Lattice Logic Slides Lattice Logic
Crystal Systems & Bravais Lattices
Why can't we tile with pentagons?
In 2D, we can fill a floor perfectly with squares, hexagons, or triangles.
"Symmetry is the key to periodicity."
Crystals are built from **translational symmetry**. Not every shape can repeat infinitely without leaving gaps.
⬛
⬢
▲
⬠ X
The Unit Cell Parameters
Lattice Vectors
1 Lengths: \( a, b, c \)
2 Angles: \( \alpha, \beta, \gamma \)
Visualizing the parallelopiped
\( \gamma \)
The **Unit Cell** is the smallest repeating unit that generates the entire crystal through translation alone.
Translation Vector:
\( \mathbf{T} = u\mathbf{a} + v\mathbf{b} + w\mathbf{c} \)
Where \( u, v, w \) are integers.
The 7 Crystal Systems
Cubic
\( a=b=c, \alpha=\beta=\gamma=90^\circ \)
Tetragonal
\( a=b\neq c, \alpha=\beta=\gamma=90^\circ \)
Orthorhombic
\( a\neq b\neq c, \alpha=\beta=\gamma=90^\circ \)
Hexagonal
\( a=b\neq c, \alpha=\beta=90, \gamma=120^\circ \)
Monoclinic
\( a\neq b\neq c, \alpha=\gamma=90\neq \beta \)
Triclinic
\( a\neq b\neq c, \alpha\neq\beta\neq\gamma \)
Trigonal (Rhombohedral)
\( a=b=c, \alpha=\beta=\gamma\neq 90^\circ \)
"A slanted cube"
The 14 Bravais Lattices
The 7 systems only tell us the **shape** of the box.
The **Bravais Lattices** tell us where the atoms can be inside the box while maintaining symmetry.
P: Primitive
I: Body-Centered
F: Face-Centered
C: Base-Centered
Primitive (P)
Nodes at corners only
Body-Centered (I)
Symmetry Hunt Worksheet Symmetry Hunt
Lattice Classification & Symmetry Analysis
Student:
Date:
Objective
Analyze the geometric parameters of various unit cells to identify their corresponding crystal systems and Bravais lattice types. For each case, identify the translational symmetry operations and calculate the effective number of atoms per unit cell.
Part 1: Identifying Systems
Given the following lattice parameters, name the Crystal System and list all possible Bravais Lattice types (P, I, F, C) associated with it.
\( a = b \neq c \)
Crystal System
Possible Bravais Types
\( a \neq b \neq c \)
Crystal System
Possible Bravais Types
\( a = b = c \)
Crystal System
Possible Bravais Types
Part 2: Effective Atom Count (\( N \))
In a 3D crystal, atoms are shared between adjacent unit cells. Use the formula: \[ N = N_{interior} + \frac{N_{face}}{2} + \frac{N_{edge}}{4} + \frac{N_{corner}}{8} \] Calculate \( N \) for the following standard Bravais Lattices:
Body-Centered Cubic (I)
Body-Centered Cubic
Show your work for the calculation of \( N \):
Total Atoms (\( N \)):
Face-Centered Cubic (F)
Face-Centered Cubic
Show your work for the calculation of \( N \):
Total Atoms (\( N \)):
Part 3: Symmetry Constraint
Explain why a "Face-Centered Tetragonal" lattice is not one of the 14 Bravais lattices. Hint: Can you redefine this structure as a smaller, simpler Body-Centered Tetragonal unit cell?
Atomic Tetris Slides Atomic Tetris
Packing Efficiency & Voids
The Kepler Conjecture
In 1611, Johannes Kepler conjectured that the most efficient way to pack spheres is in a pyramid shape.
It took nearly **400 years** to mathematically prove it.
74.048%
Maximum Efficiency
Atomic Packing Factor (APF)
\[ \text{APF} = \frac{N \times V_{\text{atom}}}{V_{\text{unit cell}}} \]
\( N \)
\( V_{\text{atom}} \)
\( V_{\text{cell}} \)
Crucial Step: Relating the lattice parameter **\( a \)** to the atomic radius **\( R \)**.
