Physics / Condensed Matter Solid State Physics I 100% Free Open Access
Chapter 1 • Theory & Derivations

Crystal Structure, Symmetry & X-Ray Diffraction

Comprehensive foundation of crystallography: crystalline state, 14 3D Bravais lattices, primitive and conventional unit cells, Wigner-Seitz construction, Miller indices, atomic packing fractions, reciprocal lattice vectors, first Brillouin zone, Laue equations, and experimental X-ray diffraction methods.

1.1The Crystalline State, Space Lattices & 14 Bravais Lattices

1. The Crystalline State of Condensed Matter

Matter in the solid state displays a broad dichotomy in structural organization: amorphous solids (e.g., vitreous silica, amorphous polymers), where positional correlations decay rapidly beyond nearest-neighbor atomic separations exhibiting only short-range order, and crystalline solids (e.g., metals, diamond, semiconductor silicon, rock-salt), where atomic constituents reside in regular, periodic spatial arrays extending across macroscopic microscopic dimensions of $10^8$ unit intervals, characterized by rigorous long-range translational order.

2. Mathematical Definition of an Ideal Space Lattice

An ideal spatial lattice is a purely geometrical abstraction: an infinite three-dimensional periodic array of mathematical points in Euclidean space, where the physical and chemical environment surrounding any arbitrary lattice point $\vec{R}'$ is strictly indistinguishable from that around any other point $\vec{R}$. The translational position vector $\vec{R}$ connecting the arbitrary origin to any point in the lattice is defined uniquely as an integer linear combination of three linearly independent primitive basis vectors $\vec{a}_1, \vec{a}_2, \vec{a}_3$:

$$\vec{R} = n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3 \vec{a}_3 \quad (n_1, n_2, n_3 \in \mathbb{Z})$$

A physical crystal structure is synthesized mathematically by convolving the geometrical space lattice with an identical group of atoms called the basis (or motif) situated at each lattice point:

$$\text{Crystal Structure} = \text{Space Lattice} + \text{Basis}$$

If the basis comprises $j = 1, 2, \dots, s$ atoms with respective atomic numbers $Z_j$, their spatial coordinates within the unit cell relative to the origin of that cell are defined by fractional vectors:

$$\vec{r}_j = x_j \vec{a}_1 + y_j \vec{a}_2 + z_j \vec{a}_3 \quad (0 \le x_j, y_j, z_j < 1)$$

3. The 14 Three-Dimensional Bravais Lattices

In three spatial dimensions, spatial translational invariance combined with point group rotational and reflection symmetries yields precisely 14 distinct Bravais lattices, partitioned among 7 crystal systems characterized by the axial lengths ($a, b, c$) and interaxial angles ($\alpha, \beta, \gamma$):

  1. Cubic System ($a = b = c, \alpha = \beta = \gamma = 90^{\circ}$): Simple Cubic ($P$), Body-Centered Cubic ($I$), Face-Centered Cubic ($F$).
  2. Tetragonal System ($a = b \neq c, \alpha = \beta = \gamma = 90^{\circ}$): Simple Tetragonal ($P$), Body-Centered Tetragonal ($I$).
  3. Orthorhombic System ($a \neq b \neq c, \alpha = \beta = \gamma = 90^{\circ}$): Simple ($P$), Base-Centered ($C$), Body-Centered ($I$), Face-Centered ($F$).
  4. Hexagonal System ($a = b \neq c, \alpha = \beta = 90^{\circ}, \gamma = 120^{\circ}$): Simple Hexagonal ($P$).
  5. Trigonal / Rhombohedral System ($a = b = c, \alpha = \beta = \gamma < 120^{\circ} \neq 90^{\circ}$): Primitive Rhombohedral ($R$).
  6. Monoclinic System ($a \neq b \neq c, \alpha = \gamma = 90^{\circ} \neq \beta$): Simple Monoclinic ($P$), Base-Centered Monoclinic ($C$).
  7. Triclinic System ($a \neq b \neq c, \alpha \neq \beta \neq \gamma \neq 90^{\circ}$): Primitive Triclinic ($P$).

