longer be in phase, and the conditions would no longer allow for constructive interference. Applying the geometric relationship below yield's Bragg's law. This states the essential condition necessary for diffraction to occur.
$BC=CD=dsin\theta$
Bragg's Law: $n\gamma =2dsin\theta$
$n$- order of reflection, number of wavelengths in the path difference between rays scattered by adjacent planes
If $\theta$is slightly different from the Bragg condition, each successive ray 1',2',3',...will be slightly phase shifted with respect to previous ray as illustrated in Figure 8.7. The amplitudes of the waves would sum to zero since every plane will have another plane at some depth which will scatter out of phase by $18{0}^{○}.$
Figure 8.7: Successive phase shifts in reflected rays
What about the “n” in $n\lambda =2dsin\theta$?
The “n” value , or order of reflection, may take on any integral value as long as sin$\theta$ does not exceed one in the Bragg's law equation. For fixed values of d and $\lambda$, various n values (n=1,2,3...) correspond to different angles of incidence ($\theta ={\theta }_{1},{\theta }_{2},{\theta }_{3}...\right)$ at which diffraction may occur.
For a first-order reflection (n=1) the path length and phase difference of the scattered X-rays 1' and 2' in Figure 8.6 would differ by one wavelength. For a second-order reflection (n=2) the path length and phase difference of the scattered rays would differ by two wavelengths. Thus, the rays scattered by all the planes of atoms are completely in phase, leading to constructive interference.
1st order n=1 $\to$$\delta =\lambda$ between consecutive planes spaced by “d”
2nd order n = 2 $\to$$\delta =2\lambda$between consecutive planes spaced by “d”
Another way of looking at the reflection order:
$\lambda =2\frac{d}{n}sin\theta$$\frac{d}{n}=\frac{{d}_{hkl}}{n}={d}_{nhnknl}$
where hkl corresponds to the 1st order or fundamental reflection planes. Thus, the Bragg equation may be rearranged in the following form, in which the “n” term is dropped and the ${d}_{hkl}$ does not need to be the lowest set of indices for a given plane (i.e. ${d}_{hkl}$=110,220,etc.):
$\lambda =2{d}_{hkl}sin\theta$or $\lambda =2dsin{\theta }_{hkl}$
If the diffracted intensity is plotted as a function of sin$\theta$, peaks corresponding to crystallographic planes would be visible as shown in Figure 8.8.
Figure 8.8: Diffracted intensity
Figure 8.9: NaCl diffraction
The (111) diffraction condition ($\theta ={\theta }_{111}\right)$ allows for constructive interference from the 1' and 3' waves and the 2' and 4' waves , as their path length difference measures exactly one wavelength. Note that rays 1'&3' from $N{a}^{+}$are in phase with each other, but $18{0}^{○}$out of phase with rays 2'&4' from $C{l}^{-}.$
For $\theta ={\theta }_{222},$ the overall path length becomes longer and all rays 1',2',3',4' from all planes $N{a}^{+}$and $C{l}^{-}$are in phase. This leads to constructive interference and higher intensity peaks shown in Figure 8.10. Note that this is a simplified model, which solely accounts for the path length difference effect in scattering from different planes which are not identical to each other. Other effects will be discussed later in the text.
Figure 8.10: NaCl diffraction peaks
Observing the Bragg Angles tells us about unit cell dimensions, namely the d-spacing between planes which can be used to derive the lattice constant. Furthermore, observing relative intensities as shown in the NaCl structure for different ${\theta }_{hkl}$ tells us about the structure within the unit cell. Namely, that there were a pair of ions ($N{a}^{+}$and $C{l}^{-}$) associated with an FCC lattice point.
On the other hand, in a simple FCC metal (111),(333), etc., would not have been weak.
Hint: Look for weak and absent reflections

### 8.5 Absent (or forbidden) Reflections

Figure below shows a two dimensional projection side view of (100) planes in a simple FCC metal. The white circles represent atoms that are half of a lattice spacing out of the plane.
Figure 8.11: FCC (100)
The plane of atoms half way between the planes will scatter at $18{0}^{○}$out of phase, canceling out the waves reflecting at the 100 planes. Therefore, the $\theta ={\theta }_{100}$ Bragg diffraction condition leads to destructive interference, and the (100) reflection is absent or not allowed. For simple cubic, the atoms halfway between the planes would not be present.
However, all (200) planes scatter in phase and the (200) reflection would be observed. This reinforces the fact that in an FCC structure, the following conditions are true:
• $\left(200\right),\left(400\right),\left(600\right),$etc. allowed.
• $\left(100\right),\left(300\right),\left(500\right),$etc. forbidden
• $\therefore$hkl all even or hkl all odd (unmixed) for FCC
How would Bragg diffraction show that a crystal is cubic? i.e., not orthorhombic
Realize that for the cubic structure the following relationship between the d-spacing of the 110 and the 100 planes is true: $\frac{sin{\theta }_{100}}{sin{\theta }_{110}}=\frac{{d}_{110}}{{d}_{100}}=\frac{1}{\sqrt{2}}$ .
By determining the reflection angle for the (100) and (110) planes and then taking the ratio of the sines, the cubic structure can be confirmed if it is equal to $\frac{1}{\sqrt{2}}$.
This same procedure may be used to determine if a crystal structure is FCC. Since (110) and (100) are forbidden for FCC, then (220) and (200) can be used instead.

## 9 Absorption/Emission

### 9.1 Spectrometers

The following three X-ray sources will be discussed in this text:
• X-ray tube e.g., Cu target
• Radioactive Source e.g., $F{e}^{55}$
• Synchrotron e.g., ${e}^{-}$storage ring
An X-ray tube must contain three vital components, namely a source of electrons, a high accelerating voltage to move these electrons, and a metal target on which these highly-accelerated electrons will collide. Figure 9.1(a) below illustrates the cross section of an X-ray filament tube. This consists of an evacuated glass casing that insulates the anode at one end from the cathode at another. The cathode is a tungsten filament and the anode is a water-cooled block of copper which also contains the desired target metal at one end. A lead of the high-voltage transformer, detailed in Figure 9.1(b) is connected to the filament and the other is connected to ground. The target is grounded by the cooling water connection. In operation, the filament is heated with a current and emits electrons that are drawn to the target. Around the filament is a metal focusing cup which is kept at a high negative voltage close to that of the filament and tends to focus the electrons on a focal spot on the target. X-rays are emitted from this focal spot in all directions and escape through the window in the tube usually made from beryllium.
(a) (b)
Figure 9.1: X-ray tube
Figure 9.2, below illustrates an X-ray radiation spectrum, plotting X-ray intensity as a function of energy.

Bremsstrahlung radiation is a continuous X-ray spectrum from X-ray tube, which is the smooth curve in the intensity plot in Figure 9.2. This occurs as a result of the deceleration, or braking of electrons in the target. The accelerating (or decelerating) of any charge q, an electron in this case, generates electromagnetic radiation.
Figure 9.3 illustrates a charge q, with acceleration in the $\stackrel{‾}{a}$ direction. If you look in the $\alpha$ direction relative to its acceleration vector, you will have electromagnetic radiation at point P whose distance from charged particle q is indicated by $\stackrel{‾}{r}.$ The electric field $\stackrel{‾}{E}$ of this radiation is pointing in the transverse direction to the radius $\stackrel{‾}{r}$. The following formula yields the magnitude of the electric field generated by the accelerated charge:
$|\stackrel{‾}{E}|=\frac{qasin\alpha }{4\pi {\epsilon }_{0}{c}^{2}r}$The following relationships for the electric field also hold true:
• $\stackrel{‾}{E}\perp \stackrel{‾}{r}$ (transverse wave)
• $\stackrel{‾}{E}\perp \left(\stackrel{‾}{a}×\stackrel{‾}{r}\right)$
Figure 9.3: Radiation from an accelerated charge
In the case of X-ray tube Bremsstrahlung radiation, the electrons accelerate (negatively) from the tungsten filament cathode to the grounded anode. Electrons accelerating in the negative $\stackrel{‾}{z}$ direction as illustrated in Figure 9.4 create an azimuthally symmetric dipole radiation field. In Figure 9.4, the angle $\theta$ is the complement of the angle $\alpha$ in Figure 9.3. Note that the magnitude of $\stackrel{‾}{r}$ does not change, and $\theta$ is the factor that affects the intensity of the generated $\stackrel{‾}{E}$ field. Because the intensity is proportional to the magnitude of the electric field squared, it is also proportional to ${sin}^{2}\alpha$ or ${cos}^{2}\theta .$ This is consistent with what is observed in Figure 9.4 and the three-dimensional representation in Figure 9.5, since the intensity at a minimum when when $\theta$ is $9{0}^{○}$ and at a maximum when $\theta$ goes to ${0}^{○}$.
Figure 9.4: Intensity of $\stackrel{‾}{E}$ field with varying $\theta$
$sin\alpha =cos\theta$
$I\propto |\stackrel{‾}{E}|\propto {cos}^{2}\theta$
Max intensity $\perp \stackrel{ˆ}{z}$

Synchrotron radiation occurs when electrons are accelerated in a curved path. The electron experiences centripetal acceleration in the radial direction, producing the dipole radiation pattern is of the same toroid shape as illustrated in Figures 9.4 and 9.5. The minimum intensity is observed in the radial direction and maximized perpendicular. However, there is a high relativistic effect due to the electron's speed, such that the dipole radiation field is distorted in the lab frame of reference creating an extremely narrow cone of radiation in the forward direction of the electron's tangential acceleration.
Figure 9.6: (A) Relativistic effect; (B) Extremely narrow forward directed cone of radiation
Recall that Bremsstrahlung radiation is white, or polychromatic, radiation resulting from the deceleration of electrons following inelastic collisions with target atoms. The kinetic energy analysis of this process is as follows:

The maximum energy loss is the kinetic energy (KE) of the electron at the vacuum target interface. This is the case in which the electron loses all of its energy from one inelastic collision.
$\Delta {E}_{Max}=K{E}_{{e}^{-}}=\frac{1}{2}m{v}^{2}=hv=\frac{hc}{\lambda }$
The maximum energy of the photon, by the de Broglie relationship $\frac{hc}{\lambda }$, is directly related to the wavelength of the X-ray tube. Note that this is a maximum case, and the range of energies will be anywhere from this maximum value to zero.This maximum kinetic energy is therefore directly related to the set potential of the X-ray tube, as shown below.
KE [keV] = V [kV]
Typically, If X-ray tube high voltage, HV = 50kV $\to$Max photon energy = 50keV
The maximum photon energy in the continuous Bremsstrahlung spectrum will show 50kV at its maximum along with lower values.
de Broglie wavelength relationship: $\lambda =\frac{hc}{E}$$\lambda \left[\stackrel{˚}{A}\right]=\frac{12.40}{E\left[keV\right]}$$\to$short wavelength limit ${\lambda }_{SWL}\left[\stackrel{˚}{A}\right]=\frac{12.40}{V\left[kV\right]}$
Therefore, in the Bremsstrahlung spectrum there will be a continuum of wavelengths showing an intensity pattern as well as a short wavelength limit (SWL) corresponding to the maximum available energy. This is due to the fact that the wavelength and energy are inversely proportional.
Figure 9.7 below shows a radiation spectrum from a copper X-ray tube, for various set potentials (25kV, 20kV, 15kV, etc.). The smooth curve indicates the continuous or Bremsstrahlung radiation curve, with the short wavelength limit (SWL) corresponding to the leftmost point on a given curve. The spikes are indicative of characteristic radiation, but
Note: for V=20kV $\to$${\lambda }_{SWL}=0.62\stackrel{˚}{A}$
The integrated intensity, or area under the curves in Figure 9.7 is proportional to the voltage of the X-ray cubed multiplied by the the atomic number of the target and the current through the filament. Typically the the voltage will be increased rather than the current, as this increases the intensity by three orders of magnitude.
$I$$\propto$${V}^{3}{Z}_{target}i$
The following is a comparison of the spectra from different synchrotons on a log scale. Notice the Bremsstrahlung spectrum located near the bottom of the plot.
Figure 9.8: On-Axis Brilliance vs Photon Energy (KeV)

### 9.4 Characteristic Spectrum

Raising the voltage on an X-ray tube above a certain critical value for a given target metal yields sharp, narrow intensity peaks, or fluorescent lines superimposed on the continuous radiation spectrum. The characteristic spectrum is comprised of several of these fluorescent lines (i.e. K,L,M,etc.) that are specific to the target in use. In order for fluorescent lines from the target to occur, the electron beam must have enough kinetic energy to overcome the binding energy of an inner core electron and knock it out of the target. The transition of a higher energy electron to fill this hole leads to the production of an X-ray which is characteristic of the given target element. Figure 9.10 illustrates these fluorescent peaks on a continuous background radiation spectrum for a copper target sample at 30 kV.
For fluorescence: $K{E}_{{e}^{-}}=eV\ge {W}_{K}$(or$\ge {W}_{L}$)
where $K{E}_{{e}^{-}}$is electron energy & ${W}_{k}\phantom{\rule{6px}{0ex}}or\phantom{\rule{6px}{0ex}}{W}_{L}$is the binding energy of target electrons.
Typically, the K fluorescence lines comprised of $K{\alpha }_{1}$, $K{\alpha }_{2}$ and $K{\beta }_{1}$ are useful in X-ray diffraction. The $K{\alpha }_{1}$ and $K{\alpha }_{2}$ lines, referred to as the $K\alpha$ doublet , often have very close wavelengths and are treated as a single line. In addition, the subscript in the $K{\beta }_{1}$ is usually dropped. The K indicates the principal quantum number of an electron which corresponds to its energy level in the electron cloud. A useful diagram of the electron energy states for copper is shown in Figure 9.10, with different principal energy levels (n=1=K and n=2=L) followed by an integer number that indicates the sub-shell within a given energy level (i.e. L1, L2, etc.). When an electron in the K shell is knocked out , the atom is left in a high-energy state. The target metal atom becomes grounded when a higher energy electron, such as one from the L shell or the M shell, drops down to fill the K electron-hole. Filling a K electron with an L-shell electron is more probable than filling it with M-shell electron, therefore yielding a higher $K\alpha$ (corresponding to the L shell transition) intensity peak relative to the $K\beta$ (corresponding to the M shell transition) intensity peak. Note that the K radiation has a definite wavelength for a given target atom and that it is impossible to excite one K fluorescence line without also exciting the other.
Figure 9.9: Cu characteristic spectrum at 30 kV. Note that the short wavelength limit is at 0.4 due to this 30 kV condition.
In Figure 9.9 above, note that the intensity of the $K\alpha$ peak is one hundred times stronger than the background radiation curve. Furthermore, the $K\alpha$ doublet differ in intensity by a factor of two, and the $K{\alpha }_{1}$and $K{\beta }_{1}$ differ by approximately a factor of five.
$\frac{{I}_{K\alpha }}{{I}_{Cont}}\approx 100$
$\frac{{I}_{K{\alpha }_{1}}}{{I}_{K{\alpha }_{2}}}=2$$\frac{{I}_{K{\alpha }_{1}}}{{I}_{K{\beta }_{1}}}\approxeq 5$
The arrows in Figure 9.10 below demonstrate the $K{\alpha }_{1}$$K{\alpha }_{2}$and $K{\beta }_{1}$ transitions, with their lengths representing the difference in the energies of the two states. This energy difference also equals the energy of the outgoing fluorescent photon. Note that an arrow does not go from the 1s to the 2s electron energy state. Because the electron originally in the 1s state had an angular momentum of 1, the angular momentum of the photon produced must equal 1 by its conservation law. The photon's angular momentum is derived from the difference in the angular momentum of the two electron energy states before and after the transition. The 1s and 2s states do not differ in angular momentum by 1, thus the transition is not favored.
Figure 9.10: Copper energy transitions
Note that in Figure 9.10 there are twice the number of electrons in the L3 state than there are in the L2 state. Due to this, the $K{\alpha }_{1}$electron transition from the L3 energy level to K energy level is twice as likely to occur as the $K{\alpha }_{2}$ transition from the L2 energy state to the K energy state. This is consistent with the double intensity of the $K{\alpha }_{1}$ fluorescence peak relative to the $K{\alpha }_{2}$ fluorescence peak shown above.
The following calculations determine the characteristic $K\alpha$ and $K\beta$ fluorescence energies for a copper target and their corresponding wavelengths:
${E}_{\gamma }$ $\lambda$
${E}_{K{\alpha }_{1}}={W}_{K}-{W}_{{L}_{3}}=8.05keV,\phantom{\rule{0ex}{0ex}}1.540\stackrel{˚}{A}$
${E}_{K{\alpha }_{2}}={W}_{K}-{W}_{{L}_{2}}=8.03keV,\phantom{\rule{0ex}{0ex}}1.544\stackrel{˚}{A}$
${E}_{K{\beta }_{1}}={W}_{K}-{W}_{{M}_{3}}=8.19keV,\phantom{\rule{0ex}{0ex}}1.392\stackrel{˚}{A}$
($\lambda =\frac{hc}{E}\to \lambda \left[\stackrel{˚}{A}\right]=\frac{12.4}{E\left[keV\right]}$)

### 9.5 Auger Effect

The Auger effect is a non-radiative process that competes with the radiative X-ray emission process to fill an inner electron hole created by the accelerated-electron bombardment of the target. Rather than producing a photon, this process ejects a valence energy level electron known as the Auger electron. An energy level diagram for Auger electron emission from a copper atom is illustrated in Figure 9.11. This process involves three discrete electron energy levels: the first is the K energy level of the electron hole, the second is the L1 or higher energy level from which an electron drops down to fill the K-hole, and the last is the L3 valence energy level from which the Auger electron is emitted. The energy released from the transition of the L1 level electron the lower energy level of the K-hole is transmitted to the L3 valence shell Auger electron, allowing for its ejection from the atom. The kinetic energy of the Auger electron, therefore, is equal to the difference between its binding energy and the energy difference between the L1 and K energy levels involved in the filling of the electron hole.
Note that the creation of the K hole leads an L1 electron to drop down and fill the hole as shown by the arrow indicating the movement of the hole from the K to L1 energy level. Following this, the energy is transmitted to the L3 valence electron which becomes the ejected Auger electron, as shown by its arrow pointing outside of the electron's energy levels. The energy of this ejected L3 electron is equal to the binding energy difference between the K and L1 energy levels and the binding energy of the L3 that must be overcome for its emission from the copper atom.
Example:
Copper K ${L}_{1}$${L}_{2}$Auger electron emission follows creation of k-hole
• K-hole is created in copper atom via accelerated electron bombardment
• L1 electron drops down to fill K hole, creating useful energy
• Useful transition energy is transferred to L3 electron, overcomes the binding energy and ejects the L3 electron from the atom
Energy analysis, where ${E}_{e}$ is the energy of the Auger electron:
${E}_{e}={W}_{K}-{W}_{L1}-{W}_{L3}$
Figure 9.11: Cu K ${L}_{1}$${L}_{3}$Auger
${E}_{e}=8.98-1.10-0.93=6.95keV$
Fluorescence yield: ${\omega }_{k}$= probability for K hole being filled by radiative process
K Auger Probability = 1-${\omega }_{k}$
The “K” line in Figure 9.12 below shows this relationship, in which a higher atomic number corresponds to a higher fluorescence yield probability for filling a K electron hole. The L-hole electron fluorescence probability is also shown in this diagram as the line labeled “L.”
Figure 9.12: Fluorescence yield vs atomic number
Note that because electrons cannot through air very easily like X-rays, the Auger spectrometer is located inside a vacuum chamber. Furthermore, it is much more surface sensitive than X-ray spectrometers.

