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CRYSTAL STRUCTURES & PROPERTIES - Coggle Diagram
CRYSTAL STRUCTURES & PROPERTIES
01 Crystallographic Planes · slides 64–72
What does (hkl) mean?
Miller indices (h k l) identify the orientation of a crystal plane
h, k, l correspond to reciprocal intercepts with the x, y, z axes
Write plane indices in parentheses: (hkl)
Five steps: find Miller indices
Choose an origin; if plane crosses it, use a parallel plane or new origin
Read x, y, z intercepts in lattice units a, b, c
Take reciprocals: 1/x, 1/y, 1/z
Clear fractions to obtain smallest whole-number ratio
Write the final three indices as (hkl)
Rules to remember
Parallel to an axis → intercept = ∞ → reciprocal = 0
Negative intercept → write a bar over the relevant Miller index
Example: intercepts (1, −1, 1) → plane (1 −1 1)
Exam examples
(101): intercepts x=1, y=∞, z=1 → parallel to y
(221): intercepts x=1/2, y=1/2, z=1
Plane A: intercepts (1,1,∞) → reciprocals (1,1,0) → (110)
When given a plane drawing, first identify origin and intercepts
Why orientation matters
Silicon wafer: (111) discussed for epitaxial growth
Silicon wafer: (100) discussed for photolithography/device processing
Diamond: lattice orientation can influence biomedical surface interactions
02 Linear & Planar Density · slides 74–77
Linear density (LD)
Definition: atoms with centers on a crystallographic direction per unit length
Formula: LD = number of atoms centered on direction / direction length
Units: atoms/nm (or nm⁻¹)
Example in slide: LD[110] = 2/(4R) = 1/(2R)
Planar density (PD)
Definition: atoms with centers on a crystallographic plane per unit area
Formula: PD = number of atoms centered on plane / plane area
Units: atoms/nm² (or nm⁻²)
Example in slide: PD(110) = 2/(8√2 R²) = 1/(4√2 R²)
How to solve exam calculations
Determine the requested [uvw] direction or (hkl) plane
Count atom centers lying on that line or plane
Determine line length or plane area using R and geometry
Substitute values and state inverse-length or inverse-area units
Why these densities matter
Higher packing density relates to the slip process in metal crystals
Slip: atoms change position along slip planes under deformation
Slide 77 shows a crystal before and after slip
03 Single vs Polycrystals · slides 79–85
Single crystal
One uninterrupted periodic atomic arrangement through the whole specimen
Single crystals have one overall crystallographic orientation
Properties depend on direction → anisotropic
Examples & uses of single crystals
Single crystal semiconductor ingots sliced into wafers
Diamond single crystals for abrasives
Single crystal turbine blades
Quartz: some crystal planes fracture more readily than others
Polycrystalline material
Most engineering materials consist of many crystalline grains
Each grain is itself a single crystal
Grains differ in crystallographic orientation
Grain boundary = interface between adjacent grains
Slides give grain sizes of about 1 nm to 2 cm
How polycrystals form
a. Small crystallite nuclei appear
b. Crystallites grow and meet adjacent crystallites
c. Irregular-shaped grains appear on completion of solidification
d. Grain boundaries are visible under a microscope
Isotropy vs anisotropy
Isotropic = property does not change with measurement direction
Anisotropic = property changes with direction
Randomly oriented polycrystal grains → typically isotropic overall
Textured / aligned grains → polycrystal can be anisotropic
Elastic modulus E: BCC iron example
Single crystal: E along edge ≈ 125 GPa
Single crystal: E along diagonal ≈ 273 GPa
Random polycrystalline iron: E ≈ 210 GPa
Grain boundaries appear dark in optical microscope images
04 Diffraction & X-Rays · slides 87–94
Why study X-ray diffraction (XRD)?