1.2Primitive Cells, Wigner-Seitz Construction & Symmetry Operations

1. Primitive vs. Conventional Unit Cells

A unit cell is any volume of space that, when translated by the full set of lattice vectors $\vec{R} = \sum_i n_i \vec{a}_i$, completely fills all space without overlapping or leaving voids. A unit cell is categorized as:

  • Primitive Cell: A unit cell having minimum volume that contains precisely one net lattice point ($N_{pts} = 1$). Its volume is calculated as the scalar triple product:
    $$V_c = |\vec{a}_1 \cdot (\vec{a}_2 \times \vec{a}_3)|$$
  • Conventional (Non-Primitive) Cell: A larger unit cell chosen intentionally to preserve and manifest the full rotational and reflection symmetry of the crystal system (e.g., BCC contains 2 lattice points, FCC contains 4 lattice points).

2. The Wigner-Seitz Primitive Cell Construction

The Wigner-Seitz cell is an invariant geometric construction providing a canonical primitive cell that exhibits the full point group symmetry of the Bravais lattice. The algorithm proceeds as follows:

  1. Select an arbitrary lattice point as the origin $\vec{0}$.
  2. Draw straight line vectors connecting this origin to all neighboring lattice points $\vec{R}_i$.
  3. Construct planes that perpendicularly bisect each of these vectors at $\vec{R}_i / 2$.
  4. The smallest closed polyhedron enclosing the origin bounded by these bisecting planes constitutes the Wigner-Seitz cell.

For an FCC lattice, the Wigner-Seitz cell is a rhombic dodecahedron (12 rhombic faces). For a BCC lattice, the Wigner-Seitz cell is a truncated octahedron (8 hexagonal faces and 6 square faces).

3. Crystallographic Symmetry Operations

The symmetry of a crystal consists of operations that map the periodic spatial arrangement onto itself:

  • Point Group Operations: Operations leaving at least one point invariant, comprising rotations $C_n = 2\pi/n$ (where $n \in \{1, 2, 3, 4, 6\}$ due to the crystallographic restriction theorem), reflections $\sigma$, inversion $i$, and roto-inversions $S_n$. There exist precisely 32 crystallographic point groups.
  • Space Group Operations: Combinations of point group operations with fractional and primitive lattice translations, including non-symmorphic glide planes (reflection plus fractional translation) and screw axes (rotation plus translation along the axis). In 3D space, there exist precisely 230 crystallographic space groups.

1.3Miller Indices & Interplanar Spacing of Crystal Planes

1. Definition and Determination of Miller Indices $(hkl)$

A crystal plane is characterized by a set of three coprime integers $(hkl)$, known as its Miller indices, which specify its spatial orientation relative to the primitive or conventional crystal axes $\vec{a}_1, \vec{a}_2, \vec{a}_3$:

  1. Determine the intercepts of the plane along the three crystallographic axes in units of lattice constants: $x_1 a, x_2 b, x_3 c$.
  2. Take the reciprocals of these fractional intercepts: $1/x_1, 1/x_2, 1/x_3$.
  3. Clear fractions by multiplying by their least common denominator to obtain the smallest coprime triplet of integers $(h, k, l)$. If an intercept is negative, say $-x_1$, the corresponding index is written with an overbar as $(\bar{h}kl)$.

2. Mathematical Derivation of Interplanar Spacing $d_{hkl}$

Consider a family of parallel equidistant planes designated by Miller indices $(hkl)$. The distance of the first plane from the coordinate origin along the normal unit vector $\hat{n}$ is the interplanar spacing $d_{hkl}$.