$p+{e}^{-}\to n+u$ (neutron + neutrino)
$F{e}^{55}+{e}^{-}\to \phantom{\rule{6px}{0ex}}\left(M{n}^{55}\right)$ with K hole
$↙↓↘$
$\left(M{n}^{55}\right)$ with ${L}_{3}$hole + $K{\alpha }_{1}$fluorescence
Example:
$F{e}^{55}$Electron Capture
Figure 9.13: $K\alpha$ fluorescence in manganese following $F{e}^{55}$ electron capture
${E}_{\gamma }={W}_{K}-{W}_{L3}=5.90keV$ $←K{\alpha }_{1}$
The $C{d}^{109}$ isotope of cadmium undergoes a similar process, whereby the electron capture leads to its conversion into $A{g}^{109}$ and characteristic fluorescence lines of Ag may be observed on the radiation spectrum.
$C{d}^{109}+{e}^{-}\to \left(A{g}^{109}\right)$

### 9.7 X-ray Absorption

${I}_{absorbed}={I}_{o}-{I}_{x}$
${I}_{o}$= incident intensity
${I}_{x}$= transmitted intensity
The constant of proportionality relating the absorbed intensity to the distance traversed is$\mu$, or the linear absorption coefficient. The plot in Figure 9 below shows this exponential relationship, in which the incident intensity ${I}_{0}$ decays at a rate of $\mu$ times the distance $x$ through the material.
Differential form: $\frac{-dI}{I}=\mu dx$ ,$\phantom{\rule{0ex}{0ex}}\mu$= linear absorption coefficient
${\int }_{{I}_{0}}^{{I}_{x}}\frac{dI}{I}=-{\int }_{0}^{x}\mu dx\to ln\left({I}_{x}\right)-ln\left({I}_{0}\right)=-\mu x$
Exponential Relationship:
${I}_{x}={I}_{0}{e}^{-\mu x}$
Figure 9.15: Transmitted intensity vs thickness traversed
$\mu \propto \rho \to \frac{\mu }{\rho }$ constant of material, independent of state (solid,liquid,gas)
$\frac{\mu }{\rho }=f\left(Z,{E}_{\gamma }\right)\equiv$mass absorption coefficient
${\mu }^{-1}$= absorption length
For more complicated structures such as chemical compounds or mixtures, the mass absorption coefficient of the substance is the weighted average of the mass absorption coefficient of its constituent elements. A general formula for this is shown below. The mass absorption coefficient for the mixture of air is calculated in Table 9, using tabulated mass absorption coefficients of nitrogen, oxygen, and argon along with their respective weight percents in air.
$\frac{\mu }{\rho }=$${\sum }_{i=1}^{n}{w}_{i}\left(\frac{\mu }{\rho }{\right)}_{i}$ ${w}_{i}$=weight fraction
Mass absorption coefficient for variable composition
Example:
Air=76% ${N}_{2}$ +23% ${O}_{2}$+1.3% Ar
 i ${w}_{1}$ $\left(\frac{\mu }{\rho }\right)$ $\frac{c{m}^{2}}{g}$ ${w}_{i}×\left(\frac{\mu }{\rho }{\right)}_{i}$ ${N}_{2}$ 0.76 7.14 5.53 ${O}_{2}$ 0.23 11.0 2.53 Ar 0.013 120.0 1.56 (C&S Appendix 8) 9.52$c{m}^{2}/g$=$\left(\frac{\mu }{\rho }{\right)}_{Air}$
@ STP ${\rho }_{Air}=0.0013g/c{m}^{3}$ ,${\mu }_{Air}=0.0124c{m}^{-1}$,$\frac{1}{\mu }=81cm$
Table 9: X-ray absorption in Air at ${E}_{\gamma }$=8.05 keV (Cu $K\alpha$)
From these calculations, it is determined that a copper $K\alpha$ (8.05 keV) X-ray beam is attenuated to $\frac{1}{e}=0.37=\frac{{I}_{x}}{{I}_{0}}$after traveling through 81 cm of air.
Figures 9.16 and 9.17 below plot the linear absorption coefficients and X-ray transmission for beryllium as a function of the photon energy. Recall that a beryllium window is used in an X-ray filament tube.
Figure 9.16: Beryllium linear absorption coefficient

### 9.8 X-ray Transmission Filters

Figure 9.18: Ni foil filter and Cu X-ray tube
Transmission filters make use of the X-ray absorption edge of a particular element. In the second plot of Figure 9.19 below, the absorption edge for nickel can be observed. This discontinuity corresponds to the energy required to eject an inner core electron from an atom of the material. At energies just below this edge the $\frac{\mu }{\rho }$ mass absorption coefficient is significantly less than the mass absorption coefficient at energies slightly above the edge. This translates into higher transmission for x-rays with energies below the absorption edge line on the energy axis. By this process the initial x-ray spectrum, plotted as ${I}_{o}$ at the top of the figure, undergoes significant reduction in the intensity of the $K\beta$ line as shown in the final plot of transmitted intensity, ${I}_{x}$. Note that the high energy background radiation is also significantly reduced in the final plot.
Figure 9.19: Ni transmission filter
Recall the X-ray transmission intensity formula, ${I}_{x}={I}_{0}{e}^{-\mu x}$.
For an 8 mil Ni Filter, (1mil=0.001”=25.4$\mu m$) the intensity of the $K\alpha$ fluorescence is five hundred times greater than the $K\beta$ fluorescence. Recall that without a filter, the $K\alpha$ transmitted intensity was five times greater than the$K\beta$ intensity; thus, the nickel filter only transmits $\frac{1}{100}$ of the original $K\beta$ X-ray intensity.
8 mil (0.8 mils = $20\phantom{\rule{6px}{0ex}}\mu m$) Ni Filter:
$x=20\phantom{\rule{6px}{0ex}}\mu m$ $\frac{I\left(K\alpha \right)}{I\left(K\beta \right)}=500$
In general, a filter with an atomic number that is one less than the target metal in the X-ray tube is used and the thickness X is chosen such that $\frac{{I}_{K\alpha }}{{I}_{K\beta }}=\frac{1}{500}$. This Z-1 relationship is utilized so that the absorption edge lies just above the desired $K\alpha$ and below the $K\beta$ to be filtered. The mass-absorption law ${I}_{x}={I}_{0}{e}^{-\left(\mu /\rho \right)\rho x}$ may used to calculate the thickness of the filter.
Generally, use a $Z-1$ filter:
Cu$\to$Ni Filter
Ag$\to$Pd Filter
Mo$\to$Zr $←$(Z-2)
A thicker filter will better suppress the $K\beta$ component, but this will also result in unwanted suppression of the $K\alpha$ component. Thus, it is useful to follow the $\frac{1}{500}$ intensity ratio.
Figure 9.20 below illustrates a general absorption spectra for an atom, by plotting the log of the mass absorption coefficient against increasing energy. Note the presence of the absorption edge previously discussed, which is labeled as the the “K-edge.” In addition to the K-edge, other discontinuity spikes corresponding to the L electron shell binding energies are present. The location of the K-edge, and its associated K-branch, further to the right on the energy axis is due to the fact that this inner electron shell has the highest binding energy in a given atomic element.
The overall absorption, labeled ${\left(\frac{\mu }{\rho }\right)}_{tot}$ is equal to the addition of the absorption from the different electron energy level branches, i.e. ${\left(\frac{\mu }{\rho }\right)}_{K},{\left(\frac{\mu }{\rho }\right)}_{{L}_{1}},$etc. Note that a single increase in the energy unit or in the atomic number have a cube effect on the mass absorption coefficient. For instance, if the X-ray energy is increased from 10keV to 20keV, the x-ray absorption will decrease by one eighth. Furthermore, if the atomic species is increased by two from, say, nickel to beryllium, then eight times the absorption occurs. Since the mass absorption coefficient $\left(\frac{\mu }{\rho }\right)$ is inversely proportional to the cube of the energy, it follows that it the wavelength of the X-ray cubed is directly proportional to the mass absorption coefficient. These relationships are summarized in Figure 9.20 below.
Figure 9.20: X-ray absorption spectra

### 9.9 Photoelectric effect

Figure 9.21: Photoelectric effect
In order for photoelectric absorption to occur, the photon's energy must be greater than the binding energy of the K-shell electron. The energy of this emitted photoelectron is given by the difference of the incident photon's energy, $h\nu$ and the binding energy overcome in the electrons emission. This is followed by an X-ray emission, allowing for the atom to return to its ground state.
${E}_{{e}^{-}}=hv-{W}_{K}$
In addition to conservation of energy, the angular momentum in this process is conserved. That is, the incident photon's intrinsic angular momentum of one and the 1s electrons angular momentum of one are combined into a dipole angular distribution in the emitted photoelectron. That is, there is a high probability that the emitted electron will be on along the z-axis perpendicular to the incident direction of the photon, and a minimal probability of its emission in the x and y direction.
Dipole selection rule for electronic transition:$\Delta l=±1$

### 9.10 X ray scattered by an electron

Figure 9.23: Thomson Scattering
The oscillating $\stackrel{‾}{E}$ field of the incoming X-ray is given by the equation: ${\stackrel{‾}{E}}_{0}\left(t\right)={\stackrel{‾}{E}}_{0}cos\left[2\pi \left(x-vt\right)/\lambda \right].$
The force on this electron is given by its charge multiplied by the magnitude of the incident electric field.
$\stackrel{‾}{F}=-e\stackrel{‾}{E}={m}_{e}\stackrel{‾}{a}$
Therefore, the acceleration of the electron is given by the following equation:
$\stackrel{‾}{a}\left(t\right)=\frac{-e\stackrel{‾}{{E}_{0}}\left(t\right)}{{m}_{e}}$
This oscillating electron emits electromagnetic radiation in all directions, or a scattered wave. The wavelength of this scattered wave ${\lambda }_{s}$ is equal to the wavelength of the incident X-ray wave ${\lambda }_{0}.$Therefore, this type of scattering is coherent, or elastic.
Oscillating E-field ${\stackrel{‾}{E}}_{0}\left(t\right)={\stackrel{‾}{E}}_{0}cos\left[2\pi \left(x-vt\right)/\lambda \right]$
$↓$
Oscillating Charge Particle $\left({e}^{-}\right)$
$↓$
E-M Radiation (oscillating fields) ${\lambda }_{s}={\lambda }_{0}$

### 9.11 Theory for Radiation Generating by Accelerated Charge

The electric field generated by an accelerated charge $q$ at a given position $\stackrel{‾}{R}$ away from the charge is illustrated in Figure 9.24 below. The acceleration vector is indicated as $\stackrel{‾}{a}$ in the vertical direction, with the $\stackrel{‾}{R}$ position being at angle $\alpha$ relative to the charge $q$'s acceleration vector.
Figure 9.24: E-field at $\stackrel{‾}{R}$
1. Direction of the electric field: $\stackrel{‾}{E}\perp \stackrel{‾}{R}$ and in-plane of $\stackrel{‾}{R}$ and $\stackrel{‾}{a}$
2. Magnitude of the electric field: $|\stackrel{‾}{E}|=\frac{1}{4\pi {\epsilon }_{0}}\frac{qasin\alpha }{{c}^{2}R}=\frac{1}{4\pi {\epsilon }_{0}}\frac{{e}^{2}}{{m}_{e}{c}^{2}}\frac{{E}_{0}sin\alpha }{R}$ Note that the electric field gets weaker as you increase the distance $\stackrel{‾}{R}$ from the charged particle.
$|\stackrel{‾}{E}|={r}_{e}\frac{{E}_{0}sin\alpha }{R}$ ${r}_{e}=2.818×1{0}^{-5}\stackrel{˚}{A}$ Classical ${e}^{-}$radius
${r}_{e}=\frac{1}{4\pi {\epsilon }_{0}}\frac{{e}^{2}}{{m}_{e}{c}^{2}}$
The classical Thomson scattering equation, as shown below, gives the intensity of the scattered x-rays from the electron as the square of the magnitude of the electric field generated by the oscillating electron. Note that the intensity is a maximum in the direction perpendicular to the acceleration $\alpha =90{}^{○}.$
${I}_{e}\propto |\stackrel{‾}{E}{|}^{2}={E}_{0}^{2}\frac{{r}_{e}^{2}}{{R}^{2}}si{n}^{2}\alpha$
$\frac{{I}_{e}}{{I}_{0}}=\frac{{r}_{e}^{2}}{{R}^{2}}{sin}^{2}\alpha$
${I}_{e}\equiv$Intensity of scattered X-rays from ${e}^{-}$at $\stackrel{‾}{R}$
${I}_{0}=$Intensity of incident X-rays
$I=\frac{c}{4\pi }|\stackrel{‾}{E}{|}^{2}\equiv \left[energy\cdot are{a}^{-1}\cdot tim{e}^{-1}\right]$
Why ignore Thomson scattering from the (charged) nucleus?
Relative to the scattering of electrons, the scattering intensity from protons in the nucleus of an atom is negligible, as shown in the following ratio:
$\frac{{I}_{p}}{{I}_{e}}=\left(\frac{{m}_{e}}{{m}_{p}}{\right)}^{2}=3×1{0}^{-7}$

## 10 Scattering

### 10.1 X-ray Scattering & Polarization

Consider an incident X-ray traveling in $\stackrel{ˆ}{x}$direction and scattered from an electron $"{e}^{-}$” whose position is at some arbitrary origin “O” as shown in Figure 10.1 below. The incident X-ray, $\stackrel{ˆ}{{s}_{0}}$, has an electric field that is polarized in some direction on the y-z plane. We can therefore represent this electric field ${\stackrel{‾}{E}}_{0}$ of the incident X-ray in two vector components - one in the y-direction $\left({\stackrel{‾}{E}}_{0y}\right)$ and one in the z-direction $\left({\stackrel{‾}{E}}_{0z}\right)$. Note that the angle $2\theta$ is the complement of the angle $\alpha$ used in the Thomson scattering equation and illustrated in Figure 9.24.
${\stackrel{‾}{E}}_{0}={E}_{oy}\stackrel{ˆ}{y}+{E}_{oz}\stackrel{ˆ}{z}$
Electric field of incident X-ray traveling in the $\stackrel{ˆ}{x}$ direction
Figure 10.1: Scattered X-ray
For an unpolarized incident x-ray from an x-ray tube, the electric field is taken to be evenly distributed in the $\stackrel{ˆ}{y}$ direction and in the $\stackrel{ˆ}{z}$ direction, thus each vector component is equated to half of the total electric field. Recall that the square of the electric field vectors measures the amplitude of the wave, which is proportional to the intensity of the wave. It follows that the intensity in the $\stackrel{ˆ}{y}$ direction is equal to the intensity in the $\stackrel{ˆ}{z}$ direction and half of the total intensity.
${E}_{oy}^{2}={E}_{oz}^{2}=\frac{1}{2}{E}_{o}^{2}$ i.e. ${I}_{oy}={I}_{oz}=\frac{1}{2}{I}_{o}$
The following procedure to determines the scattered intensity ${I}_{e}$ from a given electron at the position P in Figure 10.1 above, which is a distance$\stackrel{‾}{r}$ away from the electron in the x-z plane. If the electric field is in the $\stackrel{ˆ}{y}$ direction and perpendicular to the scattering plane (the scattering plane being x-z in Figure 10.1):
${\stackrel{‾}{E}}_{0}={E}_{o}\stackrel{ˆ}{y}$ ($\sigma$ polarization case, ${\stackrel{‾}{E}}_{0}\perp$scattering plane)
Here, $\alpha =\frac{\pi }{2}$, thus yielding maximum intensity as expected by the ${sin}^{2}\alpha$ term in the classical Thomson scattering equation.
$\begin{array}{cc}{I}_{e}^{\perp }={I}_{o}\frac{{r}_{e}^{2}}{{R}^{2}}& \left(10.1\right)\end{array}$
If the electric field is in the $\stackrel{ˆ}{z}$ direction and parallel to the x-z scattering plane:
Here, $\alpha =\frac{\pi }{2}-2\theta$, yielding an intensity which varies depending on the $2\theta$ term.
${I}_{e}^{\parallel }={I}_{0}\frac{{r}_{e}^{2}}{{R}^{2}}{cos}^{2}2\theta$
Figure 10.2 below plots this ${cos}^{2}2\theta$ relationship with a varying $2\theta$ value that is present in the intensity function for the $\pi$ polarization case.
Figure 10.2: Scattered intensity -$\pi$polarization case
For unpolarized e.g. characteristic radiation from an X-ray tube, half of the X-rays are $\sigma$ polarized and half are $\pi$ polarized. To obtain the total scattered intensity at P, sum the intensities of the $\sigma$ polarized and $\pi$ polarized components.
${I}_{e}=\frac{1}{2}{I}_{e}^{\perp }+\frac{1}{2}{I}_{e}^{\parallel }={I}_{0}\frac{{r}_{e}^{2}}{{R}^{2}}\left(\frac{1+{cos}^{2}2\theta }{2}\right)$
$↑$polarization factor
Note that this relationship is independent of $\lambda$ and dependent on $\frac{1}{{R}^{2}}$.
At $2\theta =0$ (forward scattering) and R=1cm
$\frac{{I}_{e}}{{I}_{0}}=\frac{{r}_{e}^{2}}{{R}^{2}}=\left(2.818×1{0}^{-13}{\right)}^{2}=7.94×1{0}^{-26}←$ very weak, a small fraction of the incident beam's intensity
At $2\theta =\frac{\pi }{2}$
$\frac{{I}_{e}}{{I}_{0}}=\frac{1}{2}\left(\frac{{I}_{e}}{{I}_{0}}{\right)}_{2\theta =0}$only $\sigma$polarization X-rays are scattered in $2\theta =\frac{\pi }{2}$ direction.

#### 10.1.1 Barkla

British researcher Charles Barkla performed a double scattering experiment to prove that scattering at $\frac{\pi }{2}$ produces a polarized beam. In this experiment, as shown in Figure 10.3 below, unpolarized light interacts with a charged particle at O, and a scattered beam OP is polarized in the $\stackrel{ˆ}{y}$ direction. The polarized beam is incident on the particle at P, and the scattered beam from P may be detected at a point Q in the x'y' plane. The intensity of this scattered beam at Q is measured as a function of the $\phi$ direction angle, and varies as ${cos}^{2}\phi$ as shown in Figure 10.4. The intensity pattern of this final scattered beam matches the intensity pattern observed when the incident ray scattered is polarized, thus proving that $\frac{\pi }{2}$ scattering produces a polarized beam. Essentially, in this experiment we have an x-ray polarizer at O and a x-ray polarimeter at Q.
Figure 10.3: Barkla's experiment

### 10.2 Thomson Scattering by two electrons

Recall the intensity of a scattered wave from a free electron:
${I}_{e}={I}_{0}\frac{{r}_{e}^{2}}{{R}^{2}}\left(\frac{1+{cos}^{2}2\theta }{2}\right)$ unpolarized X-ray beam
${r}_{e}=2.818×1{0}^{-13}cm$
Note: $\lambda$independent & $\frac{1}{{R}^{2}}$dependent
As shown in Figure 10.5 below, we can define the following parameters:
${\stackrel{‾}{S}}_{0}\equiv$incident direction unit vector
$\stackrel{‾}{S}\equiv$scattered direction unit vector
$2\theta =$scattering angle
Figure 10.5: Define ${\stackrel{‾}{S}}_{0},\stackrel{‾}{S},2\theta$
Figure 10.6: Scattering from two free electrons
The two scattered rays will have a phase difference $\delta =\left({x}_{1}+{x}_{2}\right)\frac{2\pi }{\lambda }$
Where ${x}_{1}=\stackrel{‾}{r}\cdot \stackrel{‾}{{S}_{0}}$ , and ${x}_{2}=\stackrel{‾}{r}\cdot \stackrel{‾}{S}$
Equivalently, $\delta =-\frac{2\pi }{\lambda }\left(\stackrel{‾}{S}-\stackrel{‾}{{S}_{0}\right)}\stackrel{‾}{r}$
Note that in the forward scattering direction, as shown in Figure 10.7 below, the angle $2\theta$ is 0, so the scattered direction is equal to incident direction and there is no phase difference.
Figure 10.7: Forward scattering direction
The scattering waves from each electron at a detector located a distance R in $\stackrel{‾}{S}$ direction are represented by the following equations:
${E}_{1}\left(R,t\right)={E}_{1}cos\left[\frac{2}{\pi }R-\omega t\right]={E}_{1}cos\phi$
${E}_{2}\left(R,t\right)={E}_{2}cos\left[\frac{2\pi R}{\lambda }+\delta -\omega t\right]={E}_{2}cos\left(\phi +\delta \right)$
Note that the second electron's scattered wave has a phase shift $\delta$ added into its cosine function.
Since the detector is at a distance which is much greater than the distance between the two electrons, the amplitude of the electric fields scattered from both electrons are taken to be equal. Furthermore, the cosine factors of the equations may be condensed using Euler's formula.
${E}_{2}={E}_{1}←R\gg r$
use ${e}^{i\phi }=cos\phi +isin\phi$
${e}^{i\left(\phi +\delta \right)}={e}^{i\phi }{e}^{i\delta }$
At the detector, the total field is equal to the sum of the scattered field amplitudes from each electron. Note that Euler's formula is used in place of the cosine terms in the first line below. The intensity is determined by dotting the total field, labeled $\epsilon$ below, by its complex conjugate ${\epsilon }^{*}.$ Recall that the scattered intensity from one electron was simply ${E}^{2}.$ Now, this ${E}^{2}$ term includes a cosine term that accounts for the fact that the electrons are at two different positions and a certain $2\theta$ angle that can lead to scattering waves at different phases. Notice that if $\delta =0,2\pi$ or any even multiple of $\pi$ then, the scattered waves are in phase and four times the intensity is observed. On the other hand, if $\delta$ is an odd multiple of $\pi ,$ then the scattered waves are out of phase, and no intensity is detected. Of course, there are instances in between, such as $\delta =\frac{\pi }{2}$, in which the intensity is in between the two extremes due to interference effects in the scattered waves.