Determine atomic and molecular arrangements in solids
Identify crystal structures and material phases
Support development and analysis of new materials
Diffraction phenomenon
Waves interact with regularly spaced scattering obstacles
Obstacle spacing is comparable to incoming wavelength λ
Scattered waves interfere depending on their relative phases
Three interference outcomes
Constructive: waves in phase → amplitudes reinforce (strong signal)
Destructive: waves out of phase → amplitudes cancel
Partial constructive: amplitudes reinforce incompletely
Why X-rays work
X-rays are high-energy electromagnetic radiation
Their short wavelengths are comparable to atomic spacings
Incoming X-rays diffract from crystallographic planes
Terms from the diagram
Monochromatic = one wavelength
Coherent = stable phase relationship
d(hkl) = spacing between successive parallel (hkl) planes
θ = Bragg angle; detector records diffraction peaks at 2θ
05 Bragg’s Law & Plane Spacing · slides 95–97
Bragg’s Law: MEMORISE
nλ = 2 d(hkl) sin θ
n = reflection order (usually 1 in the slide examples)
λ = X-ray wavelength
d(hkl) = interplanar distance
θ = angle between X-ray beam and the lattice plane
What the equation tells you
Extra path travelled by reflected rays = 2 d sin θ
Constructive reflection occurs when extra path = integer × wavelength
If Bragg’s Law is not met → very weak/nonconstructive beam
Useful rearrangements
d(hkl) = nλ / (2 sin θ)
θ = sin⁻¹[nλ / (2 d(hkl))]
Diffraction angle reported in XRD = 2θ
Convert 2θ → θ by dividing by 2
Cubic crystal interplanar spacing
d(hkl) = a / √(h² + k² + l²)
a = cubic lattice parameter / unit-cell edge length
h,k,l = Miller indices (e.g. 220 gives √8)
Rearrange: a = d(hkl) √(h² + k² + l²)
Combined exam equation
sin²θ = λ²(h² + k² + l²) / (4a²)
For fixed a and λ, sin²θ is proportional to h²+k²+l²
06 XRD Reflection Rules · slides 98–100
BCC: allowed vs forbidden peaks
ALLOWED when h + k + l is even
ABSENT when h + k + l is odd
Common early reflections: (110), (200), (211), (220), (310)
Their h²+k²+l² values: 2, 4, 6, 8, 10
FCC: allowed vs forbidden peaks
ALLOWED when h,k,l are all odd or all even
ABSENT when parity is mixed (odd and even together)
Common early reflections: (111), (200), (220), (311), (222)
Their h²+k²+l² values: 3, 4, 8, 11, 12
Fast BCC/FCC decision from two peaks
Find sin²θ(first peak) / sin²θ(second peak)
BCC: 2/4 = 0.50
FCC: 3/4 = 0.75
Exam clue: ratio near 0.50 means BCC; near 0.75 means FCC
HCP note (slide 98)
Slide 98: HCP absent when (h = 2k) = 3n and l is odd (n integer)
Slide 98 lists diffraction for all other HCP cases
07 Powder XRD & Patterns · slides 101–105
Powder diffraction method
Sample: powder or polycrystalline solid
Contains many randomly oriented crystallites
Properly oriented planes diffract and produce peaks
How the experiment is recorded
X-ray source shines on specimen
Detector rotates / measures X-ray intensity
Output: intensity (vertical) against 2θ (horizontal)
Peaks correspond to specific diffraction angles
Interpret XRD spectra
Peak positions (2θ) and intensities characterize elements or phases
Compare patterns to identify material composition
Compare different mixture fractions or relative amounts
Compare before and after sintering / material processing
Quick interpretation reminder
Use peak positions to calculate d and infer crystal structure
Use reflection rules to test BCC or FCC
Use comparison spectra to study phase changes
08 Worked XRD Problems · slides 106–110
Example 13: BCC Fe (220)
Given: a = 0.2866 nm, λ = 0.1790 nm, n = 1
Miller indices (220): h²+k²+l² = 4+4+0 = 8
d(220) = 0.2866/√8 = 0.1013 nm
sin θ = (1×0.1790)/(2×0.1013) ≈ 0.884
θ ≈ 62.13° → 2θ ≈ 124.26°
Example 14: unknown cubic metal
Given: λ = 0.154 nm
Peaks 2θ = 40°, 58°, 73°, 86.8°, 100.4°, 114.7°
First peak θ=20° gives sin²θ ≈ 0.1170
Second peak θ=29° gives sin²θ ≈ 0.2350
Ratio = 0.1170/0.2350 ≈ 0.498 ≈ 0.50 → BCC
Find a from sin²θ = λ²(h²+k²+l²)/(4a²)
Result from slide: a ≈ 0.318 nm
Identify metal: tungsten (W)
4-step solving strategy
Convert the given 2θ angles to θ
Compare sin²θ ratios with BCC (0.50) or FCC (0.75)
Use the corresponding allowed (hkl) to find a
Compare lattice parameter with known metals
Topic summary · slide 111
Crystalline = ordered; amorphous = no long-range regular order
Seven crystal systems and 14 Bravais lattices
Atomic arrangement changes packing factor and ductility
Crystallographic points, planes, directions and XRD are key skills