In a Cubic crystal system ($a = b = c, \alpha = \beta = \gamma = 90^{\circ}$), the equation of a plane intersecting the axes at $a/h, a/k, a/l$ is:

$$\frac{x}{a/h} + \frac{y}{a/k} + \frac{z}{a/l} = 1 \implies hx + ky + lz = a$$

The perpendicular distance from the origin $(0,0,0)$ to this plane is given by analytical geometry:

$$d_{hkl} = \frac{|h(0) + k(0) + l(0) - a|}{\sqrt{h^2 + k^2 + l^2}} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$$

For a general Orthorhombic system ($a \neq b \neq c, \alpha = \beta = \gamma = 90^{\circ}$):

$$\frac{1}{d_{hkl}^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2}$$

For a Hexagonal system ($a = b \neq c, \alpha = \beta = 90^{\circ}, \gamma = 120^{\circ}$):

$$\frac{1}{d_{hkl}^2} = \frac{4}{3}\left(\frac{h^2 + hk + k^2}{a^2}\right) + \frac{l^2}{c^2}$$

1.4Simple Crystal Structures & Atomic Packing Factors

1. Atomic Packing Fraction (APF) Formalism

The Atomic Packing Fraction (APF) quantifies the volumetric efficiency with which hard spherical atoms of radius $R$ fill the unit cell volume $V_{cell}$:

$$\text{APF} = \frac{N_{\text{eff}} \times V_{\text{atom}}}{V_{\text{cell}}} = \frac{N_{\text{eff}} \times \frac{4}{3}\pi R^3}{V_{\text{cell}}}$$

where $N_{\text{eff}}$ is the effective number of whole atoms inside the conventional unit cell.

2. Detailed Comparison of Canonical Metallic and Covalent Structures

  1. Simple Cubic (SC):
    • $N_{\text{eff}} = 8 \times (1/8) = 1$.
    • Touching condition along edge: $a = 2R \implies R = a/2$.
    • $\text{APF} = \frac{1 \times \frac{4}{3}\pi (a/2)^3}{a^3} = \frac{\pi}{6} \approx 0.5236$ ($52.4\%$). Coordination number $CN = 6$.
  2. Body-Centered Cubic (BCC):
    • $N_{\text{eff}} = 8 \times (1/8) + 1 = 2$.
    • Touching condition along body diagonal: $4R = \sqrt{3}a \implies R = \frac{\sqrt{3}}{4}a$.
    • $\text{APF} = \frac{2 \times \frac{4}{3}\pi \left(\frac{\sqrt{3}}{4}a\right)^3}{a^3} = \frac{\sqrt{3}\pi}{8} \approx 0.6802$ ($68.0\%$). Coordination number $CN = 8$.
  3. Face-Centered Cubic (FCC):
    • $N_{\text{eff}} = 8 \times (1/8) + 6 \times (1/2) = 4$.
    • Touching condition along face diagonal: $4R = \sqrt{2}a \implies R = \frac{\sqrt{2}}{4}a$.
    • $\text{APF} = \frac{4 \times \frac{4}{3}\pi \left(\frac{\sqrt{2}}{4}a\right)^3}{a^3} = \frac{\sqrt{2}\pi}{6} \approx 0.7405$ ($74.1\%$). Coordination number $CN = 12$.
  4. Hexagonal Close-Packed (HCP):
    • Stacking sequence: $ABABAB\dots$ Ideal axial ratio: $c/a = \sqrt{8/3} \approx 1.633$.
    • $N_{\text{eff}} = 12 \times (1/6) + 2 \times (1/2) + 3 = 6$.
    • $\text{APF} = \frac{\pi}{3\sqrt{2}} \approx 0.7405$ ($74.1\%$). Coordination number $CN = 12$.
  5. Diamond Cubic Structure:
    • FCC lattice with a two-atom basis: $(0,0,0)$ and $\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right)$.
    • $N_{\text{eff}} = 8$ atoms. Touching condition along quarter body diagonal: $8R = \sqrt{3}a \implies R = \frac{\sqrt{3}}{8}a$.
    • $\text{APF} = \frac{\sqrt{3}\pi}{16} \approx 0.3401$ ($34.0\%$). Coordination number $CN = 4$ (tetrahedral $sp^3$ coordination).
  6. Ionic Structures (NaCl, CsCl, ZnS Zincblende):
    • NaCl: FCC Bravais lattice of $\text{Cl}^-$ with $\text{Na}^+$ at $(1/2, 0, 0)$; $CN = 6:6$, $4$ formula units/cell.
    • CsCl: Simple cubic Bravais lattice with $\text{Cs}^+$ at $(0,0,0)$ and $\text{Cl}^-$ at $(1/2, 1/2, 1/2)$; $CN = 8:8$, $1$ formula unit/cell.
    • ZnS (Zincblende): FCC lattice of $\text{S}^{2-}$ with $\text{Zn}^{2+}$ occupying half the tetrahedral interstitial sites; $CN = 4:4$.