$\epsilon \left(R,t\right)={E}_{1}{e}^{i\phi }+{E}_{2}{e}^{i\phi }{e}^{i\delta }={E}_{1}{e}^{i\phi }\left(1+{e}^{i\delta }\right)$
$I\propto {\epsilon }^{2}=|\epsilon {|}^{2}=\epsilon \cdot {\epsilon }^{*}={E}_{1}^{2}\left(1+{e}^{i\delta }\right)\left(1+{e}^{-i\delta }\right)={E}_{1}^{2}\left(2+{e}^{i\delta }+{e}^{-i\delta }\right)=2{E}_{1}^{2}\left(1+cos\delta \right)$
${\epsilon }^{2}=4{E}_{1}^{2}$ for $\delta =0,2\pi ,...$
${\epsilon }^{2}=0$ for $\delta =\pi ,3\pi ,...$
${\epsilon }^{2}=2{E}_{1}^{2}$for $\delta =±\frac{\pi }{2},...$
$\to {x}_{1}+{x}_{2}=-\stackrel{‾}{r}\cdot \left(\stackrel{‾}{S}-{\stackrel{‾}{S}}_{o}\right)=n\lambda$

#### 10.2.1 Review of Thomson scattering from one and two electrons

Thomson scattering from one free electron
${I}_{e}={I}_{0}\frac{{r}_{e}^{2}}{{R}^{2}}\left(\frac{1+co{s}^{2}2\theta }{2}\right)$
The phase of scattered wave 2 is relative to scattered wave 1.
Figure 10.8: Scattering from two free electrons
${\delta }_{2}=\left({x}_{1}+{x}_{2}\right)\frac{2\pi }{\lambda }=\frac{-2\pi }{\lambda }\left(\stackrel{‾}{S}-{\stackrel{‾}{S}}_{0}\right)\stackrel{‾}{r}$
The total field is the sum of the field generated by each of the two electrons.
${\stackrel{‾}{\epsilon }}_{Total}={\stackrel{‾}{\epsilon }}_{1}+{\stackrel{‾}{\epsilon }}_{2}={e}^{i\phi }\left({\stackrel{‾}{E}}_{1}{e}^{i{\delta }_{1}}+{\stackrel{‾}{E}}_{2}{e}^{i{\delta }_{2}}\right)$
where $\phi =\frac{2\pi R}{\lambda }-\omega t$
The phase difference of scattered wave 1 is zero:${\delta }_{1}=0$
Where R is the distance of the detector and r is the distance between the electrons, $R\gg r$$\to$${E}_{2}={E}_{1}={E}_{e}$
In general, for N free electrons, the total electric field generated is the sum of the scattered fields of all of the electrons. This is based on the superposition principle, in which the sum of the individual forces yields an equivalent total force. The detector “sees” the sum of all of these individual electric force fields, having equal amplitude but individual phase factors resulting from varying positions and $2\theta$ angles. Recall that only in the forward direction will this phase difference equal zero.
${\stackrel{‾}{\epsilon }}_{tot}={\sum }_{n=1}^{N}{\stackrel{‾}{\epsilon }}_{n}={\stackrel{‾}{E}}_{e}{e}^{i\phi }{\sum }_{n}{e}^{i{\delta }_{n}}$
Thus, the total intensity is equal to the square of this total electric field, where in the equation below, the first terms in front of the summation equal the scattering from one electron. This is abbreviated as ${I}_{e}$ as seen in the second line below.
${I}_{tot}\propto |{\epsilon }_{tot}{|}^{2}={E}_{e}^{2}|{\sum }_{n=1}^{n}{e}^{i{\delta }_{n}}{|}^{2}=\frac{{r}_{e}^{2}}{{R}^{2}}\left(\frac{1+co{s}^{2}2\theta }{2}\right)|{\sum }_{n=1}^{N}{e}^{i{\delta }_{n}}{|}^{2}$
${I}_{tot}={I}_{e}|{\sum }_{n=1}^{N}{e}^{i{\delta }_{n}}{|}^{2}$
If the scattered direction is zero, then the total intensity will be the intensity of one electron times ${N}^{2}$ , where N is the total number of electrons. The intensity of the scattered wave is ${N}^{2}$ times, rather than $N$ times, the intensity of an individual electron, because of the interference of coherently scattered waves (waves of equal wavelength). Furthermore, if the $2\theta$ angle is not zero, then the intensity is less than ${N}^{2}$.
$2\theta =0\to \frac{{I}_{tot}}{{I}_{e}}={N}^{2}$$2\theta \ne 0\to \le {N}^{2}$
Note if the scattered waves had slightly different wavelengths ${\lambda }_{n}>{\lambda }_{0}$ this is called modified or incoherent scattering. The waves will not interfere and the intensity will be affected by a factor of $N$ as opposed to ${N}^{2}$ in classical Thomson scattering. $\to$$\left({I}_{tot}{\right)}_{mod}=N\left({I}_{e}{\right)}_{mod}$

#### 10.2.2 X-ray Scattering from atom with Z electrons in spherically symmetrical distributions

According to quantum mechanics, the electrons involved in scattering are not at discrete points in space, but have a probability distribution that describes their location around an atom. This probability of finding an electron in a given shell is given by a wave function and may be pictorially represented by an electron-cloud rather than set electron locations, as illustrated Figure 10.9 below. The electron density in a given atom, therefore, is the sum of the individual probability densities of its Z amount of electrons. This the sum is equal to the sum of each electron's wave functions multiplied by the wave function's complex conjugate, as shown below.
${\rho }_{atom}={\sum }_{1}^{Z}{\rho }_{e}=\sum {\psi }_{e}{\psi }_{e}^{*}$
H-like wave function$↑$
${\rho }_{e}=|{\psi }_{e}{|}^{2}$

#### 10.2.3 Defining Atomic Scattering Factor

In Figure 10.9 below, an incident x-ray beam, represented by the blue arrow pointing to the right and positioned to the left of the red electron cloud, is incident on some differential volume multiplied by the electron density to yield a differential charge value.
Figure 10.9: Scattering from ${e}^{-}$cloud
$f=\frac{{E}_{atom}}{{E}_{{e}^{-}}}$
The atomic scattering factor is calculated by the following integral which sums the scattering contributions from each differential volume while accounting for phase differences $\left({e}^{i\delta }={e}^{i\frac{2\pi }{\lambda }\left(\stackrel{‾}{S}-{\stackrel{‾}{S}}_{0}\right)\stackrel{‾}{r}}\right)$over the entire atomic volume.
$f={\int }_{atom}^{atom}\rho \left(\stackrel{‾}{r}\right){e}^{i\frac{2\pi }{\lambda }\left(\stackrel{‾}{S}-{\stackrel{‾}{S}}_{0}\right)\stackrel{‾}{r}}dV$
Recall that for N electrons, the magnitude of the electric field scattered in the forward direction should equal N, since there is no phase shift between the scattered waves. This is consistent with the formula below, as the ${e}^{i\delta }$ term, where $\delta$ is the phase shift, goes to one, and $f=Z.$ The atomic scattering factor, therefore, is dictated by the $\theta$ angle describing the phase shift between scattered waves of multiple electrons.
Essentially, as waves scattered by the atom's electrons become more out of phase ($\theta$ increases), the atomic scattering factor decreases ($f$ decreases) due to the partially destructive interference in scattered waves. In addition to the $\theta$angle, the atomic scattering factor also depends on the wavelength. At a fixed $\theta$ value, for instance, a decrease in wavelength will make path differences larger relative to the wavelength, thus causing greater interference between scattered waves.

#### 10.2.4 Calculating the Atomic Scattering Factor

Figure 10.10 below illustrates an incident x-ray beam ${\stackrel{‾}{S}}_{0}$, interacting with an atom O and scattering in the $\stackrel{‾}{S}$ direction. The following formula may be used to calculate the amplitude of the scattering for one electron:
${f}_{e}=\int {\rho }_{e}{e}^{i\frac{2\pi }{\lambda }\left(\stackrel{‾}{S}-{\stackrel{‾}{S}}_{0}\right)\stackrel{‾}{r}}dV$
Here, ${\rho }_{e}$ is the charge density for the electron, and the exponential term is what relates the direction of the incident beam ${\stackrel{‾}{S}}_{0}$ to that of the scattered beam $\stackrel{‾}{S}.$
Figure 10.10: Atomic Scattering
$\left(\stackrel{‾}{S}-{\stackrel{‾}{S}}_{0}\right)\stackrel{‾}{r}=|\stackrel{‾}{S}-{\stackrel{‾}{S}}_{0}||\stackrel{‾}{r}|cos\phi =2sin\theta rcos\phi$
Note: ${\int }_{0}^{\pi }{e}^{iqrcos\phi }sin\phi d\phi$
Let $u=qrcos\phi$ $\frac{du}{dq}=-qrsin\phi$
$\phi =0\to u=qr$
$\phi =\pi \to u=-qr$
$\begin{array}{c}⇒{\int }_{0}^{\pi }{e}^{iqrcos\phi }sin\phi d\phi =-\frac{1}{qr}{\int }_{-qr}^{qr}{e}^{iu}du\\ =-\frac{1}{qr}\left[{\int }_{-qr}^{qr}cosudu+i{\int }_{-qr}^{qr}sinudu\right]=-\frac{2sin\left(qr\right)}{qr}\end{array}$
Let $q=4\pi sin\theta /\lambda$ Azaroff $q\equiv k$
${f}_{e}=2\pi {\int }_{r=0}^{\infty }{\int }_{\phi =0}^{\pi }\rho \left(r\right){e}^{iqrcos\phi }{r}^{2}sin\phi d\phi dr$
If $\rho \left(\stackrel{‾}{r}\right)=\rho \left(r\right)$ (spherical symmetric)
${f}_{e}=4\pi {\int }_{r=0}^{\infty }\rho \left(r\right){r}^{2}\frac{sinqr}{qr}dr$ for spherically symmetrical charge distribution
Example: Lithium
Suppose that the individual electron densities for a neutral lithium atom, of atomic number 3 and electron configuration$\left(1{s}^{2}2{s}^{1}\right)$, are given by the following hydrogen-like expression:
${\rho }_{e}=\frac{{e}^{\left(-2r/a\right)}}{\pi {a}^{2}}$
Where for one K electron${a}_{K}=0.20\stackrel{˚}{A}$ and for one L electron ${a}_{L}=1.60\stackrel{˚}{A}$
The scattering factor for each electron may be calculated from the following equation:
Furthermore, the total scattering factor for lithium is obtained by summing the scattering factors of the two electrons in its K shell and the one electron in its L shell. ${f}_{Li}=2{f}_{eK}+{f}_{eL}$
Figure 10.11: Li scattering factor
Figure 10.12:

### 10.3 Compton Scattering

Compton scattering, also known as modified or incoherent scattering is the effect in which scattered waves have slightly different wavelengths than their incident x-ray wavelengths. This occurs when x-rays encounter loosely bound electrons, and may be explained by considering the incident beam to be comprised of a stream of photons of energy $\hslash {\omega }_{o}$, as illustrated in Figure 10.13 below. When one of these photons strikes a loosely bound electron that is initially at rest, energy and momentum are transferred from the photon to the electron. Due to this transfer of energy and momentum, the scattered x-ray photon has lower energy and therefore a larger wavelength. Note that another important characteristic of Compton scattering is that the phase of the scattered wave has no direct relationship to the phase of the incident beam, thus making this scattering incoherent and preventing interference effects.
Figure 10.13: Compton Scattering
The following equations combine the principles of conservation of energy and momentum in the photon-electron collision to yield a function of the change in wavelength of the x-ray beam with respect to the scattering angle, 2$\theta .$ Note that the theory of Compton scattering expresses the energy and momentum of the photons and electrons in relativistic values. These relativistic values are necessary because the photons scattered are massless and because the energy transferred to the electron is measured relative to its rest energy.
Conservation of energy:
$ħ{\omega }_{0}=ħ\omega +\frac{1}{2}{m}_{e}{v}_{e}^{2}$assuming initial ${e}^{-}$at rest
Conservation of momentum
$\frac{ħ{\omega }_{0}}{c}=\frac{ħw}{c}cos2\theta +{m}_{e}{v}_{e}cos\alpha$ $←$x direction
$0=\frac{ħ\omega }{c}sin2\theta -{m}_{e}{v}_{e}sin\alpha$ $←$y direction
Eliminate $\alpha$ and ${v}_{e}$ $\lambda =\frac{2\pi c}{\omega }$ $ħ=\frac{h}{2\pi }$
$\lambda ={\lambda }_{0}=\Delta \lambda =\frac{h}{{m}_{e}c}\left(1-cos2\theta \right)$
$\Delta \lambda \left(\stackrel{˚}{A}\right)=0.0243\left(1-cos2\theta \right)$
Note that the change in wavelength in the forward direction is 0 and the change in wavelength in the backward direction $2\theta =180{}^{○}$ is about 0.05$\stackrel{˚}{A}.$
$\Delta \lambda =0$ at $2\theta =0{}^{○}$
$\Delta \lambda =0.0486\stackrel{˚}{A}$ at $2\theta =180{}^{○}$
The change in wavelength may be related to the change in energy by the following equations:
$\lambda -{\lambda }_{0}=\Delta \lambda =\frac{h}{{m}_{e}c}\left(1-cos2\theta \right)$  $\Delta E=\frac{{E}_{\gamma }^{2}\left(1-cos2\theta \right)}{{m}_{e}{c}^{2}+{E}_{\gamma }\left(1-cos2\theta \right)}$
where ${m}_{e}{c}^{2}=511keV$ (rest energy of electron)
Figure 10.14:
Ex: If $2\theta =90{}^{○}$ ${E}_{\gamma }=18.5KeV$ $\Delta E=0.65KeV$
Figure 10.15: Photon Energy
Note that Compton scattering will not occur if the change in energy is less than the binding energy for the electron.$\Delta E<{E}_{B}$
For hydrogen, a one electron atom, the total intensity according to the Thomson scattering equation is equal to the summed intensities of the modified (Compton) and unmodified (Bragg) scattered wave.
${I}_{e}={I}_{unmod}+{I}_{mod}$
${I}_{e}={I}_{0}\left(\frac{{r}_{e}}{R}{\right)}^{2}{sin}^{2}\alpha \to Thomson$
${I}_{unmod}=Coherent\to Bragg$ ${I}_{mod}=Incoherent\to Compton$
1= $\frac{{I}_{unmod}}{{I}_{e}}+\frac{{I}_{mod}}{{I}_{e}}$ in electron units
$1={f}_{e}^{2}+\left({I}_{eu}{\right)}_{mod}$ $⇒$$\left({I}_{eu}{\right)}_{mod}=1-{f}_{e}^{2}$
Figure 10.16: Hydrogen scattered intensities
The scattering intensity for atoms with atomic numbers greater than one, such as lithium (Z=3), varies from the simple hydrogen model shown above. Again, the modified scattering intensity for each electron is given by $1-{f}_{e}^{2}$. Recall that no interference effects can occur for incoherent scattering due to wavelength variations after collisions; thus, the total incoherent scattering intensity is simply given by the sum of the incoherent scattering intensities of the individual electrons.
$\left({I}_{eu}{\right)}_{mod}=2\left(1-{f}_{e}^{2}{\right)}_{K}+\left(1-{f}_{e}^{2}{\right)}_{L}=Z-{\sum }_{n}\left({f}_{e}{\right)}_{n}^{2}$
Furthermore, the unmodified scattering intensity is equal to the square of the atomic scattering factor, since $f=\frac{{E}_{atom}}{{E}_{e}}$, and the intensity is the square of the magnitude of the electric field. Recall that for coherently scattered rays, the most efficient scattering is in the forward direction, where $f=Z.$ Since the unmodified intensity varies as ${f}^{2},$ its magnitude will be less than or equal to ${Z}^{2}.$ Furthermore, the modified scattering intensity will be less than or equal to the atomic number of a given element. For lighter elements and for a low wavelength, the modified scattering intensity will be significant relative to the total scattering intensity.
$\left({I}_{eu}{\right)}_{unmod}={f}^{2}\le {Z}^{2}$
$\left({I}_{eu}{\right)}_{mod}\le Z\phantom{\rule{0ex}{0ex}}←$significant for low Z and low $\lambda$
Figure 10.17 below demonstrates the relationship between scattering intensities and $\frac{sin\theta }{\lambda }$ for atoms with $Z>1.$
Figure 10.17: Scattering intensities for elements of Z>1

#### 10.3.1 Anomalous Dispersion - $\lambda$dependence of $f$

$f={f}_{o}\left(\frac{sin\theta }{\lambda }\right)+\Delta f\text{'}\left(\lambda \right)+i\Delta f"\left(\lambda \right)$
${f}_{o}\to$Atomic scattering factor ${f}_{o}\left(\stackrel{‾}{q}\right)={\int }_{atom}^{atom}\rho \left(\stackrel{‾}{r}\right){e}^{i\stackrel{‾}{q}\cdot \stackrel{‾}{r}}d\stackrel{‾}{r}$
scattering factor is Fourier transform of atom's electron distribution, and vice versa
$q=\frac{2\pi }{\lambda }\left(\stackrel{‾}{S}-{\stackrel{‾}{S}}_{o}\right)$$q=\frac{4\pi sin\theta }{\lambda }$
The ${f}_{o}$term is what was previously taken to be the atomic scattering factor, and the $\Delta f\text{'}$ and $i\Delta f"$ correction terms account for the anomalous dispersion effect. The latter two terms are dependent on the element and the wavelength of the incident x-ray.
If the frequency of the incident x-ray beam is much greater than the natural resonance frequency of the electron (i.e. if the energy of the x-ray is much greater than the binding energy for a K shell electron) the scattered wave will be $\pi$ out of phase with respect to the incident wave, similar to the $\pi$ phase lag in Thomson scattering by free electrons. However, if the energy of the incident beam is close to the electron's binding energy, the scattered wave will deviate from this $\pi$ phase shift with respect to the incident wave. Therefore, the in-phase rays scattered from the K electrons destructively interfere with the out of phase x-rays scattered from outer electrons.