Structural Comparison
BCC
CN = 8
Diagonal: \(\sqrt{3}a = 4R\)
Atoms/Cell: \( 2 \)
APF = 0.68
FCC
CN = 12
Diagonal: \(\sqrt{2}a = 4R\)
Atoms/Cell: \( 4 \)
APF = 0.74
The Space Between
Even in "close-packed" structures (74%), there is **26% empty space**.
These **Interstitial Voids** are where smaller impurity atoms (like Carbon in Iron) live.
Octahedral Voids
Surrounded by 6 atoms. Larger radius ratio.
Tetrahedral Voids
Surrounded by 4 atoms. Smaller radius ratio.
Void Volume Worksheet Void Volume
Packing Efficiency & Coordination Geometry
Student:
Date:
1. Step-by-Step APF Derivation
Derive the Atomic Packing Factor (APF) for a Face-Centered Cubic (FCC) lattice. Follow the logic of the geometry.
A. Relate \( a \) to \( R \)
Consider the face diagonal where spheres touch. Draw a sketch and use the Pythagorean theorem.
Result: \( a = \) ____________________
B. Calculate the Volume Ratio
Substitute your expression for \( a \) into the APF formula: \[ \text{APF} = \frac{4 \times (\frac{4}{3}\pi R^3)}{a^3} \]
Result (Percentage): ____________________ %
2. The Geometry of Voids
The "Radius Ratio Rule" helps determine if a smaller atom (radius \( r \)) can fit into a void created by larger atoms (radius \( R \)) without distorting the lattice.
The Tetrahedral Void
CN = 4
The tetrahedral void exists at coordinates like \( (\frac{1}{4}, \frac{1}{4}, \frac{1}{4}) \). The geometric limit for the radius ratio is: \[ \frac{r}{R} = \sqrt{\frac{3}{2}} - 1 \approx 0.225 \]
Problem
If a host metal has an atomic radius \( R = 1.44 \, \text{Å} \), what is the maximum radius of an impurity atom that can fit in the tetrahedral void?
Answer: \( r_{\text{max}} = \) ___________ \( \text{Å} \)
The Octahedral Void
CN = 6
The octahedral void is larger. Its geometric limit is: \[ \frac{r}{R} = \sqrt{2} - 1 \approx 0.414 \]
Reflection
Carbon atoms (\( r = 0.77 \, \text{Å} \)) fit into BCC Iron (\( R = 1.24 \, \text{Å} \)). Which void type do they prefer and why?
Consider both size and the number of available sites.
3. Scaling Up
FCC Lattice Count
For an FCC unit cell containing \( N=4 \) atoms, identify the total number of voids per unit cell.
Octahedral:
Tetrahedral:
Material Impact
How does the presence of interstitial atoms affect the mobility of dislocations in the lattice? (Short answer)
Vector Slice Slides Vector Slice
Miller Indices & Crystal Planes
How do we name a slice?
In a lattice, properties like **stiffness, conductivity, and growth rate** vary depending on the direction.
We need a mathematical shorthand to identify specific planes of atoms.
"Miller Indices allow us to describe 3D orientation with three simple integers."
(hkl)
The Miller Algorithm
01
Intercepts
Identify where the plane crosses the axes: \( x, y, z \).
If parallel to an axis, intercept is \(\infty\).
02
Reciprocals
Take the reciprocal of each intercept: \( 1/x, 1/y, 1/z \).
Reciprocal of \(\infty\) is \(0\).
03
Integerize
Multiply by the common denominator to clear fractions.
Result: (h k l)
Standard Notation
(hkl)
A Specific Plane
Parentheses indicate a single plane.
{hkl}
Family of Planes
Curly braces indicate all planes equivalent by symmetry.
[uvw]
A Vector Direction
Square brackets indicate a direction vector.
The Bar Notation
For negative intercepts, we place a bar over the number.
(\( \bar{1} \) 0 0)
Pronounced "one-bar zero zero"
Interplanar Spacing (\( d_{hkl} \))
For Cubic Systems Only:
\[ d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \]
The distance between successive parallel planes.