1.5Reciprocal Lattice, Brillouin Zones & X-Ray Diffraction

1. Rigorous Definition of the Reciprocal Lattice

Given direct lattice primitive basis vectors $\vec{a}_1, \vec{a}_2, \vec{a}_3$ with unit cell volume $V_c = \vec{a}_1 \cdot (\vec{a}_2 \times \vec{a}_3)$, the corresponding reciprocal lattice primitive vectors $\vec{b}_1, \vec{b}_2, \vec{b}_3$ are defined uniquely by the orthogonality relation:

$$\vec{a}_i \cdot \vec{b}_j = 2\pi \delta_{ij}$$

Explicit vector formulas for the reciprocal basis vectors are:

$$\vec{b}_1 = 2\pi \frac{\vec{a}_2 \times \vec{a}_3}{V_c}, \quad \vec{b}_2 = 2\pi \frac{\vec{a}_3 \times \vec{a}_1}{V_c}, \quad \vec{b}_3 = 2\pi \frac{\vec{a}_1 \times \vec{a}_2}{V_c}$$

An arbitrary reciprocal lattice vector $\vec{G}$ is expressed in terms of integer Miller components $(h, k, l)$:

$$\vec{G}_{hkl} = h \vec{b}_1 + k \vec{b}_2 + l \vec{b}_3$$

Fundamental Theorems of the Reciprocal Lattice:

  1. The reciprocal lattice vector $\vec{G}_{hkl}$ is normal to the family of crystal planes $(hkl)$ in direct space.
  2. The magnitude of $\vec{G}_{hkl}$ is inversely proportional to the interplanar spacing $d_{hkl}$:
    $$|\vec{G}_{hkl}| = \frac{2\pi}{d_{hkl}} \implies d_{hkl} = \frac{2\pi}{|\vec{G}_{hkl}|}$$
  3. The reciprocal lattice of an FCC direct lattice is a BCC reciprocal lattice, and vice-versa.

2. The First Brillouin Zone

The First Brillouin Zone (1st BZ) is the Wigner-Seitz primitive cell of the reciprocal lattice. It contains all wavevectors $\vec{k}$ that can propagate through the periodic crystal without undergoing elastic Bragg reflection from the lattice planes. The zone boundaries are defined by the Bragg condition:

$$\vec{k} \cdot \left(\frac{1}{2}\vec{G}\right) = \left(\frac{1}{2}\vec{G}\right)^2 \implies 2\vec{k} \cdot \vec{G} + G^2 = 0$$

3. Von Laue Diffraction Equations & Bragg's Law

When an incident plane wave of X-rays with wavevector $\vec{k}$ ($|\vec{k}| = 2\pi/\lambda$) scatters elastically from a crystal to wavevector $\vec{k}'$ ($|\vec{k}'| = |\vec{k}|$), the scattering wavevector transfer is $\Delta \vec{k} = \vec{k}' - \vec{k}$. Constructive interference across the entire crystal occurs if and only if the Laue condition is satisfied:

$$\Delta \vec{k} = \vec{G}_{hkl}$$

Taking the magnitude squared: $|\vec{k}' - \vec{k}|^2 = G^2 \implies k^2 + k'^2 - 2 k k' \cos(180^{\circ} - 2\theta) = G^2$. Since $k' = k = 2\pi/\lambda$ and $G = 2\pi/d_{hkl}$:

$$2k^2(1 - \cos(\pi - 2\theta)) = 4k^2 \sin^2\theta = G^2 \implies 2\left(\frac{2\pi}{\lambda}\right)\sin\theta = \frac{2\pi}{d_{hkl}} \implies 2d_{hkl}\sin\theta = \lambda$$