#### 10.3.2 Hönl's Correction Factors

As shown in Figure 10.18 below, the $\Delta f$ real component is $18{0}^{○}$out of phase with the normally scattered radiation, ${f}_{o}$, and affects the amplitude of the wave. Furthermore, the $\Delta f"$ imaginary component is $9{0}^{○}$ out of phase with respect to ${f}_{o}$ and affects the phase of the scattered wave.
Figure 10.18: Honl's correction factors
Figure 10.19 below demonstrates the effect of increasing energy on the $\Delta f\text{'}$ and $\Delta f"$ correction factors. Note that the $\Delta f"$ imaginary component becomes highly positive slightly above the absorption edge. Conversely, the real $\Delta f\text{'}$ component becomes largely negative close to the absorption edge. This proximity to the absorption edge indicates that the frequency of the incident ray is close to the electron's natural frequency. Since the absorption edge is characteristic of a given element, the correction factors in $f={f}_{o}+\Delta f\text{'}+i\Delta f"$ can give chemical sensitivity to the x-ray scattering process. These ${f}_{o},\Delta f\text{'},\Delta f"$ terms are tabulated in the International Tables of Crystallography Vol. IV. In addition, since the intensity is proportional to $ff*$ - the scattering factor times its complex conjugate - the scattered intensity may be manipulated by using a synchrotron radiation source with a tunable ${E}_{\gamma },$ or wavelength of the incident x-ray beam.
Figure 10.19: Correction factors as a function of incident x-ray energy
Evidently, if the incident x-ray is not close to the absorption edge for an element, i.e. ,${E}_{K}$ or ${E}_{L}$ etc., then the correction factors will be negligible with respect to ${f}_{o}$. In 361, we will ignore anomalous dispersion effects unless specifically stated.

#### 10.3.3 X-ray Scattering by Many Atoms

For one atom scattered E-field amplitude is
${E}_{a}=f{E}_{e}=f{E}_{o}\frac{{r}_{e}}{R}sin\alpha$$\phantom{\rule{10px}{0ex}}sin\alpha =1$ for $\sigma$-polarized case; $sin\alpha =cos2\theta$ for $\pi$-polarized case
Elastic Coherent Scattering $\to f$
${r}_{e}=2.818×1{0}^{-13}cm$
$f={f}_{o}\left(\frac{sin\theta }{\lambda }\right)\to \Delta f\text{'}=\Delta f"=0$
${f}_{o}\left(0\right)=$
For unpolarized ${E}_{o}$ ${E}_{ox}={E}_{oz}=\frac{{E}_{o}}{\sqrt{2}}$

## 11 Kinematic Theory

Consider the scattering from a collection of N atoms, as shown in Figure 11.1 below. Note the incident direction $\stackrel{‾}{S}$ and scattering direction $\stackrel{‾}{S}.$
Figure 11.1: Scattering from collection of atoms
If we consider the scattering from atoms “0” and “3,” their scattered wave vectors will be parallel and of the same wavelength, but the scattered waves will be out of phase with each other due to a separation between the two atoms, indicated by ${\stackrel{‾}{R}}_{3}.$ The difference between the incident vector and the scattered wave vector dotted with the distance between the two atoms yields this extra path length, as shown in the following expression:
Extra path length ${\stackrel{‾}{R}}_{m}\cdot \left(\stackrel{‾}{S}-{\stackrel{‾}{S}}_{o}\right)$
The path length difference normalized by the wavelength gives the scattering vector.
Scattering vector $\stackrel{‾}{Q}=\frac{\stackrel{‾}{S}-{\stackrel{‾}{S}}_{o}}{\lambda }$
Multiplying the scattering vector by $2\pi$ and taking its dot product with ${\stackrel{‾}{R}}_{m}$ yields the path difference will be in terms of the phase angle. If this phase angle is an integer multiple of $2\pi$, then perfect constructive interference occurs. The following equation sums the total scattered field by N atoms, in which the exponential terms make up a phase factor that keeps track of the phase difference for all of the atoms in a sample:
$\frac{{E}_{Total}}{{E}_{e}}={\sum }_{m=0}^{N-1}{f}_{m}{e}^{2\pi i\stackrel{‾}{Q}\cdot {\stackrel{‾}{R}}_{m}}$
Note that ${f}_{m}$ is the amplitude of scattering by an individual atom, which will be the same for all atoms of the same type and distinct for different atoms. The summed fields are normalized by the scattering amplitude that is expected from one electron according to classical Thomson scattering- $\frac{{E}_{Total}}{{E}_{e}}$.
${f}_{1}={f}_{2}:$ If all atoms are same type, otherwise different
The scattered intensity in electron units, or the scattered intensity of N atoms as compared to the scattered intensity of a single electron is given by the following equation:
${I}_{\left(eu\right)}=|\frac{{E}_{Total}}{{E}_{e}}{|}^{2}=\left[{\sum }_{m=0}^{N-1}{f}_{m}{e}^{2\pi i\stackrel{‾}{Q}\cdot {\stackrel{‾}{R}}_{m}}\right]\left[{\sum }_{m=0}^{N-1}{f}_{m}{e}^{-2\pi i\stackrel{‾}{Q}\cdot {\stackrel{‾}{R}}_{m}}\right]$
Here, the normalized total electric field equation is multiplied by its complex conjugate to yield the intensity as given by $|E{|}^{2}$.
In the following step, the ${\stackrel{‾}{R}}_{m}$ is subdivided into ${\stackrel{‾}{R}}_{m}$ and ${\stackrel{‾}{R}}_{n}$ allowing for a grouping into a double summation. Here the ${f}_{m}$and ${f}_{n}$ factors account for possibilities of interference of scattered waves between two atoms.
$={\sum }_{m=0}^{N-1}{\sum }_{n=0}^{N-1}{f}_{m}{f}_{n}{e}^{2\pi \stackrel{‾}{Q}\cdot \left({\stackrel{‾}{R}}_{m}-{\stackrel{‾}{R}}_{n}\right)}$
${I}_{eu}={\sum }_{m=0}^{N-1}{\sum }_{n=0}^{N-1}{f}_{m}{f}_{n}{e}^{2\pi i\stackrel{‾}{Q}\cdot {\stackrel{‾}{r}}_{mn}}$
In the following vector diagram for the ${\stackrel{‾}{R}}_{m}-{\stackrel{‾}{R}}_{n}$ term, the origin is conveniently chosen to be at the location of the green “0” atom. The difference of ${\stackrel{‾}{R}}_{3}$ and ${\stackrel{‾}{R}}_{4}$ gives the inter-atomic distance between atoms 3 and 4, labeled ${\stackrel{‾}{r}}_{34}$.
Figure 11.2: ${\stackrel{‾}{R}}_{m}-{\stackrel{‾}{R}}_{n}$
Scattered intensity for N atoms
Now, the ${\stackrel{‾}{r}}_{mn}$ inter-atomic spacing vector replaces the ${\stackrel{‾}{R}}_{m}-{\stackrel{‾}{R}}_{n}$ term previously used.
$\frac{{I}_{Total}}{{I}_{e}}={I}_{eu}={\sum }_{n=0}^{N-1}{f}_{m}{f}_{n}{e}^{2\pi i\stackrel{‾}{Q}\cdot {\stackrel{‾}{r}}_{mn}}$
From this double summation, there are ${N}^{2}$ terms that are of the type ${f}_{m}\cdot {f}_{n}$
There are N terms of the type: ${f}_{m}^{2}$ m=0,1,2...N-1
This term is like the scattering intensity from a single isolated atom, in which no interference effects occur. Here ${\stackrel{‾}{r}}_{mn}$ is 0, and the exponential term goes to 1.
There are $\frac{N\left(N-1\right)}{2}$ of the type: ${f}_{m}{f}_{n}\left({e}^{2\pi i\stackrel{‾}{Q}\cdot {\stackrel{‾}{r}}_{mn}}+{e}^{-2\pi i\stackrel{‾}{Q}\cdot {\stackrel{‾}{r}}_{mn}}\right)=2{f}_{m}{f}_{n}cos\left(2\pi \stackrel{‾}{Q}\cdot {\stackrel{‾}{r}}_{mn}\right)$
These are like cross terms, where ${f}_{m}\ne {f}_{n}.$
This is the general equation for kinematical scattering from any type of atomic collection - i.e. gas, solid, liquid, crystal, amorphous, etc. The following conditions are specific to to gases, liquids and crystals:
• gases & liquids $←$require proper averaging of $\stackrel{‾}{Q}\cdot {\stackrel{‾}{r}}_{mn}$ (angle averaging) by small angle x-ray scattering
• crystals are periodic $←$can use symmetry to reduce ${N}^{2}$ terms
Figure 11.3: X-ray plane wave incident on crystal
Figure 11.4: Scattering vector
The crystal contains $M={M}_{1}{M}_{2}{M}_{3}$ unit cells, in which ${M}_{1}=\stackrel{‾}{a},\phantom{\rule{6px}{0ex}}{M}_{2}=\stackrel{‾}{b},\phantom{\rule{6px}{0ex}}{M}_{3}=\stackrel{‾}{c}$. Each unit cell contains N atoms, which may or may not be of different types. Therefore, there are MN total atoms in the unit cell.
Each atom in the unit cell is located by a vector ${\stackrel{‾}{r}}_{n}$ , as shown in Figure 11.5 below. The atom is illustrated as a black dot in the unit cell block.
Figure 11.5: Atom located by ${\stackrel{‾}{r}}_{n}$
Assuming the crystal is parallelepiped, any unit cell's origin may be located by ${m}_{1}\stackrel{‾}{a}+{m}_{2}\stackrel{‾}{b}+{m}_{3}\stackrel{‾}{c}$, where $0\le {m}_{i}\le {M}_{i}-1$ is an integer. This is given by translational symmetry of unit cells.
Therefore, any atom within entire crystal is located by the following position vector:
${\stackrel{‾}{R}}_{{m}_{1}{m}_{2}{m}_{3}}^{n}={m}_{1}\stackrel{‾}{a}+{m}_{2}\stackrel{‾}{b}+{m}_{3}\stackrel{‾}{c}+{\stackrel{‾}{r}}_{n}$
$↖↑↗$$↖$
u.c. locationatom location in u.c.
The first three indices locate the specific unit cell containing the atom, using a translational symmetry operator. The ${\stackrel{‾}{r}}_{n}$term indicates the atom's position within that specific unit cell.
Figure 11.6 below illustrates the use of these four indices to locate an atom within a crystal. Note that the location of the origin is selected and arbitrary.
Figure 11.6: Location of atom within entire crystal (${\stackrel{‾}{R}}_{{m}_{1}{m}_{2}{m}_{3}}^{n}$)

#### 11.0.1 The scattered electric field from the entire crystal relative to the scattered field form a single electron is given by the following summation:

$\frac{{E}_{Total}}{{E}_{e}}={\sum }_{{m}_{1}=0}^{{M}_{1}-1}{\sum }_{{m}_{2}=0}^{{M}_{2}-1}{\sum }_{{m}_{3}=0}^{{M}_{3}-1}{\sum }_{n=0}^{N-1}{f}_{n}{e}^{2\pi i\stackrel{‾}{Q}\cdot {\stackrel{‾}{R}}_{{m}_{1}{m}_{2}{m}_{3}}^{n}}$
$={\sum }_{{m}_{1}=0}^{{M}_{1}-1}{\sum }_{{m}_{2}=0}^{{M}_{2}-1}{\sum }_{{m}_{3}=0}^{{M}_{3}-1}{\sum }_{n=0}^{N-1}{f}_{n}{e}^{2\pi i\stackrel{‾}{Q}\cdot \left({m}_{1}\stackrel{‾}{a}+{m}_{2}\stackrel{‾}{b}+{m}_{3}\stackrel{‾}{c}\right)}{e}^{2\pi \stackrel{‾}{Q}\cdot {\stackrel{‾}{r}}_{n}}$
The four indices yield four summations, where only the last summation term is related to the exponential phase factor.
The structure factor, given by the following summation, is the amplitude of the scattered electric field from a single unit cell:
$F={\sum }_{n=0}^{N-1}{f}_{n}{e}^{2\pi i\stackrel{‾}{Q}\cdot {\stackrel{‾}{r}}_{n}}$
This structure factor accounts for internal interference effects from N atoms in the unit cell. Adding the structure factor to the calculation of the total scattered field yields the following triple summation:
$\frac{{E}_{Total}}{{E}_{e}}=F{\sum }_{{m}_{1}=0}^{{M}_{1}-1}{e}^{2\pi i\stackrel{‾}{Q}\cdot {m}_{1}\stackrel{‾}{a}}{\sum }_{{m}_{2}=0}^{{M}_{2}-1}{e}^{2\pi i\stackrel{‾}{Q}\cdot {m}_{2}\stackrel{‾}{b}}{\sum }_{{m}_{3}=0}^{{M}_{3}-1}{e}^{2\pi i\stackrel{‾}{Q}\cdot {m}_{3}\stackrel{‾}{c}}$
Consider the following simplification:
${\sum }_{{m}_{1}=0}^{{M}_{1}-1}{e}^{2\pi i\stackrel{‾}{Q}\cdot {m}_{1}\stackrel{‾}{a}}={\sum }_{m=0}^{M-1}{x}^{m}$Geometric series $\to$ converges
$=\frac{{x}^{M}-1}{x-1}=\frac{{e}^{2\pi i\stackrel{‾}{Q}\cdot \stackrel{‾}{a}{M}_{1}}-1}{{e}^{2\pi i\stackrel{‾}{Q}\cdot \stackrel{‾}{a}}-1}=\frac{{e}^{2i\eta M}-1}{{e}^{2i\eta }-1}$
$=\frac{{e}^{i\eta M}\left({e}^{i\eta M}-{e}^{-i\eta M}\right)}{{e}^{i\eta }\left({e}^{i\eta }-{e}^{-i\eta }\right)}=\frac{{e}^{i\eta M}sin\eta M}{{e}^{i\eta }sin\eta }$
$=\frac{sin\left(\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{a}{M}_{1}\right){e}^{\pi i\stackrel{‾}{Q}\cdot \stackrel{‾}{a}{M}_{1}}}{sin\left(\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{a}\right){e}^{\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{a}}}=\frac{sin\left(\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{a}{M}_{1}\right)}{sin\left(\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{a}\right)}{e}^{\pi i\stackrel{‾}{Q}\cdot \stackrel{‾}{a}\left({M}_{1}-1\right)}$
In one dimension, the total scattering intensity for M unit cells relative to the scattering intensity from one intensity is given by the following:
${\stackrel{‾}{I}}_{eu}={|\frac{{E}_{total}}{{E}_{e}}|}^{2}=|F{|}^{2}\frac{si{n}^{2}\left(\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{a}M\right)}{si{n}^{2}\left(\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{a}\right)}$
1D Interference Function
The scattering from a single unit cell is given by the structure factor squared, and the second term accounts for interference from M repeated unit cells.
In three dimensions, each M factor is multiplied by its complex conjugate
${\stackrel{‾}{I}}_{eu}={|\frac{{E}_{total}}{{E}_{e}}|}^{2}=F{F}^{*}{|\frac{{e}^{2\pi i\stackrel{‾}{Q}\cdot \stackrel{‾}{a}{M}_{1}}-1}{{e}^{2\pi i\stackrel{‾}{Q}\cdot \stackrel{‾}{a}}-1}\cdot \frac{{e}^{2\pi i\stackrel{‾}{Q}\cdot \stackrel{‾}{b}{M}_{2}}-1}{{e}^{2\pi i\stackrel{‾}{Q}\cdot \stackrel{‾}{b}}-1}\cdot \frac{{e}^{2\pi i\stackrel{‾}{Q}\cdot \stackrel{‾}{c}{M}_{3}}-1}{{e}^{2\pi i\stackrel{‾}{Q}\cdot \stackrel{‾}{c}}-1}|}^{2}$
yielding the following three dimensional interference function. This is a rigorous theory for explaining x-ray scattering from a 3-D periodic lattice. Note that the three terms account for repeated unit cells along three axes.
${\stackrel{‾}{I}}_{eu}=|F{|}^{2}\frac{si{n}^{2}\left({M}_{1}\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{a}\right)}{si{n}^{2}\left(\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{a}\right)}\cdot \frac{si{n}^{2}\left({M}_{2}\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{b}\right)}{si{n}^{2}\left(\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{b}\right)}\cdot \frac{si{n}^{2}\left({M}_{3}\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{c}\right)}{si{n}^{2}\left(\pi \stackrel{‾}{Q}\cdot c\right)}$
3D Interference Function
Considering the 1D Interference Function
$\frac{{sin}^{2}Mx}{{sin}^{2}x}={M}^{2}$ when $x=±h\pi$ $h=0,1,2,...$
Both the numerator and denominator vary, with the numerator having M times more oscillations than the denominator. In addition, both have periodic zeros at the same integer values of $\pi$, this causes the function to go to ${M}^{2}$ at these values. Physically, this condition is when $\stackrel{‾}{Q}\cdot \stackrel{‾}{a}=±h$
Recall that $\stackrel{‾}{Q}\cdot \stackrel{‾}{a}$ is extra length in units of $\lambda$for scattering by two atoms separated by $\stackrel{‾}{a.}$ This makes sense with previous considerations of Bragg's law, in which an integer multiple wavelength difference in paths yields a diffraction peak.
In Figure 11.7 below, $sinx,$ ${sin}^{2}x$, and $\frac{1}{5}{sin}^{2}x$ are plotted against $x$. The green ${sin}^{2}x$ curve represents the denominator, and the red curve represents the ${sin}^{2}Mx$ numerator where M=5. Note that the red curve has five times the oscillations of the green curve and the two have three common zeros in this range of $x.$
Figure 11.7: 1D interference oscillations
In the one dimensional interference function, the red function in Figure 11.7 is divided by the green function to yield the plot in Figure 11.8.