Higher index planes are **closer together** and have **lower atomic density**.
Index Investigation Worksheet Index Investigation
Visualizing and Calculating Crystal Planes
Student:
Date:
1. Identifying Indices
For each unit cell below, identify the Miller Indices (hkl) of the shaded plane. Use the standard origin \( (0,0,0) \) at the back-bottom-left corner.
Plane A
Intercepts:
Reciprocals:
Miller Indices:
Plane B
Intercepts:
Reciprocals:
Miller Indices:
2. Sketching Planes
Within each unit cell, sketch the indicated plane. Label the intercepts on the axes.
(1 1 0)
(0 2 0)
3. Interplanar Spacing Calculations
Copper has an FCC structure with a lattice parameter \( a = 3.61 \, \text{Å} \). Calculate the interplanar spacing \( d_{hkl} \) for the following three planes. Rank them from largest spacing to smallest.
(1 0 0)
\( d = \) _______________
(1 1 0)
\( d = \) _______________
(1 1 1)
\( d = \) _______________
Conclusion & Physical Significance:
Which plane has the highest atomic packing density in the FCC system? How does this relate to the calculated \( d_{hkl} \)?
Reciprocal Reality Slides Reciprocal Reality
The Fourier Transform of Matter
The Inverse Relationship
In physics, position and momentum are linked by a **Fourier Transform**.
Small distances in real space \(\rightarrow\) Large distances in reciprocal space.
"We don't see atoms; we see the diffraction pattern they create. That pattern lives in the Reciprocal Lattice."
Real Space
Reciprocal Space
Constructing the Reciprocal Lattice
For real lattice vectors \( \mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3 \):
\( \mathbf{b}_1 = 2\pi \frac{\mathbf{a}_2 \times \mathbf{a}_3}{\mathbf{a}_1 \cdot (\mathbf{a}_2 \times \mathbf{a}_3)} \)
Each reciprocal vector is perpendicular to two real-space vectors.
Key Property
\( \mathbf{a}_i \cdot \mathbf{b}_j = 2\pi \delta_{ij} \)
The dot product is \(2\pi\) if indices match, and zero otherwise.
From Planes to Points
Real Space
A crystal is a set of parallel planes (hkl).
Complex Planes
Reciprocal Space
The entire set of planes collapses into a **single point** defined by the reciprocal lattice vector:
\( \mathbf{G} = h\mathbf{b}_1 + k\mathbf{b}_2 + l\mathbf{b}_3 \)
Length of \( \mathbf{G} \) is proportional to \( 1/d_{hkl} \).
The Brillouin Zone
The **Wigner-Seitz cell** of the reciprocal lattice.
It represents the unique region of momentum space for electrons and phonons.
Dual Relationship:
Reciprocal of BCC is FCC.
Reciprocal of FCC is BCC.
1st BZ
Fourier Flip Worksheet Fourier Flip
Reciprocal Vector Construction & Zone Analysis
Student:
Date:
1. Reciprocal Vector Algebra
Consider a simple monoclinic lattice with real-space vectors: \[ \mathbf{a}_1 = a\mathbf{\hat{i}}, \quad \mathbf{a}_2 = b\mathbf{\hat{j}}, \quad \mathbf{a}_3 = c\cos\beta\mathbf{\hat{i}} + c\sin\beta\mathbf{\hat{k}} \] Show the derivation for the reciprocal vector \( \mathbf{b}_2 \).
Step 1: Calculate the Cross Product \( \mathbf{a}_3 \times \mathbf{a}_1 \)
Step 2: Calculate the Unit Cell Volume \( V = \mathbf{a}_1 \cdot (\mathbf{a}_2 \times \mathbf{a}_3) \)
Step 3: State the Final Reciprocal Vector \( \mathbf{b}_2 \)
\( \mathbf{b}_2 = \) __________________________________
2. Symmetry and Duality
A. Mathematically prove that the reciprocal vector \( \mathbf{b}_1 \) is perpendicular to the real-space planes defined by \( \mathbf{a}_2 \) and \( \mathbf{a}_3 \).