Generalizing to order $n$, this yields Bragg's Law of X-Ray Diffraction:

$$2d_{hkl} \sin\theta = n\lambda$$

4. Experimental X-Ray Diffraction Techniques

  1. Laue Method: Uses polychromatic (white) continuous X-ray radiation on a stationary single crystal. Each crystal plane $(hkl)$ selects a specific wavelength satisfying $2d\sin\theta = \lambda$. Used to determine crystal orientation and symmetry.
  2. Rotating Crystal Method: Uses monochromatic X-ray radiation on a single crystal rotating about a crystallographic axis. Different planes pass through the Bragg angle $\theta$ sequentially, forming layer lines on a cylindrical film.
  3. Powder Diffraction (Debye-Scherrer) Method: Uses monochromatic X-rays incident upon a finely powdered polycrystalline specimen containing millions of randomly oriented crystallites. Diffraction emerges as concentric cones of half-angle $2\theta$, yielding characteristic diffraction rings used for phase identification.
EXAM SUCCESS WORKSHOP

Solved University Examination Problems

Step-by-step mathematical solutions to classic university honors examination questions.

SOLVED PROBLEM 1.1

Interplanar Spacing and Bragg Angle Calculation for Silicon (111)

Silicon crystallizes in the diamond cubic structure with a conventional lattice parameter of $a = 5.431 \text{ Å}$. Monochromatic $\text{Cu } K_\alpha$ X-rays with wavelength $\lambda = 1.5406 \text{ Å}$ are directed at a single-crystal silicon wafer. (a) Calculate the interplanar spacing $d_{111}$ for the (111) planes. (b) Determine the first-order ($n=1$) Bragg diffraction angle $\theta_{111}$ and the total scattering deflection angle $2\theta$.

RIGOROUS DERIVATION & EXAM SOLUTION
Step 1: Calculate Interplanar Spacing d_111
d_{111} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} = \frac{5.431\text{ \AA}}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{5.431}{\sqrt{3}}\text{ \AA} \approx 3.1356\text{ \AA}

Apply the cubic interplanar spacing formula for Miller indices (h, k, l) = (1, 1, 1).

Step 2: Apply Bragg's Law to Find Diffraction Angle theta
\sin\theta_{111} = \frac{n\lambda}{2d_{111}} = \frac{1 \times 1.5406\text{ \AA}}{2 \times 3.1356\text{ \AA}} = \frac{1.5406}{6.2712} \approx 0.24566

Substitute n=1, lambda = 1.5406 Angstroms, and d_111 into Bragg's equation 2 d sin(theta) = n lambda.

Step 3: Evaluate Arcsin and Deflection Angle 2 theta
\theta_{111} = \arcsin(0.24566) = 14.22^{\circ} \implies 2\theta = 28.44^{\circ}

Calculate the Bragg angle theta and double it to determine the detector deflection angle 2 theta measured in standard powder and single-crystal diffractometers.

Final Answer & Physical Insight

d_{111} = 3.136 \text{ Å}, \quad \theta_{111} = 14.22^{\circ}, \quad 2\theta = 28.44^{\circ}

SOLVED PROBLEM 1.2

Reciprocal Lattice Volume and Primitive Vectors for FCC Copper

Copper crystallizes in a face-centered cubic (FCC) lattice with lattice constant $a = 3.615 \text{ Å}$. (a) Write down the primitive translation vectors of the FCC direct lattice. (b) Calculate the volume of the primitive direct unit cell $V_c$. (c) Derive the primitive reciprocal lattice vectors $\vec{b}_1, \vec{b}_2, \vec{b}_3$ and evaluate the volume of the First Brillouin Zone $V_{BZ}$.

RIGOROUS DERIVATION & EXAM SOLUTION
Step 1: Direct Primitive Basis Vectors and Unit Cell Volume
\vec{a}_1 = \frac{a}{2}(\hat{y} + \hat{z}), \quad \vec{a}_2 = \frac{a}{2}(\hat{z} + \hat{x}), \quad \vec{a}_3 = \frac{a}{2}(\hat{x} + \hat{y})

The primitive vectors of FCC connect the origin to three adjacent face centers. The primitive cell volume is one-fourth of the conventional cubic cell volume:

Step 2: Calculate Numerical Value of Direct Primitive Volume V_c
V_c = \frac{a^3}{4} = \frac{(3.615\text{ \AA})^3}{4} = \frac{47.241\text{ \AA}^3}{4} \approx 11.810\text{ \AA}^3 = 1.181 \times 10^{-29}\text{ m}^3

Evaluate V_c = a^3 / 4 using the lattice parameter of copper.