#### 11.0.2 Note that the one dimensional interference function is multiplied by a $\frac{1}{{M}^{2}}$ factor to yield the following normalized function that has a maximum of one:

$I\left(x\right)=\frac{si{n}^{2}Mx}{{M}^{2}si{n}^{2}x}$
Figure11.8 illustrates the unity amplitude diffraction peaks that occur when the numerator and denominator have common zeros, at $x=±h\pi$ with $h=0,1,2$, etc. There is a zeroth order peak that occurs when all waves are scattered in phase and in the forward direction. In addition, there are always $M-2$ subsidiary peaks between any two diffraction peaks, with the other two peaks hidden at half of the width of the large diffraction peaks at either side. These subsidiary peaks are due to the fact that the numerator, ${sin}^{2}Mx,$ gives zero intensity when $Mx=\pi$. The width of these peaks is also due to the $\frac{\pi }{M}$ periodicity of the numerator.
Figure 11.8: Normalized 1D interference function, where M=6
What is $x$ in terms of the angle $\theta$?
The quantity $x$ is the scattering vector $\overline{Q}$ multiplied by $\pi$ , and projected in the direction of the lattice constant, $\overline{a}$. If the lattice constant and the scattering vector are in the same direction , this simplifies to the following expression:
$x=\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{a}=\pi Qa=\pi \frac{|\stackrel{‾}{S}-{\stackrel{‾}{S}}_{0}|}{\lambda }a=\pi \frac{2sin\theta }{\lambda }a$
Figure 11.9: Physical relationship between x-ray directions, and $\stackrel{‾}{a}$ - used in describing “$x$
At $x=\frac{\pi }{M}=\frac{2\pi sin{\theta }_{min}}{\lambda }a\to \frac{sin{\theta }_{min}}{\lambda }=\frac{1}{2Ma}\to {\theta }_{min}={sin}^{-1}\left[\frac{\lambda }{2Ma}\right]$
Here, ${\theta }_{min}$ represents the first appearing minimum of $\theta$ in the periodic intensity pattern.
For a very large M, $sin\theta$ can be set equal to $\theta ,$ and the following expression can be used to determine the full width, full max diffraction peak width as illustrated in Figure 11.10. Note that this is simply twice the value of ${\theta }_{min}.$
$\Delta \theta =2{\theta }_{min}=\frac{\lambda }{Ma}$ for h=0
Therefore by measuring the width of the entire diffraction peak, the value $M$ may be determined, which indicates the number of unit cells in the direction of $\stackrel{‾}{a}$. Multiplying $M$ by $a$ yields the thickness of the crystal in the direction of $a.$ Notice that $\Delta \theta$ and $Ma$ are inversely proportional, therefore as the width of a crystal increases the $\Delta \theta$, or width of the peak, gets smaller.
Figure 11.10: $\Delta \theta$ (FWHM) of peaks shown in Figure 11.8
Note the following correlation with Bragg's Law:
$\lambda =2dsin\theta$, in this case $\lambda =2asin\theta$
$\frac{2asin\theta }{\lambda }=1\to \stackrel{‾}{Q}\cdot \stackrel{‾}{a}=1=h$
Furthermore, for all other peaks ($h\ne 0$)
$FWHM=\Delta \theta =\frac{\lambda }{Macos\theta }$
For all peaks, therefore, the width is inversely proportional to the size of the crystal. Therefore, by looking at the width of the peak, one can see the size of the crystal along the $\stackrel{‾}{a}$ direction.
• for $large$ M, the area under the intensity curve becomes :
$Area={M}^{2}=\Delta \theta =\frac{M\lambda }{acos\theta }$
Figure 11.11: Delta functions, where $\stackrel{‾}{Q}$ is the scattering vector and $\stackrel{‾}{a}$ is the scattering section

## 12 Diffraction

Recall the the following 3D interference function for crystal, where $M$ gives the number of unit cells:
${\stackrel{‾}{I}}_{eu}=|F{|}^{2}\frac{{sin}^{2}\left({M}_{1}\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{a}\right)}{{sin}^{2}\left(\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{a}\right)}\cdot \frac{{sin}^{2}\left({M}_{2}\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{b}\right)}{{sin}^{2}\left(\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{b}\right)}\cdot \frac{{sin}^{2}\left({M}_{3}\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{c}\right)}{{sin}^{2}\left(\pi \stackrel{‾}{Q}\cdot c\right)}$, $\phantom{\rule{10px}{0ex}}{M}^{2}=\left({M}_{1}{M}_{2}{M}_{3}{\right)}^{2}$ M = # of unit cells

### 12.1 Laue condition for diffraction

In three dimensions, the diffraction condition is satisfied and the intensity reaches a maximum when all three fractional factors after the $|{F}^{2}|$ term in the interference function simultaneously go to zero, that is when the following three conditions are met:
$\stackrel{‾}{Q}\cdot \stackrel{‾}{a}=h$,$\stackrel{‾}{Q}\cdot \stackrel{‾}{b}=k$ ,$\stackrel{‾}{Q}\cdot \stackrel{‾}{c}=l$ where $h,k,l$ are integers
This is known as the Laue condition for diffraction. Note that the $h,k,l$ indices are the same indices used in describing the planes of a crystal in real space.
The three scalar quantities describing $\stackrel{‾}{Q}$ in the Laue condition are equivalent to the following vector equation, which also describes the reciprocal lattice vector that points perpendicular to the $hkl$ planes in real space with a length of $\frac{1}{{d}_{hkl}}:$
$\stackrel{‾}{Q}=h{\stackrel{‾}{a}}^{*}+k{\stackrel{‾}{b}}^{*}+l{\stackrel{‾}{c}}^{*}={\stackrel{‾}{r}}_{hkl}^{*}$
Consider the scalar equation $|\stackrel{‾}{Q}|=|{\stackrel{‾}{r}}_{hkl}^{*}|$. Recall that $|\stackrel{‾}{Q}|$ is equivalent to $\frac{2sin\theta }{\lambda }$ and that $|{\stackrel{‾}{r}}_{hkl}^{*}|$ is equivalent to the inverse of the inter-planar spacing, which leads to Bragg's Law:
Bragg's Law: $\frac{2sin\theta }{\lambda }=\frac{1}{{d}_{hkl}}$$\to$${d}_{hkl}=\frac{d}{n}$
Therefore, the $h,k,l$ diffraction peak maxima occurs at:
$\stackrel{‾}{Q}=\frac{\stackrel{‾}{S}}{\lambda }-\frac{{\stackrel{‾}{S}}_{0}}{\lambda }={\stackrel{‾}{r}}_{hkl}^{*}$
$\stackrel{‾}{Q}$- Scattering vector, $\frac{\stackrel{‾}{S}}{\lambda }$- Scattered wave vector,$\frac{{\stackrel{‾}{S}}_{0}}{\lambda }$- Incident wave vector ,${\stackrel{‾}{r}}_{hkl}^{*}$- Reciprocal lattice vector
This relationship is illustrated in Figure 12.1 below. The scattering vector may be seen as the instrument being used in the diffraction experiment, since the incident direction, the wavelength, the location of the crystal, and the orientation of the detector as indicated by $2\theta$ may be selected before performing the experiment. The ${\stackrel{‾}{r}}_{hkl}^{*}$ describes the sample, which is a crystal consisting of a discrete set of planes in discrete directions indexed by ${r}_{hkl}$ in real space and ${\stackrel{‾}{r}}_{hkl}^{*}$ in reciprocal space. Therefore, as described by the Laue condition, when the instrument $\stackrel{‾}{Q}$ coincides with an ${\stackrel{‾}{r}}_{hkl}^{*}$ vector that is part of a set of discrete vectors pertaining to the crystal, a diffraction peak will appear.
Figure 12.1: Diffraction peak condition

#### 12.1.1 Vector Representation in Reciprocal Space

In order to picture a diffraction experiment in reciprocal space, recall that the crystal planes as referenced by $hkl$ Miller indices are transformed into $hkl$ reciprocal lattice points in reciprocal space.
Figure12.2 illustrates the $012$ reflection in the $h=0$ layer of a cubic unit cell. In this example incident beam ${I}_{o}$ is normalized by the wavelength into $\frac{{\stackrel{‾}{S}}_{o}}{\lambda }$ , and has a fixed direction and wavelength. The beam interacts with the sample at point C, and the scattered wave is in the direction of $\frac{\stackrel{‾}{S}}{\lambda }.$ The incident wave vector and the scattered wave vector form a locus of points that is known as the Ewald sphere. A cross section of the Ewald sphere with radius $\frac{1}{\lambda }$ is illustrated in Figure 12.2 .
Ex. 012 reflection, cubic h=0 layer
Figure 12.2: Ewald Sphere

#### 12.1.2 Ewald Sphere (Radius=$\frac{1}{\lambda }\right)$

If incident direction (${\stackrel{‾}{S}}_{0}\right)$ is fixed and $\lambda$fixed
1. Sphere does not move.
2. $\frac{{\stackrel{‾}{S}}_{0}}{\lambda }$ always points from center (C) to 000.
3. Rotate crystal, represented by reciprocal lattice, about 000.
4. $\frac{\stackrel{‾}{S}}{\lambda }$ points form C to any points on sphere.
5. Diffraction occurs when an ($hkl$) reciprocal lattice point coincides with the surface of the Ewald sphere.
6. For X-ray wavelengths multiple reflections are rare. This probability increases with higher energies and a larger Ewald sphere.

### 12.2 Single Crystal Diffractometer

Consider a silicon wafer
How would you use a $\theta -2\theta$ diffractometer and a monochromatic beam to determine if its face is (111) or (001)?
Figure 12.3: Si wafer
1. Rotate detector to $2\theta =0$
2. Rotate sample to $\omega =0$, i.e., surface $\parallel$to beam
In reciprocal space, move ${\stackrel{‾}{r}}_{hhh}^{*}$ into diffractometer plane, i.e., rotate crystal $\to {\stackrel{‾}{r}}_{hhh}^{*}\perp {\stackrel{‾}{S}}_{0}\to {\stackrel{‾}{r}}_{hhh}^{*}\perp 2\theta$axis
3. Do coupled $\theta -2\theta$ scan
4. Looking at the diffraction peaks, the silicon wafer face is found to be (111), since a (001) face would yield diffraction peaks of the order (00$l$), with $l=4,\phantom{\rule{6px}{0ex}}8,\phantom{\rule{6px}{0ex}}12,\phantom{\rule{6px}{0ex}}...$.
Figure 12.4: $\theta -2\theta$ scan of Si wafer
Note: The 222 is missing because ${F}_{222}=0$ for Si.

### 12.3 Diffraction Methods

10 summarizes the Laue method, rotating crystal method, and powder method based on the variability of $\lambda$ and $\theta$. The rotating crystal method, discussed in the previous section, uses a monochromatic beam of fixed wavelength and has a variable $\theta$ for scanning diffraction peaks in series. The Laue method has a fixed $\theta$ and a variable wavelength since the incident beam is continuous rather than monochromatic and allows for various diffraction peaks to appear at once. The powder method has a monochromatic beam of fixed wavelength and a variable $\theta$ since the crystallites in the beam have random orientations.
 Diffraction Methods $\lambda$ $\theta$ Laue Method Variable Fixed Rotating Crystal Method Fixed Variable Powder Method Fixed Variable
Table 10: Diffraction methods

#### 12.3.1 Laue Method - white beam / single crystal

For the Laue method, there is an incident white beam consisting of a continuum of wavelengths which interacts with a crystal as shown in Figure 12.5 below.The crystal consists of sets of planes, indexed in reciprocal space by the ${\stackrel{‾}{r}}_{hkl}^{*}$ vector perpendicular to the planes in real space. Satisfying Bragg's law based on these crystal planes creates a Laue spot on the film located downstream from the sample. This film location is used in the transmission Laue experiment setup.
Transmission Laue
$\theta <2\theta <90{}^{○}$
${0}^{○}<\theta <45{}^{○}$
Figure 12.5: Transmission Laue
In the back reflection geometry, the beam comes through a hole in the film or 2D area detector and hits the crystal planes in the single crystal sample as illustrated in Figure 12.6. The crystal selects a particular wavelength that matches up to a particular reflection of planes and spacings such that the scattered monochromatic beam that satisfies Bragg's law. There are multiple wavelengths that are harmonics or integer multiples of each other that make up a Laue spot i.e. 220,440,880,etc. However, the diffraction intensity gets weaker for higher harmonics due to many different factors including the atomic scattering factor.
Back Reflection Laue
$9{0}^{○}<2\theta <180{}^{○}$
$4{5}^{○}<\theta <90{}^{○}$
Figure 12.6: Back reflection Laue

#### 12.3.2 Reciprocal lattice treatment of Laue method

In reciprocal space, there is a continuum of Ewald spheres due to the continuum of wavelengths in the incident white beam. Recall that the Ewald sphere radius is equal to $\frac{1}{\lambda }.$ All Ewald spheres touch the origin, as illustrated in Figure 12.7. Note that this illustration shows cross sections of the Ewald spheres , and a cut of the orthorhombic P reciprocal lattice.
The incident wave vector is in the direction of ${\stackrel{‾}{S}}_{o}$, and its length is continuum over some range due to the continuum of wavelengths in the incident beam. In the case of Figure 12.8 , the crystal is oriented in a fixed direction such that the [100] direction is coincident with the incident beam. The wavelength range of this beam is from the short wavelength limit set by the accelerating voltage to a rough upper limit of $2\stackrel{˚}{A}$ due to absorption effects.
Figure 12.7: Reciprocal lattice of Laue method
Looking back at Figure 12.7, the smaller sphere corresponds to the upper wavelength limit of lower energy and the larger sphere corresponds to the short wavelength limit of higher energy. Note that the radius of the small sphere is $AO$ and the radius of the larger sphere is $BO.$ The two radii correspond to the limits of the incident wave vector ${\stackrel{‾}{S}}_{0},$ ranging in length from $AO$ to $BO.$
• $AO<\frac{{\stackrel{‾}{S}}_{0}}{\lambda }
• $AO=\frac{1}{2\stackrel{˚}{A}}=0.5{\stackrel{˚}{A}}^{-1}$
• $BO=\frac{1}{{\lambda }_{SWL}}\approx 5{\stackrel{˚}{A}}^{-1}$
Consider the ($\stackrel{‾}{1}30\right)$ Bragg reflection whose vector diagram is illustrated in Figure 12.9. Note that the length of the normalized incident wave vector is equal to the length of the normalized scattering vector due to their equal wavelengths. This $\left(\stackrel{‾}{1}30\right)$ reflection is allowed since its vector is between the minimum and maximum Ewald spheres AO and BO in Figure 12.7. Note that the two vectors form an isosceles triangle and that the scattering vector $\stackrel{‾}{Q}$ is coincident with the reciprocal lattice vector ${\stackrel{‾}{r}}_{\stackrel{‾}{1}30}^{*}.$ Since the $2\theta$ angle is less than $9{0}^{○},$ this is the transmission Laue condition.
Figure 12.9: ($\stackrel{‾}{1}30$) reflection
$2\theta <90{}^{○}\to$Transmission Laue
Now consider the ($\stackrel{‾}{3}\stackrel{‾}{1}0$) reflection illustrated in Figure 12.10 , in which $2\theta$ is greater than $9{0}^{○}$. This is a back reflection Laue condition. Note that the incident and scattered wave vector form two legs of an isosceles triangle with their vertex at some point between the smallest and largest Ewald spheres for the experiment.
Figure 12.10: $\stackrel{‾}{3}\stackrel{‾}{1}0$ reflection
$2\theta >90{}^{○}\to$Back Reflection Laue
In general, the reciprocal lattice points outside the smallest Ewald sphere of radius $\frac{1}{2\stackrel{˚}{A}}$ and inside the largest sphere of radius $\frac{1}{{\lambda }_{SWL}}$ may produce Bragg reflections.

### 12.4 Structure Factor Examples

Recall that at the Bragg peak, the intensity relative to the intensity of one electron scatterer is derived from the three dimensional interference function. The following equation yields this relative intensity: ${I}_{\left(eu\right)}=FF*{M}^{2}$
Here, $M={M}_{1}{M}_{2}{M}_{3}$ represents the number of unit cells that are diffracted, where in three dimensions ${M}_{1},\phantom{\rule{6px}{0ex}}{M}_{2}$ and ${M}_{3}$ define the number of unit cells in the three distinct axes. The structure factor times its complex conjugate, ${F}^{2}$, affects the relative intensity of the Bragg peak and is given by summation of the individual atomic scattering factors of the atoms in the unit cell and their respective phase factors.
$F={\sum }_{n=0}^{N-1}{f}_{n}{e}^{2\pi i\stackrel{‾}{Q}\cdot {\stackrel{‾}{r}}_{n}}$
Structure Factor for unit cell with N atoms per unit cell
Recall that the individual atomic scattering factor has a form factor and correction factors where ${f}_{n}=\left({f}_{0}\left(\frac{sin\theta }{\lambda }\right)+\Delta f\text{'}+i\Delta f"{\right)}_{n}$ in general. We simplify this by assuming that ${f}_{n}=\left({f}_{0}{\right)}_{n}$ for atoms at positions ${\stackrel{‾}{r}}_{n}$
where ${\stackrel{‾}{r}}_{n}={x}_{n}\stackrel{‾}{a}+{y}_{n}\stackrel{‾}{b}+{z}_{n}\stackrel{‾}{c}$.
At the Bragg peak the scattering vector coincides with a distinct ${\stackrel{‾}{r}}_{hkl}^{*}$ vector within the crystal. $\stackrel{‾}{Q}={\stackrel{‾}{r}}_{hkl}^{*}=h{\stackrel{‾}{a}}^{*}+k{\stackrel{‾}{b}}^{*}+l{\stackrel{‾}{c}}^{*}$. This ${\stackrel{‾}{r}}_{hkl}^{*}$ may replace the scattering vector in the structure factor equation to yield the following simplified relationship for a structure factor at a Bragg peak:
${F}_{hkl}={\sum }_{n=0}^{N-1}{f}_{n}{e}^{2\pi i\left(h{x}_{n}+k{y}_{n}+l{z}_{n}\right)}$
Structure Factor at hkl Bragg Peak

#### 12.4.1 Primitive Unit Cell (P)

The simplest structure factor example is a primitive unit cell with one atom per lattice point. This atom is placed at the origin for convenience. In this case, ${F}^{2}$ is the same for all reflections and does not depend on the $h,k,l$.
1 atom/lattice point $\to$N=1 ${x}_{n}={y}_{n}={z}_{n}=0$
${F}_{hkl}=f\left(\frac{sin\theta }{\lambda }\right)=f\left(\frac{1}{2{d}_{hkl}}\right)$
$F{F}^{*}={f}^{2}$ for all $hkl$

#### 12.4.2 Body-Centered (I) Bravais Lattice

1 atom/lattice point $\to$ N=1
The 2 identical atoms are located at positions ${r}_{1}$ and ${r}_{2}$:
${r}_{1}=000$, ${r}_{2}=\frac{1}{2}\frac{1}{2}\frac{1}{2}$
Figure 12.11: Body-centered (I) Bravais lattice
The following is the structure factor calculation for the body-centered Bravais lattice:
${F}_{hkl}=f\left(1+{e}^{2\pi i\left(\frac{h}{2}+\frac{k}{2}+\frac{l}{2}\right)}\right)=f\left(1+{e}^{\pi i\left(h+k+l\right)}\right)$
If the exponential is raised to an even power, then the structure factor is equal to $2f$ and the two atoms are scattering perfectly in phase.
${F}_{hkl}=f\left(1+1\right)=2f$ for all $h+k+l=2n$ (even sum)
If the exponential is raised to an odd power, then the structure factor is zero, and the two atoms are scattering perfectly out of phase.
${F}_{hkl}=f\left(1-1\right)=0$ for all $h+k+l=2n+1$ (odd sum)
Note:
${e}^{n\pi i}=\left(-1{\right)}^{n}\left\{\begin{array}{cc}-1,& for\phantom{\rule{6px}{0ex}}odd\phantom{\rule{6px}{0ex}}n\\ 1,& for\phantom{\rule{6px}{0ex}}even\phantom{\rule{6px}{0ex}}n\end{array}$
The $l$ index has no effect on the body-centered structure factor - for instance, 111, 112, and 113 all have the same $2f$ structure factor.