B. If a real-space cubic lattice has a parameter \( a \), what is the lattice parameter \( b \) of its reciprocal lattice? Include units.
\( b = \) ____________________
3. Mapping the First Brillouin Zone
The Brillouin Zone is the Wigner-Seitz cell of the reciprocal lattice. For a 2D square reciprocal lattice with spacing \( 2\pi/a \):
Origin
Construction Space (Sketch boundaries below)
Instructions:
Draw vectors from the origin to its 4 nearest neighbors.
Construct the perpendicular bisectors for each vector.
Shade the area enclosed by these bisectors (the 1st BZ).
Label the "High Symmetry Points":
\( \Gamma \) (the origin)
\( X \) (center of a face)
\( M \) (a corner)
Reflect:
Why is the Brillouin zone significant for understanding the "Band Gap" in semiconductors?
Diffraction Mapping Slides Diffraction Mapping
Bragg's Law & Crystal Discovery
Measuring the Invisible
Visible light has a wavelength of ~500 nm. Atoms are spaced ~0.1 nm (1 Å) apart.
To "see" atoms, we need light with wavelengths comparable to the spacing: **X-rays**.
"We don't look at the atoms; we look at how the light bounces off the planes."
Incoming Photon Wave
The Geometry of Interference
For constructive interference, the **Path Difference** must be an integer multiple of the wavelength.
\[ n\lambda = 2d\sin\theta \]
Spacing = d
The Diffraction Fingerprint
(110)
(200)
(211)
10° Intensity (Counts) vs 2θ 80°
Peak Position (\( \theta \)): Tells us the interplanar spacing \( d \).
Peak Identity (\( hkl \)): Tells us the crystal structure (BCC, FCC, etc.).
Systematic Absences
Not all planes produce a diffraction peak. Destructive interference from interior atoms in centered cells (I, F) can cancel out certain reflections.
BCC Selection Rule
\( h + k + l = \text{even} \)
Example: (110) is OK, (100) is absent.
FCC Selection Rule
\( h, k, l \) all odd or all even
Example: (111) is OK, (110) is absent.
Bragg Battle Worksheet Bragg Battle
Diffraction Analysis & Selection Rules
Student:
Date:
1. Solving for Spacing
A sample of unknown cubic metal is scanned using X-ray radiation with wavelength \( \lambda = 1.541 \, \text{Å} \) (Cu K\(\alpha\)). The first diffraction peak is observed at an angle of \( 2\theta = 44.5^\circ \).
Step A: Calculate \(\theta\)
Note: The detector measures \(2\theta\). Don't forget to divide.
\(\theta = \) _______________ \(^\circ\)
Step B: Use Bragg's Law
Solve for \( d_{hkl} \) assuming first-order diffraction (\(n=1\)).
\( d_{hkl} = \) _______________ \( \text{Å} \)
Work Area
2. The Systematic Absence Test
Identify which crystal structure (BCC or FCC) is present based on the observed reflections.
Observed Peak Index (hkl) \( h+k+l \) Sum Parity (h,k,l) Allowed for BCC? Allowed for FCC? (1 1 1) 3 All Odd No Yes (2 0 0) 2 All Even Yes Yes (2 2 0) 4 All Even Yes Yes (1 1 0) _________ _________ _________ _________
Conclusion:
If the (110) peak is present in the diffraction pattern, the sample structure MUST be: ____________________
3. Scaling and Resolution
A scientist wants to measure the lattice spacing of a protein crystal where \( d \approx 50 \, \text{Å} \). Could they use the same Cu K\(\alpha\) X-rays (\( \lambda = 1.541 \, \text{Å} \))? Explain the geometric constraint regarding the term \( \sin\theta \) in Bragg's Law.
Structure Factor Challenge:
Explain physically why the (100) reflection is "absent" in a Body-Centered Cubic (BCC) lattice even though the geometry of the planes exists. Hint: Think about the phase shift of waves scattered from the body-centered atom compared to the corner atoms.