Step 3: Derive Reciprocal Vectors and Brillouin Zone Volume
\vec{b}_1 = \frac{2\pi}{a}(-\hat{x} + \hat{y} + \hat{z}), \quad \vec{b}_2 = \frac{2\pi}{a}(\hat{x} - \hat{y} + \hat{z}), \quad \vec{b}_3 = \frac{2\pi}{a}(\hat{x} + \hat{y} - \hat{z})

These are the primitive vectors of a Body-Centered Cubic (BCC) reciprocal lattice. The volume of the First Brillouin Zone is given by V_BZ = (2 pi)^3 / V_c:

Step 4: Compute First Brillouin Zone Volume V_BZ
V_{BZ} = \frac{(2\pi)^3}{V_c} = \frac{4 (2\pi)^3}{a^3} = \frac{32\pi^3}{(3.615\text{ \AA})^3} \approx \frac{992.88}{47.241}\text{ \AA}^{-3} \approx 21.017\text{ \AA}^{-3} = 2.102 \times 10^{31}\text{ m}^{-3}

The volume of the First Brillouin Zone in reciprocal space.

Final Answer & Physical Insight

V_c = 11.81 \text{ Å}^3, \quad V_{BZ} = 21.02 \text{ Å}^{-3} = 2.102 \times 10^{31} \text{ m}^{-3}

SOLVED PROBLEM 1.3

Debye-Scherrer Powder Diffraction Indexing for BCC Iron

An X-ray powder diffraction pattern of alpha-iron (BCC structure) is recorded using monochromatic radiation of $\lambda = 1.5418 \text{ Å}$. The first two diffraction peaks are observed at scattering angles $2\theta_1 = 44.67^{\circ}$ and $2\theta_2 = 65.02^{\circ}$. (a) Determine the Miller indices $(hkl)$ for these two reflection peaks taking into account the BCC selection rules ($h+k+l = \text{even}$). (b) Calculate the lattice constant $a$ of iron.

RIGOROUS DERIVATION & EXAM SOLUTION
Step 1: Extract Bragg Angles theta_1 and theta_2
\theta_1 = \frac{44.67^{\circ}}{2} = 22.335^{\circ} \implies \sin\theta_1 = \sin(22.335^{\circ}) \approx 0.38004

Convert 2 theta to theta and calculate sin(theta) for the first peak.

Step 2: Calculate sin(theta) for the Second Peak
\theta_2 = \frac{65.02^{\circ}}{2} = 32.51^{\circ} \implies \sin\theta_2 = \sin(32.51^{\circ}) \approx 0.53745

Convert 2 theta to theta and calculate sin(theta) for the second peak.

Step 3: Compare Ratio of sin^2(theta) to BCC Selection Rules
\frac{\sin^2\theta_1}{\sin^2\theta_2} = \frac{(0.38004)^2}{(0.53745)^2} = \frac{0.14443}{0.28885} \approx 0.5000 = \frac{1}{2}

In a BCC lattice, allowed reflections require h+k+l = even. The lowest permitted values of s = h^2 + k^2 + l^2 are (110) with s=2, and (200) with s=4. The ratio is 2/4 = 1/2, perfectly matching the experimental ratio.

Step 4: Calculate the Lattice Constant a
a = \frac{\lambda \sqrt{h^2 + k^2 + l^2}}{2\sin\theta_1} = \frac{1.5418\text{ \AA} \times \sqrt{2}}{2 \times 0.38004} = \frac{1.5418 \times 1.4142}{0.76008} \approx 2.8686\text{ \AA}

Using peak 1 with (hkl) = (110), solve for the lattice parameter a.

Final Answer & Physical Insight

\text{Peak 1: } (110), \quad \text{Peak 2: } (200), \quad a = 2.869 \text{ Å}