#### 12.4.3 Base Center (C) Bravais Lattice

The base centered lattices have two lattice points per unit cell.
1 atom per lattice point $\to$ N=2
The atoms are located at the following positions:
${r}_{1}=000$, ${r}_{2}=\frac{1}{2}\frac{1}{2}0$
Plugging these positions into the structure factor equation yields the following:
${F}_{hkl}=f\left(1+{e}^{\pi i\left(h+k\right)}\right)=f\left(1+\left(-1{\right)}^{h+k}\right)$
Again, there are two cases, one in which the two atoms are scattering perfectly in phase and one in which they are scattering perfectly out of phase.
${F}_{hkl}=\left\{\begin{array}{cc}2f& h+k=2n\\ 0& h+k=2n+1\end{array}$

#### 12.4.4 Face Center (F) Bravais Lattice

For the face centered lattice, if there were one atom per lattice point, there will be four atoms total in the non-primitive unit cell.
1 atom per lattice point $\to$ N=4
The atoms are located at the following positions with the unit cell:
${r}_{n}=\left\{000,\frac{1}{2}\frac{1}{2}0,0\frac{1}{2}\frac{1}{2},\frac{1}{2}0\frac{1}{2}\right\}$
Figure 12.12: Face-center (F) Bravais lattice
The face center lattice positions yield the following structure factor:
${F}_{hkl}=f\left(1+{e}^{\pi i\left(h+k\right)}+{e}^{\pi i\left(k+l\right)}+{e}^{\pi i\left(h+l\right)}\right)$
There are two distinct cases for the face center structure. That is, when the $hkl$ are unmixed (all even or all odd), then all four atoms scatter waves perfectly in phase and when the $hkl$ are mixed (even and odd) then the four atoms scatter waves that destructively interfere.
${F}_{hkl}=\left\{\begin{array}{cc}4f& h,k,l\phantom{\rule{6px}{0ex}}unmixed\\ 0& h,k,l\phantom{\rule{6px}{0ex}}mixed\end{array}$
For instance, the 022 planes would yield perfectly constructive interference and the 011 planes would yield perfectly destructive interference.
${F}_{022}=4f\left(\frac{1}{2{d}_{022}}\right)$, ${F}_{011}=0$

#### 12.4.5 Hexagonal Closed Pack (HCP)

For the hexagonal closed pack structure, there are two atoms of the same type per lattice point. Recall that the unit cell is hexagonal primitive, and there are two atoms in total. One atom is positioned at the origin 000 and the other atom is in the $\frac{1}{3}\frac{2}{3}\frac{1}{2}$ position.
Hexagonal-P Bravais Lattice
Ex. Zn, Ti, Mg
Figure 12.13: HCP structure
Using the hexagonal closed pack atom locations yields the following structure factor:
${F}_{hkl}=f\left[1+{e}^{2\pi i\left(\frac{h}{3}+\frac{2k}{3}+\frac{l}{2}\right)}\right]=f\left(1+{e}^{2\pi iq}\right)$, where $q=\frac{h+2k}{3}+\frac{l}{2}$
The structure factor times its complex conjugate eliminates the imaginary terms and simplifies the expression into the following:
$FF*={f}^{2}\left(2+{e}^{2\pi iq}+{e}^{-2\pi iq}\right)=2{f}^{2}\left(1+cos2\pi q\right)$
Because there are two atoms in the unit cell, we expect two extreme conditions - one in which the two scatter perfectly out of phase and one in which the two scatter perfectly out of phase. When the parameter $q$ is an integer, $cos2\pi q$ will equal $1$, producing a very strong reflection due to constructive interference. If $q$ is equal to an odd integer, then $2\pi q$ will equal $-1$, yielding a total intensity of zero.
Squaring the structure factor yields the following expression:
${|F|}^{2}=4{f}^{2}co{s}^{2}\pi q=4{f}^{2}co{s}^{2}\left[\pi \left(\frac{h+2k}{3}\right)+\frac{l}{2}\right]$
1. $l=2n$ (even) AND $h+2k=3m$
$⇒q=\left(m+n\right)\in Z\to {cos}^{2}\pi q=\left(±1{\right)}^{2}=1$
$|F{|}^{2}=4{f}^{2}$ e.g., (002), (112), ... very strong (in-phase scattering)
$l=2n$ AND $h+2k=3m±1$ $⇒q=\left(m+n\right)±\frac{1}{3}=m\text{'}±\frac{1}{3}$
$cos\pi q=cos\pi m\text{'}cos\frac{\pi }{3}±sin\pi m\text{'}sin\frac{\pi }{3}=cos\frac{\pi }{3}=\frac{1}{2}$
$|F{|}^{2}={f}^{2}$ e.g., (102), (200), (100)...weak reflection
2. $l=2n+1$ (odd) AND $h+2k=3m$
$⇒q=n+\frac{1}{2}+m=m\text{'}+\frac{1}{2}$$\to cos\pi q=cosm\text{'}\pi cos\frac{\pi }{2}-sinm\text{'}\pi sin\frac{\pi }{2}=0$
$|F{|}^{2}=0$ e.g., (001), (111), ... forbidden reflection
$l=2n+1$ AND $h+2k=3m±1$ $⇒q=n+\frac{1}{2}+m±\frac{1}{3}=m\text{'}±\frac{1}{6}$
$|F{|}^{2}=3{f}^{2}$ e.g., (103), (013), (101)... strong reflection
Therefore, there are four types of diffraction peak intensities for this hexagonal closed pack structure, ranging from forbidden reflections to very strong reflections.

#### 12.4.6 More than one atom type per unit cell

We will now analyze the case of more than one atom type per unit cell such as the rock-salt lithium fluoride structure.
LiF $\to$ NaCl structure  FCC
The following indices locate the lithium the lithium and fluorine atoms within the face centered cubic unit cell:
$L{i}^{+}$at 000 + $fct$
${F}^{-}$at $\frac{1}{2}00+fct$
$fct=\left\{000,\frac{1}{2}\frac{1}{2}0,0\frac{1}{2}\frac{1}{2},\frac{1}{2}0\frac{1}{2}\right\}$
The structure factor is given by the following equation:
${F}_{hkl}={\sum }_{n=1}^{8}{f}_{n}{e}^{2\pi i\left(h{x}_{n}+k{y}_{n}+l{z}_{n}\right)}$
${F}_{hkl}={f}_{L{i}^{+}}\left(1+{e}^{\pi i\left(h+k\right)}+{e}^{\pi i\left(k+l\right)}+{e}^{\pi i\left(h+l\right)}\right)+{f}_{{F}^{-}}{e}^{2\pi i\frac{h}{2}}\left(1+{e}^{\pi i\left(h+k\right)}+{e}^{\pi i\left(k+l\right)}+{e}^{\pi i\left(h+l\right)}\right)$
${F}_{hkl}=$
• $4\left({f}_{L{i}^{+}}+{f}_{{F}^{-}}\right)$ for $h,k,l$ all even $\to$(200), (222), etc.
• $4\left({f}_{L{i}^{+}}-{f}_{{F}^{-}}\right)$ for $h,k,l$ all odd $\to$(111), (311), etc.
• 0 for $h,k,l$ mixed $\to$(210), (001), etc.
Recall, in the lab that you saw: strong, weak, & forbidden FCC.
Kinematical scattering ignores multiple scattering due to low probabilities of re-scattering after a wave has scattered from one atom.
Note that the Laue method is useful for finding the symmetry of crystal orientations, but not useful for finding the lattice constant due to a variable wavelength. Furthermore, the Laue spot position is is independent of the length of $r*$and only dependent on direction.
Figure 12.14: Laue spot position

### 12.5 Width of Diffraction Peaks (single crystal)

Reciprocal lattice points have dimensions which are in proportional to the size of a crystal in a given direction. $\propto \frac{1}{a{M}_{1}},\frac{1}{b{M}_{2}},\frac{1}{c{M}_{3}}$
Figure 12.15: Diffraction peak
Recall the 3-D Interference Function whose ${M}_{1}$ component is given by the following:
$\frac{si{n}^{2}\left({M}_{1}\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{a}\right)}{si{n}^{2}\left(\pi \stackrel{‾}{Q}\cdot \stackrel{‾}{a}\right)}$
Figure 12.16: Widths of diffraction peaks in reciprocal space
For a $\theta -2\theta$ scan along the ${\stackrel{‾}{r}}_{100}^{*}$ direction, the diffraction peak width is proportional to the breadth of reciprocal lattice spot in ${\stackrel{‾}{r}}_{100}^{*}$ direction or $\frac{2}{a{M}_{1}}.$ The full width half max $\Delta \theta$ length is estimated by Scherrer's Formula
$\Delta \theta FWHM={\epsilon }_{1/2}\left(radians\right)={\left(\frac{ln2}{\pi }\right)}^{1/2}\frac{\lambda }{{D}_{hkl}cos\theta }=\frac{0.47\lambda }{{D}_{hkl}cos\theta }$
Scherrer's Formula
where ${D}_{hkl}$ is the crystal size (or X-ray coherence length) in the ${\stackrel{‾}{r}}_{hkl}^{*}$ direction.
e.g., ${D}_{100}=a{M}_{1}$
If crystallites, such as those in a powder sample, are on the order of a tenth of a micron, then the width of the pic will be one milliradian. $D\approx 1000\stackrel{˚}{A}\to {\epsilon }_{1/2}\approx 1mrad=0.06{}^{○}$
Therefore when measuring a diffraction pattern, you first locate the peaks which give information about the lattice constant and symmetry of the crystal. The shape of the peak is an indicator of the crystal size in different direction.
Each Bragg peak maximum is given by ${I}_{hkl}^{\left(eu\right)}={|{F}_{hkl}|}^{2}{M}^{2}$, where ${F}_{hkl}$ is affected by atomic positions in the unit cell.
Each Peak Width $\to$ crystal size
In addition, crystal size may be determined from a transverse or longitudinal scan.
${D}_{hkl}$ from $\theta -2\theta$ scan (longitudinal scan) ${D}_{\left(hkl\right)\perp }$ from $\omega$ scan (transverse scan)
What else can affect the height and width of the diffraction peaks?
• Angular divergence of incident beam - if an incident beam is very divergent
• Crystal defects - affects the width of the peak
To smooth over these two effects, we use angle integrated peak intensities to find relative values of $|F{|}^{2}$ (then compare to calculated $|F{|}^{2}$ from model).

### 12.6 Transverse scan $\left(\omega -scan\right)\equiv$ rock crystal through peak with $2\theta$ fixed

In a transverse scan, the sample is always at the center of the Ewald sphere. This experiment setup is illustrated in Figure 12.17 below. The incident beam wave vector of length $\frac{1}{\lambda }$ enters the sample and always hits the origin (000) of the reciprocal lattice on the Ewald sphere. For the $2\theta$ fixed scan, in which the angle of the sample $\omega$ is varied, the reciprocal lattice points rotate about 000, as illustrated by the black dashed lines on the left side of Figure 12.17. Since $2\theta$ is restricted to the width of the slit, the Ewald sphere intersection with lattice point at 010 will produce a line of points that does not capture the entire integrated intensity of the reciprocal lattice ellipsoid. Recall that this is a two dimensional projection, and that the line of points depicted by a horizontal red line across the blue ellipse in Figure 12.17, is actually a plane intersection of the 010 reciprocal lattice ellipsoid. If we remove the slit and do the same scan across $\omega$, the entire intensity is captured, as illustrated by the red ellipse in Figure 12.17. This is an angle integrated scan.
Figure 12.17: Transverse and Longitudinal Scan
The integrated counts or the energy in a peak is given by the following equation:
${I}_{hkl}=\frac{{I}_{0}}{\omega \text{'}}{r}_{e}^{2}{F}^{2}\frac{{\lambda }^{3}}{{V}_{uc}^{2}}\left(\frac{1+co{s}^{2}2\theta }{2sin2\theta }\right)\delta V$
$\omega \text{'}=\frac{d\omega }{dt}\equiv$ angular velocity of crystal rotation, a step rotational scan is now used
$\omega \text{'}=\frac{\Delta \omega }{\Delta t}\equiv \frac{angular\phantom{\rule{6px}{0ex}}step\phantom{\rule{6px}{0ex}}size}{time\phantom{\rule{6px}{0ex}}per\phantom{\rule{6px}{0ex}}step}\to$ step rotational scan, a faster scan yields fewer counts
${V}_{uc}=$ unit cell volume ${V}_{uc}=\left({V}_{uc}^{*}{\right)}^{-1}$
$\delta V={M}_{1}{M}_{2}{M}_{3}=$ Effective volume of irradiated sample
$\left(\frac{1+{cos}^{2}2\theta }{2}\right)=$Polarization Factor for unpolarized beam
$\left(sin2\theta {\right)}^{-1}=$ Lorentz Factor for single crystal diffraction with constant $\omega \text{'}$, $\theta \propto$ time taken to sweep Ewald sphere through a relative point (reciprocal lattice point)
Note that the tangential velocity of relative point through Ewald sphere is given by $\omega \text{'}{\stackrel{‾}{r}}_{hkl}^{*}$.
In Figure 12.18, notice that the 200 and 400 diffraction conditions are scanned at different speeds due to the geometry of the Ewald sphere . The intensity for the 200 lattice point will be greater because the Ewald sphere travels and scans more quickly through this point yielding a smaller intensity relative to the 400 lattice point.
This is another reason why ${I}_{200}>{I}_{400}$ for LiF
Figure 12.18: ${\stackrel{‾}{r}}_{200}^{*}$ vs ${\stackrel{‾}{r}}_{400}^{*}$ scan

### 12.7 Polycrystalline Aggregate (Powder sample)

In the powder method, the crystal is reduced to a fine powder of thousands of randomly oriented small single crystalline grains to be placed in a beam of monochromatic x-rays. In reciprocal space, this random orientation results in a continuum of concentric spherical shells with a range of ${r}_{hkl}^{*}$. The illustration in Figure 12.19 below illustrates a reciprocal lattice vector range from ${r}_{200}^{*}$ to ${r}_{400}^{*}$.
Figure 12.19: Ewald spheres for powder sample
The surface density of $hkl$ relative points is equal to the number of crystallites divided by the surface are a of the spherical shell.
• Surface density $\propto \frac{N}{4\pi {r}_{hkl}^{*2}}$
We assume that since the powder sample is isotropic, each sphere has uniform surface density. If a texture sample is used, a preferred direction may exist and the density will be non-uniform.

#### 12.7.1 Ewald construction for one ${r}_{hkl}^{*}$ with uniform density

The Ewald sphere for one ${r}_{hkl}^{*}$ with uniform density is illustrated in Figure 12.20 below. This blue Ewald sphere has a definite center and radius, and the monochromatic incident beam $\frac{{\stackrel{‾}{s}}_{0}}{\lambda }$ passes through its center to hit the origin of the sample's reciprocal lattice sphere at 000. The intersection of the Ewald sphere and the sample's sphere for one ${r}_{hkl}^{*}$, marked by the red dashed circle, represents the entire set of allowed scattered wave vectors $\frac{\stackrel{‾}{s}}{\lambda }$ that will emanate from the center of the Ewald sphere. This circular locus of points will produce a diffraction cone of scattered wave vectors beginning from the origin of the Ewald sphere and ending at the circular intersection of the two spheres. In a small slit powder diffractometer, only a segment of the circle is captured as the slit scans around $2\theta .$ Placing a film downstream from the sample would show the entire Bragg ring .
This satisfies the Laue condition in which the scattering vector is equal to a specific reciprocal lattice vector of the sample.
$\frac{\stackrel{‾}{s}-\stackrel{‾}{{s}_{0}}}{\lambda }={\stackrel{‾}{r}}_{hkl}^{*}$

#### 12.7.2 Calculated Diffracted Intensity (Power)

$P={I}_{0}\frac{cos\theta }{2}{r}_{e}^{2}{F}^{2}\frac{{\lambda }^{3}}{{V}_{uc}^{2}}\left(\frac{1+co{s}^{2}2\theta }{2sin2\theta }\right)N\delta V$This is for entire Bragg Ring of circumference:
$C=2\pi R\left(sin2\theta \right)$, $cos\theta \propto \frac{locus\phantom{\rule{6px}{0ex}}circle}{great\phantom{\rule{6px}{0ex}}circle\phantom{\rule{6px}{0ex}}of\phantom{\rule{6px}{0ex}}{r}^{*}}$
Figure 12.21: Spherical film of radius R
A horizontal diffractometer with $2\theta$ detector slit at R and vertical slit height $l$ records a fraction of the Bragg ring $\frac{l}{2\pi Rsin2\theta }$.

#### 12.7.3 Multiplicity Correction

E.g. Cubic:
${r}_{100}^{*}={r}_{010}^{*}={r}_{001}^{*}={r}_{\stackrel{‾}{1}00}^{*}={r}_{0\stackrel{‾}{1}0}^{*}={r}_{00\stackrel{‾}{1}}^{*}=\frac{1}{a}$
$\therefore {m}_{100}=6,$ ${m}_{h00}=6$
Note: C&S $m\to p$

#### 12.7.4 Powder Diffraction Intensity

Adding in the multiplicity correction factor, the measure of the power for a diffraction condition of a given hkl is given by the following:
$P{\text{'}}_{hkl}=\frac{{I}_{0}l}{16\pi R}{r}_{e}^{2}\frac{{\lambda }^{3}}{{V}_{uc}^{2}}{F}_{hkl}^{2}{m}_{hkl}\left(\frac{1+co{s}^{2}2\theta }{sin2\theta sin\theta }\right)Vs$
Where${V}_{s}=$ effective sample volume $N\delta V$ and $LP=\frac{1+{cos}^{2}2\theta }{sin2\theta sin\theta }=$
Lorentz-Polarization factor for powder diffraction
Note: $sin2\theta =2sin\theta cos\theta$
Why do we use a symmetric reflection geometry for powder diffraction, if the random orientation of the crystals in the powder allow for various sampling angles and diffraction conditions at once?
(i.e. $\omega =\theta =\frac{2\theta }{2}$ or $\theta -2\theta$ scan)

#### 12.7.5 Absorption Effects

Recall that the linear absorption coefficient is represented by $\mu$ and that the absorption length of a material is given by $\frac{1}{\mu }$. A low Z material like water or a hydrocarbon has a longer absorption length $\frac{1}{\mu }$ than a solid metal.
Consider a large planar sample with thickness t in reflection geometry as illustrated in Figure 12.22. The incident beam with area A across enters at an angle $\alpha$, and scatters from a differential volume at some depth z. The $2\theta$ detector is lined up to pick up the scattered intensity at the angle $\beta$. In general, $\alpha$ and $\beta$ do not have to be equal.
Figure 12.22: Planar sample in reflection geometry
the beam area produces a projected area of the slab that is $A/sin\alpha$
Therefore the volume element is given by $dV=\frac{A}{sin\alpha }dz$
The diffracted intensity at depth z is given by the following:
$d{I}_{z}=C\frac{A}{sin\alpha }dz{I}_{0}{e}^{-uz/sin\alpha }{e}^{-uz/sin\beta }$
There is an absorption effect that attenuates the beam as it enters and exits sample, which is accounted for by the following two transmission factors:
${e}^{-uz/sin\alpha }$- in transmission factor
${e}^{-uz/sin\beta }$- out transmission factor
To calculate the total intensity, we integrate over the total thickness:
$I={\int }_{0}^{t}d{I}_{z}={I}_{0}C\frac{A}{\mu }{\left(1+\frac{sin\alpha }{sin\beta }\right)}^{-1}$
Assuming the thickness is much greater than the attenuation length, or $\mu t\gg 1$ - for a typical oxide material this thickness is about 0.5 mm.
${\int }_{0}^{t}d{I}_{z}={\int }_{0}^{\infty }d{I}_{z}$ “t” drops out by assuming $\mu t\gg 1$ or $t\to \infty$
For symmetric reflection $\alpha =\beta =\theta$$\to$$\phantom{\rule{10px}{0ex}}I={I}_{0}C\frac{A}{2\mu }$ $\frac{A}{2\mu }=$ effective volume = ${V}_{eff}$
Note that the effective volume, ${V}_{eff}$ , is constant and independent of $\theta$ for $\alpha =\beta =\theta$. As you increase the angle in this geometry, you are increasing the depth but decreasing the effective area. Therefore, it is convenient to use symmetric reflection geometry for a powder diffraction experiment, in order to be able to avoid varying absorption effects from a varying effective volume.
We want as much material in front of the beam to yield a high scattering intensity, but we also want a small enough thickness to avoid too much absorption. There is an optimal thickness in this transmission geometry, $\frac{1}{\mu },$ that yields the maximum scattering compromised correctly with the absorption. Recall that $\frac{1}{\mu }$ is the absorption length.
Effective Volume: $\frac{At}{cos\theta }{e}^{\frac{-\mu t}{cos\theta }}$
Here,${V}_{eff}$ depends on $\theta$.
Figure 12.23: Transmission geometry

#### 12.7.6 Summarize: Powder diffractometer in symmetric reflection geometry

The area under the diffraction peak is proportional to:
${I}_{hkl}=|{F}_{hkl}{|}^{2}{m}_{hkl}\left(\frac{1+co{s}^{2}2\theta }{si{n}^{2}\theta cos\theta }\right){e}^{-2M}$
${F}_{hkl}$- structure factor $hkl$ and $sin\theta /\lambda$
${m}_{hkl}$- multiplicity $\left(hkl\right)$
$\left(\frac{1+{cos}^{2}2\theta }{{sin}^{2}\theta cos\theta }\right)$- Lorentz polarization factor, $\theta$
${e}^{-2M}$- temperature dependent Debye-Waller factor $sin\theta /\lambda$ accounting for atomic vibrations
Our objective is to measure the integrated intensities ${I}_{hkl}$'s and use the structure factor ${F}_{hkl}$'s to determine the crystal structure.
$LP={\left(\frac{1+{cos}^{2}2\theta }{{sin}^{2}\theta cos\theta }\right)}_{←Geometrical\phantom{\rule{6px}{0ex}}correction}^{←Polarization\phantom{\rule{6px}{0ex}}correction}$
Figure 12.24: LP vs. $\theta$

#### 12.7.7 Multiplicity Factor

${m}_{hkl}$: number of $\left(hkl\right)$ planes in the $\left\{hkl\right\}$ family for a given crystal system
Example:
0kk
 022 $0\stackrel{‾}{2}\stackrel{‾}{2}$ 202 $\stackrel{‾}{2}0\stackrel{‾}{2}$ 220 $\stackrel{‾}{2}\stackrel{‾}{2}0$ $02\stackrel{‾}{2}$ $0\stackrel{‾}{2}2$ $20\stackrel{‾}{2}$ $\stackrel{‾}{2}02$ $2\stackrel{‾}{2}0$ $\stackrel{‾}{2}2$0
Cubic $\to {m}_{0kk}=12$, since all 12 $|{r}^{*}|$'s are equal
triclinic $a\ne b\ne c$, $\alpha \ne \beta \ne \gamma$
${r}_{022}^{*}\ne {r}_{220}^{*}$ triclinic ${m}_{hkl}=2$
${r}_{022}^{*}={r}_{0\stackrel{‾}{2}\stackrel{‾}{2}}^{*}$ Since only ${r}_{\stackrel{‾}{h}\stackrel{‾}{k}\stackrel{‾}{l}}^{*}={r}_{hkl}^{*}$
See Cullity and Stock Appendix 11
 Cubic: $\frac{hkl}{48*}$ $\frac{hhl}{24}$ $\frac{0kl}{2{4}^{*}}$ $\frac{0kk}{12}$ $\frac{hhh}{8}$ $\frac{00l}{6}$ Hexagonal and Rhombohedral: $\frac{hk\cdot l}{2{4}^{*}}$ $\frac{hh\cdot l}{1{2}^{*}}$ $\frac{0k\cdot l}{1{2}^{*}}$ $\frac{hk\cdot 0}{1{2}^{*}}$ $\frac{hh\cdot 0}{6}$ $\frac{0k\cdot 0}{6}$ $\frac{00\cdot l}{2}$ Tetragonal: $\frac{hkl}{1{6}^{*}}$ $\frac{hhl}{8}$ $\frac{0kl}{8}$ $\frac{hk0}{{8}^{*}}$ $\frac{hh0}{4}$ $\frac{0k0}{4}$ $\frac{00l}{2}$ Orthorhombic: $\frac{hkl}{8}$ $\frac{0kl}{4}$ $\frac{h0l}{4}$ $\frac{hk0}{4}$ $\frac{h00}{2}$ $\frac{0k0}{2}$ $\frac{00l}{2}$ Monoclinic: $\frac{hkl}{4}$ $\frac{h0l}{2}$ $\frac{0k0}{2}$ Triclinic: $\frac{hkl}{2}$
Table 11: Multiplicity factors for powder photographs

#### 12.7.8 Indexing Powder Diffraction Patterns

We can determine ${r}_{hkl}^{*}$ lengths (not directions) from experimentally measured $\theta \text{'}s$ and a known $\lambda$ using the following relationships:
$\frac{2sin{\theta }_{hkl}}{\lambda }={r}_{hkl}^{*}=\left(\stackrel{‾}{r}*\cdot \stackrel{‾}{r}*{\right)}^{\frac{1}{2}}$
$=\left[{h}^{2}a{*}^{2}+{k}^{2}b{*}^{2}+{l}^{2}c{*}^{2}+2hk{a}^{*}b*cos\gamma *+2klb*c*cos\alpha *+2hl{a}^{*}c*cos\beta *{\right]}^{\frac{1}{2}}$
First index the series of peaks (i.e. determine $\left(hkl{\right)}_{n}$), then determine the lattice constants of the Bravais lattice : $a,b,c,\alpha ,\beta ,\gamma$
Initially check for cubic, the simplest case, where $a=b=c,\phantom{\rule{6px}{0ex}}\alpha =\beta =\gamma =90{}^{○}$ .
Cubic: $\underset{measured}{\underset{⏟}{r{*}^{2}}}=\underset{unknown}{\underset{⏟}{a{*}^{2}\left[{h}^{2}+{k}^{2}+{l}^{2}\right]}}$ subject to Bravais lattice conditions.
 ${h}^{2}+{k}^{2}+{l}^{2}$ $hkl$ P F I Diamond 1 100 x 2 110 x x 3 111 x x x 4 200 x x x 5 210 x 6 211 x x 7 8 220 x x x x 9 300,221 x,x 10 310 x x 11 311 x x x 12 222 x x x 13 320 x 14 321 x x 15 16 400 x x x x
Table 12: Determining unit cells from ${h}^{2}+{k}^{2}+{l}^{2}$
If $\left\{{r}^{*}{}_{i}^{2}\right\}=$
$C\left\{1,2,3,4,5,6,8,...\right\}\to$ Cubic-P
$C\left\{1,\frac{4}{3},\frac{8}{3},\frac{11}{3},4,...\right\}\to$ Cubic-F
$C\left\{1,2,3,4,5,6,7,8\right\}\to$ Cubic-I
$C\left\{1,\frac{8}{3},\frac{11}{3},\frac{16}{3},\frac{19}{3},...\right\}\to$ Diamond
Note that we need to measure ${r}^{*}=2sin\vartheta /\lambda$ for at least the first seven peaks to distinguish between Cubic-P and Cubic-I.
Tetragonal and hexagonal (“uniaxial”)
Tetragonal: ${r}_{hkl}^{*2}={a}^{*2}\left({h}^{2}+{k}^{2}\right)+{c}^{*2}{l}^{2}={a}^{*2}\left[\left({h}^{2}+{k}^{2}\right)+{\left(\frac{{c}^{*}}{{a}^{*}}\right)}^{2}{l}^{2}\right]$
Hexagonal: $cos\gamma *=\frac{1}{2}$
${r}_{hkl}^{*2}={a}^{*2}\left({h}^{2}+{k}^{2}+hk\right)+{c}^{*2}{l}^{2}={a}^{*2}\left[\left({h}^{2}+{k}^{2}+hk\right)+{\left(\frac{{c}^{*}}{{a}^{*}}\right)}^{2}{l}^{2}\right]$
${\left(\frac{{c}^{*}}{{a}^{*}}\right)}^{2}={\left(\frac{a}{c}\right)}^{2}$ is unknown
is 1st peak ?:
100 or 001 (P)
110 or 002 (Tetragonal I)
100 or 002 (HCP)
${sin}^{2}\theta =\frac{{\lambda }^{2}}{4}{r}_{hkl}^{*2}=A\left({h}^{2}+{k}^{2}\right)+C{l}^{2}←Tetragonal$
$A\left({h}^{2}+{k}^{2}+hk\right)+C{l}^{2}←Hexagonal$

#### 12.7.9 Example 1

$\lambda =1.542\stackrel{˚}{A}$
 Line No. ${sin}^{2}\theta$ $\frac{\left({sin}^{2}\theta {\right)}_{n}}{\left({sin}^{2}\theta {\right)}_{1}}$ ${h}^{2}+{k}^{2}+{l}^{2}$ $hkl$ 1 0.0603 1 3/3 3 111 2 0.1610 2.67 8/3 8 220 3 0.221 3.67 11/3 11 311 4 0.322 5.34 16/3 16 400 5 0.383 6.35 19/3 19 331 6 0.484 8.03 24/3 24 422 7 0.545 9.04 27/3 27 333 8 0.645 10.7 32/3 32 440
Table 13: Indexing Powder Patterns, Example 1
Ratios $\to$Cubic
$\frac{\left({sin}^{2}\theta {\right)}_{n}}{\left({sin}^{2}\theta {\right)}_{1}}=\frac{\left({h}^{2}+{k}^{2}+{l}^{2}{\right)}_{n}}{\left({h}^{2}+{k}^{2}+{l}^{2}{\right)}_{1}}$
unmixed hkl: Cubic F
$h+k+l\ne 4n±2$: diamond cubic
What is “a”?
${r}_{400}^{*}=16a{*}^{2}=\frac{16}{{a}^{2}}=\frac{4{sin}^{2}{\theta }_{400}}{{\lambda }^{2}}$
$a=\frac{4\lambda }{2sin{\theta }_{400}}=\frac{2×1.542\stackrel{˚}{A}}{\left(0.322{\right)}^{1/2}}=5.434\stackrel{˚}{A}$
Silicon a=$5.431\stackrel{˚}{A}$

#### 12.7.10 Example 2

$\lambda =1.542\stackrel{˚}{A}$
 Line No. ${sin}^{2}\theta$ $\left({sin}^{2}\theta {\right)}_{n}$ $\left({sin}^{2}\theta {\right)}_{1}$ $nA+mC$ $\left({sin}^{2}{\right)}_{calc}$ hkl 1 0.0806 1.00 A+C 0.0806 101 2 0.0975 1.21 4C 0.0976 002 3 0.1122 1.39 2A 0.1124 110 4 0.210 2.61 2A+4C 0.210 112 5 0.226 2.80 4A 0.225 200 6 0.274 3.40 A+9C 0.276 103 7 0.305 3.78 5A+C 0.306 211 8 0.321 3.98 4A+4C 0.323 202
Table 14: Indexing powder patterns, example 2
Ratios $\to$not cubic
Note Relations:
(4)=(2)+(3)
(5)=2(3)
(7)=(5)+(1)
(8)=4(1)
Assume Tetragonal:
${sin}^{2}\theta =A\left({h}^{2}+{k}^{2}\right)+C{l}^{2}$
Tetragonal P$\to$(1) = 100 or 001
No peak at 2(1) $\to$not P
2(1) =110, if (1) = 100
Tetragonal I: (1) =110,101, or 002
Try (1)=101 $\to {sin}^{2}\theta =A+C$
${sin}^{2}\theta =A\left({h}^{2}+{k}^{2}{\right)}_{n}+C\left({l}^{2}{\right)}_{n}$
hk 00 10 11 20 21 22 30 31 32 ... $l$ 0 1 2 3 4 ...
${h}^{2}+{k}^{2}$0 14 5 8 9 10 13 ...${l}^{2}$0 1 4 9 16 ...
If $\left(1\right)=101=A+C\to \left(2\right)=002=4C$ *
Try:
$.\begin{array}{c}\left(2\right)\to 4C=0.0975\\ \left(1\right)\to A+C=0.083\end{array}\right\}A=0.0562,C=0.0244$
Note $\left(3\right)=2A\to 110=A\left({1}^{2}+{1}^{2}\right)+C\left({0}^{2}\right)$
Check to see if each $\left({sin}^{2}\theta {\right)}_{n}=A\left({h}^{2}+{k}^{2}{\right)}_{n}+C{l}_{n}^{2}$ for A = 0.0652 and C = 0.0244
Conclusion: Tetragonal I
Lattice constant: a=? c=?
$\left(2\right)\to 2c*=\frac{2}{c}=\frac{2sin{\theta }_{002}}{\lambda }⇒c=\frac{\lambda }{sin{\theta }_{002}}=\frac{1.542\stackrel{˚}{A}}{\left(0.0975{\right)}^{\frac{1}{2}}}=4.94\stackrel{˚}{A}$
$\left(3\right)\to \sqrt{2}a*=\frac{\sqrt{2}}{a}=\frac{2sin{\theta }_{110}}{\lambda }⇒a=\frac{\lambda }{\sqrt{2}sin{\theta }_{110}}=3.62\stackrel{˚}{A}$Indium
Procedure for hexagonal similar to tetragonal except:
$\left({sin}^{2}\theta {\right)}_{n}=A\left({h}^{2}+{k}^{2}+hk{\right)}_{n}+{l}^{2}C$
where ${h}^{2}+{k}^{2}+hk=\left\{0,1,3,4,7,9,12,13,...\right\}$
Indexing difficulty increases as crystal symmetry decreases: need computer
Note: When $K{\alpha }_{1}$ and $K{\alpha }_{2}$ are unresolved (at lower $2\theta$), use avg ${\lambda }_{K\alpha }=2/3{\lambda }_{K\alpha 1}+1/3{\lambda }_{K{\alpha }_{2}}=1.5418\stackrel{˚}{A}$
At higher $2\theta ,$$K{\alpha }_{1}$ and $K{\alpha }_{2}$ may be resolved, use correct ${\lambda }_{K\alpha 1}$ and ${\lambda }_{K\alpha 2}$, not ${\lambda }_{K\alpha }$

#### 12.7.11 Quantitative Analysis

Recall: ${I}_{hkl}=C|{F}_{hkl}{|}^{2}\frac{{m}_{hkl}}{{V}_{C}^{2}}LP\frac{A}{2\mu }$ , ignore ${e}^{-2M}$
Consider a 2 phase powder with:
${v}_{A}=$ volume fraction of phase A, $1-{v}_{A}={v}_{B}=$ volume fraction of phase B
Phase A diffraction peaks ${I}_{A}=C{\left[|{F}_{hkl}{|}^{2}\frac{{m}_{hkl}}{{V}_{C}^{2}}LP\right]}_{A}{v}_{A}\frac{A}{2\mu }=C{K}_{A}{v}_{A}\frac{A}{2\mu }$
${K}_{A}$ varies with $hkl$
$\mu$ varies with ${v}_{A}$
Phase B diffraction peaks
${I}_{B}=C{K}_{B}\left(1-{v}_{A}\right)\frac{A}{2\mu }$
$\mu$ is the same
Hence, $\begin{array}{cc}\frac{{I}_{A}}{{I}_{B}}=\frac{{K}_{A}}{{K}_{B}}\frac{{v}_{A}}{\left(1-{v}_{A}\right)}& \left(12.1\right)\end{array}$
To determine volume fraction for mixture of 2 powder crystalline phases A&B
$\begin{array}{cc}\underset{}{\underset{⏟}{\frac{{I}_{A}}{{I}_{B}}}}=\stackrel{}{\stackrel{⏞}{\frac{{K}_{A}}{{K}_{B}}}}\underset{}{\underset{⏟}{\frac{{v}_{A}}{\left(1-{v}_{A}\right)}}}& \left(12.2\right)\end{array}$
The relative intensities of peaks for each phase should be checked to ensure an ideally random powder sample. Texture will cause errors in quantitative analysis.
In the analysis of the integrated area under the diffraction peaks, the Debye-Waller factor, ${e}^{-2{M}_{A}};$ ${e}^{-2{M}_{B}}$ , may be included if known. The two different phases are made up of atoms in two different lattices with different vibrational amplitudes. The vibrational amplitude $M$ is given by the following:
$\begin{array}{cc}M=\frac{2{\pi }^{2}⟨{u}_{hkl}^{2}⟩}{{d}_{hkl}^{2}}& \left(12.3\right)\end{array}$Here $\sqrt{⟨{u}_{hkl}^{2}⟩}$ is the root-mean-square (rms) vibrational amplitude in the hkl direction. These vibrations are typically isotropic with a magnitude of $\approx$0.1$\stackrel{˚}{A}$. An increase in temperature increases the vibrational amplitude of atoms.

### 12.8 Rotating Crystal Method

In the rotating crystal method, a single crystal is mounted with an axis that is normal to a monochromatic beam. A cylindrical film encloses the single crystal, which records a diffraction pattern as the crystal is rotated. This setup is illustrated in Figure 12.25. Whereas older films required chemical exposure to develop, x-ray sensitive image plates are used today and diffraction patterns are read out by a laser.
In this procedure, the crystal is rotated smoothly about a chosen axis that coincides with the axis of the film cylinder and different planes cause diffraction peaks at points in the rotation. Diffraction peaks occur when a set of lattice planes make the correct Bragg angle for reflection. In the setup illustrated in Figure 12.25, prior to inserting the crystal in the camera, the c-axis of the crystal's unit cell was determined and the crystal is then rotated about this axis. Due to this rotation about the c-axis, the spots appear in layers with heights according to constant $l$ index values. All of the center spots are $hk0$ .
The rotating crystal method helps determine the structure of a crystal in an organized manner, and is useful for methods where a single crystal can be attained as well as when the $c$ and ${c}^{*}$ axes of a crystal's unit cell coincide with each other.

#### 12.8.1 Rotating Crystal Method

1. Monochromatic X-ray beam diffraction from single crystal surrounded by film cylinder.
2. The crystal continually rotated about the axis that coincides with film cylinder axis
3. If rotation axis = unit cell axis (e.g., $\stackrel{‾}{c}$) then reflection pattern (spots) on film form circles at heights corresponding to constant (e.g., $l$)
• This is for unit cells where ${\overline{c}}^{*}||\phantom{\rule{6px}{0ex}}\overline{c}$
Figure 12.25: Rotating crystal method
Figure 12.26 below illustrates a developed rotating-crystal film of a quartz crystal. The extra streaks around the center of the pattern are due to Brems-strahlung radiation not removed by the filter.
Figure 12.26: Rotating-crystal pattern of a quartz crystal (hexagonal) rotated about its c axis. Filtered copper radiation. (Courtesy of B.E. Warren.)

#### 12.8.2 Reciprocal Lattice Treatment of Rotating-Crystal Method

How do we picture the rotating-crystal method in reciprocal space? Visualizing the method in reciprocal space, the Ewald sphere is the instrument, and the reciprocal lattice structure is the crystal sample. An orthorhombic crystal crystal centered at the Ewald sphere may be rotated about the c-axis, as illustrated in Figure 12.27 Recall that the incident wave vector always points from the center of the Ewald sphere to the origin of the reciprocal lattice. If the crystal is rotating about the c-axis, then reciprocal lattice rotates about its ${\stackrel{‾}{c}}^{*}$ or $\stackrel{‾}{{b}_{3}}$ axis. A diffraction peak will appear when a reciprocal lattice point touches the Ewald sphere. In this camera, all diffracted beams will be detected and recorded except those that travel close to the cylinder's central axis and miss the film.
${\stackrel{‾}{b}}_{1}={\stackrel{‾}{a}}^{*}$, ${\stackrel{‾}{b}}_{2}={\stackrel{‾}{b}}^{*}$, ${\stackrel{‾}{b}}_{3}={\stackrel{‾}{c}}^{*}$
Figure 12.27: Rotating crystal method in reciprocal space
• The reciprocal lattice rotates about ${\stackrel{‾}{r}}_{100}^{*}={\stackrel{‾}{c}}^{*}$ as the crystal rotates about $\stackrel{‾}{c}.$
• Spots appear on the film when reciprocal lattice points coincide with the Ewald sphere.
• Recall the Laue condition, when $\frac{\stackrel{‾}{S}}{\lambda }-\frac{{\stackrel{‾}{S}}_{0}}{\lambda }=\stackrel{‾}{Q}={\stackrel{‾}{r}}_{hkl}^{*}$
We will divide the scattering vector into horizontal and vertical components, ${\stackrel{‾}{r}}_{H}^{*}$ and ${\stackrel{‾}{r}}_{V}^{*}.$The $\alpha$ and $\beta$ angles determine the in plane and out of plane components of ${\stackrel{‾}{r}}_{hkl}^{*}.$
After the scattered diffracted beam places various spots on the film, the film is unrolled and the specific $x$ and $y$ coordinates of these characteristic spots are noted. Using these coordinates and the radius of the cylinder, $\alpha$ and $\beta$ may be determined using the following equations:
$\begin{array}{cc}tan\beta -y/R& \left(12.4\right)\end{array}$
$\begin{array}{cc}\alpha =x/R& \left(12.5\right)\end{array}$
$\begin{array}{cc}|{\stackrel{\to }{r}}_{V}^{*}|=\left(\frac{1}{\lambda }\right)sin\beta & \left(12.6\right)\end{array}$
$\begin{array}{cc}|{\stackrel{\to }{r}}_{H}^{*}|=|h\stackrel{\to }{a}*+k\stackrel{\to }{b}*{|}^{2}=\frac{1}{{\lambda }^{2}}\left[2\left(1-cos\alpha cos\beta \right)-si{n}^{2}\beta \right]& \left(12.7\right)\end{array}$
Example: $\lambda =1.542\stackrel{˚}{A},$${a}^{*}=0.25{\stackrel{˚}{A}}^{-1}$, ${b}^{*}=0.15{\stackrel{˚}{A}}^{-1}$, ${c}^{*}=0.36{\stackrel{˚}{A}}^{-1}$, and $R=50$ mm
Figure 12.29: Convert x,y coordinates of film spot to $\alpha ,\beta$

#### 12.8.3 Ewald Construction for Rotating Crystal Method

Define angles:
$\alpha =x/R\to$ in-plane angle
$tan\beta =y/R\to$ elevation angle
Separate ${\stackrel{‾}{r}}_{hkl}^{*}={\stackrel{‾}{r}}_{H}^{*}+{\stackrel{‾}{r}}_{V}^{*}$
$|{\stackrel{\to }{r}}_{V}^{*}|=|\frac{\stackrel{‾}{S}}{\lambda }|sin\beta \to \lambda {r}_{V}^{*}=sin\beta$
Note: $cos2\theta =cos\alpha cos\beta$
${\lambda }^{2}\left({r}_{H}^{*}{\right)}^{2}=2\left(1-cos\alpha cos\beta \right)-{sin}^{2}\beta$
Bragg's law below
$\left({r}_{hkl}^{*}{\right)}^{2}=\left({r}_{H}^{*}{\right)}^{2}+\left({r}_{V}^{*}{\right)}^{2}=\frac{4{sin}^{2}\vartheta }{{\lambda }^{2}}$
Figure 12.30: Ewald sphere for rotating crystal method
A. For hexagonal ,tetragonal, cubic, orthorhombic, etc., where $\stackrel{‾}{c}$ coincides with ${\stackrel{‾}{c}}^{*}$, measuring the $y$ coordinate allows us to calculate the elevation angle, and the vertical component of the scattering vector $y\to \beta \to {r}_{V}^{*}⇒{r}_{00l}^{*}\to c$ determined directly
By remounting the crystal to rotate about $\stackrel{‾}{a}$ or $\stackrel{‾}{b}$, we can determine $a$ and $b$.
B. Indexing a reflection in layer line
For a given layer line, $\beta$ is constant.
${\stackrel{‾}{r}}_{H}^{*}=h{\stackrel{‾}{a}}^{*}+k{\stackrel{‾}{b}}^{*}$for $l=$ constant layer line
${\lambda }^{2}\left({r}_{H}^{*}{\right)}^{2}=\left[{h}^{2}\left({a}^{*}{\right)}^{2}+{k}^{2}\left({b}^{*}{\right)}^{2}+2hk{a}^{*}{b}^{*}cos{\gamma }^{*}\right]{\lambda }^{2}$
$=2\left(1-cos\alpha cos\beta \right)-{sin}^{2}\beta$
$\alpha =x/R$ $tan\beta =y/R$
$\therefore x,y\to {r}_{H}^{*},{r}_{V}^{*}$

#### 12.8.4 Bernal Chart

Graphical method for converting $x,y\to \lambda {r}_{H}^{*},\lambda {r}_{V}^{*}$
Horizontal Lines $\to$ constant $l$ for $\stackrel{‾}{c}$ axis rotating
$\lambda {r}_{H}^{*}$ lines equally incremented from 0 to 2.0. The maximum is when $2\theta =\pi$
$\lambda {r}_{H}^{*}=0\to x=0\to \alpha =0\to hk=00$
$\lambda {r}_{H}^{*}=2\to x=\pi R\to \alpha =\pi \to {r}^{*}={r}_{H}^{*}=\frac{2}{\lambda }$ for $l=0$
Start with $l=0$ layer
• Assume ${a}^{*}={b}^{*},{\gamma }^{*}=90{}^{○}\to \frac{1}{a}\left({h}^{2}+{k}^{2}{\right)}^{\frac{1}{2}}$
• Assume ${a}^{*}\ne {b}^{*},{\gamma }^{*}=90{}^{○}\to {r}_{H}^{*}={\left(\frac{{h}^{2}}{{a}^{2}}+\frac{{k}^{2}}{{b}^{2}}\right)}^{1/2}$
Figure 12.31: Schematic view of film overlain on a Bernal chart. Note that the X-ray diffraction spots lie along rows of constant $\xi$and$\zeta$. (Only a few possible reflections are indicated)
Once the lattice constants and orthorhombic unit cell have been determined, index and determine Lattice Type: P, I, C, F.
Measure X and Y from the film
R = 50 mm , $\lambda =1.542\stackrel{˚}{A}$
For $l=0$ layer, $\to \beta =0\to {r}_{H}^{*}=\frac{1}{\lambda }\sqrt{2\left(1-cos\alpha \right)}$
 X [mm] $\alpha =X/R$ $\alpha$[${}^{○}$] ${r}_{H}^{*}\left[\stackrel{˚}{{A}^{-1}}\right]$ hkl 22.4 0.488 25.67 0.288 110 23.3 0.466 26.70 0.299 020 39.6 0.792 45.38 0.500 200 40.8 0.816 46.75 0.515 130 46.6 0.932 53.40 0.583 220 48.1 0.962 55.12 0.600 040 63.1 1.262 72.31 0.765 310 64.6 1.292 74.03 0.781 240 65.5 1.31 75.06 0.790 150 74 1.48 84.80 0.875 330 76.7 1.534 87.89 0.900 060 88 1.76 100.84 1.000 400
${r}_{H}^{*}$defines radius of circle in the $l=0$ reciprocal space plane
Table 15: Conversion of X to hkl
Figure 12.33 below demonstrates the graphical solution for layer $l=0.$
Plot hk0 lattice: ${a}^{*}=0.25{\stackrel{˚}{A}}^{-1}$ & ${b}^{*}=0.15{\stackrel{˚}{A}}^{-1}$
Plot circles with radii ${r}_{H}^{*}$, shown in Figure 12.33 below.
The blue squares represent the reciprocal lattice points corresponding to the planes that caused diffraction to occur.
Figure 12.33: Graphical Solution for ${r}_{H}^{*}={h}^{2}{a}^{*2}+{k}^{2}{b}^{*2}$
Repeat this procedure for $l=1$ layer $\to \beta =ta{n}^{-1}\left(y/R\right)$
$\begin{array}{cc}{r}_{H}^{*}=\frac{1}{\lambda }\sqrt{2\left(1-cos\alpha cos\beta -si{n}^{2}\beta }& \left(12.8\right)\end{array}$

## 13 Temperature Effects

The atoms in a crystal vibrate about equilibrium lattice positions, even at ${0}^{○}$K and the vibrational amplitudes increase as temperature increases. With increasing atomic kinetic energies, the unit cell expands and the lattice constant becomes greater. This temperature effect, also known as the Debye-Waller effect, affects the area under diffraction peaks even at room temperature. It is important to account for this effect since the area under diffraction peaks gives information about the crystal structure via the structural factor.
Figure 13.1 illustrates a simple model of interatomic spacing in a crystal as a function of energy, in which the red dots indicate the average interatomic distance. Note that as temperature and energy increase, the width of the potential well widens to the right, towards a greater separation distance. An increase in temperature leads to a sampling of higher energies in the potential well and a lattice expansion as indicated by the red dots in the plot. The width bounded by the two walls of the potential also increases, allowing the atoms to take greater path lengths and have greater vibrational amplitudes.
Figure 13.1: Interatomic separation as a function of energy

### 13.1 Describe a crystal in terms of primitive unit crystal.

Considering a crystal of M primitive unit cells, the ${\stackrel{‾}{R}}_{m}\text{'}$ will replace the former lattice point position vector, which now includes an extra factor - a time dependent displacement vector for each atom.
• For simplicity sake we will assume one atom per lattice point at origin
• Let $M={M}_{1}{M}_{2}{M}_{3}=$ # of unit cells
• Let $m=\left\{{m}_{1}{m}_{2}{m}_{3}\right\}$ = unit crystal index
• Let ${\stackrel{‾}{R}}_{m}=$ lattice point position vector. Locates origin of the ${m}^{th}$ unit cell
• Let $\stackrel{‾}{R}{\text{'}}_{m}\left(t\right)={\stackrel{‾}{R}}_{m}+\underset{}{\underset{⏟}{{\stackrel{‾}{\Delta }}_{m}}}\left(t\right)$
Atoms can vibrate longitudinally or transversely, as shown in the figures below.
Figure 13.2: Longitudinal Wave
Figure 13.3: Transverse Wave
The summation of the coherently scattered wave intensities in electron units from $M$ atoms for a crystal with one atom per unit cell is given by the following:
$\begin{array}{cc}{I}_{\left(eu\right)}\left(\stackrel{‾}{Q}\right)=\left[{\sum }_{m}^{M}{f}_{m}{e}^{2\pi i\stackrel{‾}{Q}\cdot \stackrel{‾}{R}{\text{'}}_{m}}\right]\left[{\sum }_{m}^{M}{f}_{m}{e}^{-2\pi i\stackrel{‾}{Q}\cdot \stackrel{‾}{R}{\text{'}}_{m}}\right]={\sum }_{m}^{M}{\sum }_{n}^{M}{f}_{m}{f}_{n}{e}^{2\pi i\stackrel{‾}{Q}\cdot \left(\stackrel{‾}{R}{\text{'}}_{m}-\stackrel{‾}{R}{\text{'}}_{n}\right)}& \left(13.1\right)\end{array}$
Assuming all of the atoms are the same, the ${f}_{m}$ and ${f}_{n}$ terms may be taken out of the summation. Adding in the time dependent Debye-Waller factors yields the following:
$\begin{array}{cc}{I}_{\left(eu\right)}\left(\stackrel{‾}{Q}\right)={f}^{2}{\sum }_{m}{\sum }_{n}{e}^{2\pi i\stackrel{‾}{Q}\cdot \left({\stackrel{‾}{R}}_{m}-{\stackrel{‾}{R}}_{n}\right)}{e}^{2\pi i\stackrel{‾}{Q}\cdot \left({\stackrel{‾}{\Delta }}_{m}-{\stackrel{‾}{\Delta }}_{n}\right)}& \left(13.2\right)\end{array}$
Recall that $\stackrel{‾}{Q}$ is the scattering vector that is coincident with ${r}_{hkl}^{*}$ under the Bragg diffraction condition.
Figure 13.4: Scattering vector and $\Delta$
<table class='mathtable'><tr><td class='math'>\bar{Q}\cdot\bar{\Delta}_{m}=\frac{2sin\theta}{\underset{\hookrightarrow Q}{\lambda}}\Delta_{m}cos\phi_{m}</td><td>(13.3)</td></tr></table>
The vector ${u}_{m}$ is the projection of the instantaneous displacement vector along the scattering vector's direction:
$\begin{array}{cc}{u}_{m}={\Delta }_{m}cos{\theta }_{m}& \left(13.4\right)\end{array}$
Figure 13.5: ${u}_{m}$
Note that ${u}_{m}=0$ for longitudinal wave along diffraction plane.
Using $2\pi Q=\frac{4\pi sin\theta }{\lambda }=q$, the intensity equation is the following:
${I}_{\left(eu\right)}\left(\stackrel{‾}{Q}\right)={f}^{2}{\sum }_{m}{\sum }_{n}{e}^{2\pi i\stackrel{‾}{Q}\cdot \left({\stackrel{‾}{R}}_{m}-{\stackrel{‾}{R}}_{n}\right)}{e}^{iq\left({u}_{m}-{u}_{n}\right)}$
Note that ${\Delta }_{m};{u}_{m}$are instantaneous vibrations that occur in the phonon lifetime which is on the order of ~ $1{0}^{-14}$ seconds , and the intensity measurement time is > $1{0}^{-3}$seconds. Therefore, a time average of these instantaneous intensities as affected by vibrational effects is used.
The function is integrated over a period of time $T$ that is long relative to the fluctuations and is normalized by dividing by $T.$
$\begin{array}{cc}⟨f\left(t\right)⟩=\frac{1}{T}{\int }_{0}^{T}f\left(t\right)dt& \left(13.5\right)\end{array}$
The time averaged intensity is indicated by bracket notation: $\begin{array}{cc}⟨{I}_{\left(eu\right)}⟩={f}^{2}{\sum }_{m}{\sum }_{n}{e}^{2\pi i\stackrel{‾}{Q}\cdot \left({\stackrel{‾}{R}}_{m}-{\stackrel{‾}{R}}_{n}\right)}⟨{e}^{iq\left({u}_{m}-{u}_{n}\right)}⟩& \left(13.6\right)\end{array}$
Simplifying, by Maclaurin series expansion:
${e}^{ix}=1+ix-\frac{{x}^{2}}{2}-\frac{i{x}^{3}}{6}+...$
$⟨{e}^{iq\left({u}_{m}-{u}_{n}\right)}⟩=1+iq⟨{u}_{m}-{u}_{n}⟩-\frac{{q}^{2}}{2}⟨\left({u}_{m}-{u}_{n}{\right)}^{2}⟩-...$
$\approxeq 1-\frac{{q}^{2}}{2}⟨\left({u}_{m}-{u}_{n}{\right)}^{2}⟩={e}^{\frac{-{q}^{2}}{2}⟨\left({u}_{m}-{u}_{n}{\right)}^{2}⟩}={e}^{\frac{-{q}^{2}}{2}⟨{u}_{m}^{2}+{u}_{n}^{2}-2{u}_{m}{u}_{n}⟩}$
$⟨{u}_{m}^{2}⟩=⟨{u}_{n}^{2}⟩=⟨{u}^{2}⟩$$⇒$$⟨{e}^{iq\left({u}_{m}-{u}_{n}\right)}⟩\approxeq {e}^{-{q}^{2}⟨{u}^{2}-{u}_{m}{u}_{n}⟩}$
Let $⟨{u}_{m}{u}_{n}⟩=\underset{}{\underset{⏟}{{\alpha }_{mn}}}⟨{u}^{2}⟩$
which accounts for the correlation of vibrational motion between the atoms, and may range from 0 to 1.
$⟨{e}^{iq\left({u}_{m}-{u}_{n}\right)}⟩={e}^{-{q}^{2}⟨{u}^{2}⟩\left(1-{\alpha }_{mn}\right)}$
and $⟨{I}_{\left(eu\right)}⟩={\sum }_{m}{\sum }_{n}{f}^{2}{e}^{2\pi i\stackrel{‾}{Q}\cdot \left({\stackrel{‾}{R}}_{m}-{\stackrel{‾}{R}}_{n}\right)}\underset{}{\underset{⏟}{{e}^{{q}^{2}⟨{u}^{2}⟩{\alpha }_{mn}}}}$
$={e}^{-{q}^{2}⟨{u}^{2}⟩}\underset{}{\underset{⏟}{{\sum }_{m}{\sum }_{n}{f}^{2}{e}^{2\pi i\stackrel{‾}{Q}\cdot \left({\stackrel{‾}{R}}_{m}-{\stackrel{‾}{R}}_{n}\right)}}}+\underset{}{\underset{⏟}{{q}^{2}⟨{u}^{2}⟩{e}^{-{q}^{2}⟨{u}^{2}⟩}}}{\sum }_{m}{\sum }_{n}{f}^{2}{\alpha }_{mn}{e}^{2\pi i\stackrel{‾}{Q}\cdot \left({\stackrel{‾}{R}}_{m}-{\stackrel{‾}{R}}_{n}\right)}$
[3-D interference function]
With higher $2\theta$, the diffraction peak intensities decrease due to the damping term ${e}^{-{q}^{2}⟨{u}^{2}⟩}$ at the front.
$x{e}^{-x}=x\left(1-x+\frac{{x}^{2}}{2}-....\right)\approx x-{x}^{2}\approx 1-{e}^{-x}=x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{6}...$
$⇒⟨{I}_{\left(eu\right)}⟩={e}^{-{q}^{2}⟨{u}^{2}⟩}\left[CR\right]+\left(1-{e}^{-{q}^{2}⟨{u}^{2}⟩}\right){\sum }_{n}{f}^{2}{\alpha }_{mn}{e}^{2\pi i\stackrel{‾}{Q}\cdot \left({\stackrel{‾}{R}}_{m}-{\stackrel{‾}{R}}_{n}\right)}$
${q}^{2}⟨{u}^{2}⟩=\left(2\pi Q{\right)}^{2}⟨{u}^{2}⟩=2B{\left(\frac{sin\theta }{\lambda }\right)}^{2}=\frac{B{Q}^{2}}{2}=2M$
The new term, B, is sensitive to the temperature and vibrational amplitudes. Using B, the intensity can be written as follows:
$⇒⟨{I}_{\left(eu\right)}⟩=\underset{}{\underset{⏟}{{e}^{\frac{-B{Q}^{2}}{2}}}}\left[CR\right]+\stackrel{}{\stackrel{⏞}{\left(1-{e}^{-\frac{B{Q}^{2}}{2}}\right)}}\underset{}{\underset{⏟}{{\sum }_{m}{\sum }_{n}{f}^{2}{\alpha }_{mn}{e}^{2\pi i\stackrel{‾}{Q}\cdot \left({\stackrel{‾}{R}}_{m}-{\stackrel{‾}{R}}_{n}\right)}}}$
Since the correlation function is equal to unity at an atom (i.e. ${\alpha }_{mn}=1$) and only has a considerable effect at a small region around an atom, the diffraction peaks in the second summation are broadened. Recall that a smaller number of unit cells produces broader peaks than a relatively larger amount of unit cells.
Why time average?
Because measurement time for intensity (1 sec) $\gg$ phonon lifetime ($1{0}^{-14}$ sec)
Why do odd powers of $⟨{u}^{n}⟩=0$?
We assume about u = 0
The average time displacement for odd powers averages to zero, but for even powers, the area under the curve is $\pi$. This explains why even powers are present in the time average and odd powers are not.
Figure 13.6: Harmonic oscillation assumes symmetric potential well
Figure 13.7: ${\int }_{0}^{2\pi }sin\left(mx\right)dx=0$
Figure 13.8: Area = ${\int }_{0}^{2\pi }{sin}^{2}\left(mx\right)dx=\pi$
$Average=\frac{Area}{2\pi }=\frac{1}{2}$

### 13.2 Debye Theory of Lattice Dynamics/heat capacity — Corrected by Waller

The quantity B, which describes the vibration of the atoms, is described by the following equation:
$\begin{array}{cc}B=\frac{6{h}^{2}T}{{m}_{A}{k}_{B}{\theta }_{D}^{2}}Q\left(\frac{{\theta }_{D}}{T}\right)=8{\pi }^{2}⟨{u}^{2}⟩& \left(13.7\right)\end{array}$
where the variables are defined as follows:
• ${m}_{A}=$ atomic mass = $A×1.660*1{0}^{-24}g/amu$
• $h=$ Planck's constant = $6.628*1{0}^{-27}erg-sec$
• $T{=}^{○}K$ absolute temp
• ${k}_{B}=$ Boltzmann's constant = $1.380*1{0}^{-16}erg/{}^{○}K$
• ${\theta }_{D}=$Debye temperature (related to stiffness of crystal) e.g. ${\theta }_{D}^{Ge}=360{}^{○}K,$ ${\theta }_{D}^{Al}=394{}^{○}K$, ${\theta }_{D}^{Si}=625{}^{○}K$
$Q\left({\theta }_{D}/T\right)\approx 1$ for $T>{\theta }_{D}/2$ declines rapidly for $T<{\theta }_{D}/2$
Unvarying constants: $\frac{6{h}^{2}}{{m}_{A}K}=1.151×1{0}^{-12}c{m}^{2}=1.151×1{0}^{4}{\stackrel{˚}{A}}^{2}/A$