Crystal Structure
Ashraf Bastawros
www.public.iastate.edu\~bastaw\courses\Mate271.html
Week 2
Material Sciences and Engineering
MatE271
1
Goals for this unit
- Define basic terms in crystallography
- Be able to identify 7 crystal systems and 14
Bravais lattices
- To compare and contrast the structures of metal,
ceramics and polymer materials
- Explain the three most important structures for
metals
- Calculate atomic structure densities
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1
Crystal Structure and Periodicity
• Crystalline Materials
atoms are in an ordered 3-D periodic array
Single crystal or polycrystalline solids
(metals, ceramics, semiconductors, some polymers )
• Amorphous Materials
short range order
( glasses, many polymers)
• Intermediates
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Crystal Structure and Periodicity
Lattice - a 3-D array of positions in space. A
geometric framework on which to hang atoms
2-D
Unit Cell - the smallest repeat unit which defines
the crystal structure
2-D
3-D
Note - In these cases, there is one atom centered on each lattice site
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2
Simple Cubic (SC)
SC lattice and crystal structure
Lattice
a = 2R
Where:
Hang 1 atom on
each lattice point
R = atomic radius atom
a = lattice parameter
Crystal Structure
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Week 2
Axes Labels
By convention
origin - 0,0,0
z
c
α
β
a
b
y
γ
x
Lattice constant
- a, b and c are lengths of edges
- α, β, γ are angles (α is across from a, etc.)
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3
Crystal Systems (7-systems)
Cubic
a=b=c
α=β=γ=90
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Tetragonal
a=b≠c
α=β=γ=90
Orthorhombic
a≠b≠c
α=β=γ=90
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Crystal Systems (7-systems)
Rhombohedral
a=b=c
α=β=γ ≠ 90
Hexagonal
a=b≠c
α=β=90 γ=120
“Pushed over”
cube
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“Squished”
tetragonal
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4
Crystal Systems (7-systems)
Monoclinic
a≠b≠c
α=β=90, β ≠ 90
“Pushed over” orthorhombic
(in one direction)
Triclinic
a≠b≠c
α ≠ β ≠ γ ≠ 90
Orthorhombic
a≠b≠c
α=β=γ=90
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“Pushed over” orthorhombic
(in two directions)
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Fourteen Crystal (Bravais) Lattices
- Lattices are merely geometric constructs
forming a framework of points in space on (or
around) which you can position atoms
- In the simplest case, you can apply one atom
centered at each lattice point…Most
METALLIC materials fall into this class
- Represent the entire lattice with one unit cell
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5
Cubic Systems
- Based on a Cubic unit Cell
a=b=c, α=β=γ=90°
n
Simple Cubic (SC)
w One atom on each corner
w Coordination number of 6
n
Body-centered Cubic (BCC)
w One atom on each corner and one in the center
w Coordination number of 8
n
Face-centered Cubic (FCC)
w One atom on each corner and on each face
w Coordination number of 12
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Simple Cubic (SC)
- SC lattice and crystal structure
Lattice
a = 2R
Where:
Hang 1 atom on
each lattice point
R = atomic radius atom
a = lattice parameter
Crystal Structure
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6
Body Centered Cubic (BCC)
- BCC lattice
and crystal
structure
A
a =
4R
3
where:
R = atomic radius atom
a = lattice parameter
B
Staking order A-B-A-B
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Week 2
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Cubic Packing - BCC
a
√3a
a
√2 a
√2 a
a
√3a=4R
a=4R/√3
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Face Centered Cubic (FCC)
a =
4R
2
a = 2R 2
where:
R = atomic radius atom
a = lattice parameter
Stacking
Direction
[100]
B
A
Staking order A-B-A-B
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Cubic Packing - FCC
a
√2 a
a
4R=√2a
a=2R√2
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Atomic Packing Factor
- Fraction of solid sphere volume in a unit
cell
APF = volume of atoms in unit cell
total cell volume
- Example for FCC
How many atoms are in the unit cell?
What is the cell volume?
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Atomic Packing Factor - FCC
4R
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- There are 4 spheres
in the cell
- The volume of the
spheres:
4 3 16 3
4 × πR = πR
3
3
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9
Atomic Packing Factor - FCC
- What is the volume of
the cube?
n
a3 (length of side
cubed)
- What is that in terms
of R? (sphere radius)
n
n
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a = 2R√2
(2R√2)3 = 16R3√2
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Atomic Packing Factor - FCC
APF =
volume of atoms in unit cell
= Vs /Vc
total cell volume
16
πR 3
3
16 R 3 2
=
π
3 2
= 0.74
The atomic packing factor for FCC is 0.74
(74% of the space is filled)
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10
HCP Crystal Structure
This is
not a
lattice
point!
c
a
0,0,0
HCP crystal structure = Hexagonal
Bravais lattice with 2 atoms per lattice site.
1st atom at 0,0,0 (i.e. lattice point)
2nd atom at 2/3, 1/3, 1/2
Note - 2nd atom environment different than
the 1st atom
For “ideal” HCP only
c = 1.633 a
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HCP Crystal Structure
This sphere is not totally
inside unit cell although
the partial spheres all add
up so that there are two
spheres per unit cell!
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ABA vs ABC Packing
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FCC and HCP Compared
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Structure of Selected Metals
Metal
Aluminum
Chromium
Cobalt
Copper
Gold
Lead
Crystal
Structure
FCC
BCC
HCP
FCC
FCC
FCC
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Atomic
Radius (nm)
0.1431
0.1249
0.1253
0.1278
0.1442
0.1750
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Example Problem
o If you know the crystal structure, the atomic
radius and the atomic weight, you can
calculate the density of a particular material.
o Copper has an atomic radius 0.128 nm an
FCC crystal structure and an atomic weight of
63.5 g/mol. Calculate its density.
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13
Crystalline & Noncrystalline Materials
o Single crystals
• repeated arrangement of atoms extends
throughout the specimen
• all unit cells have the same orientation
• exist in nature
• can also be grown (Si)
• without external constraints, will have flat,
regular faces
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Polycrystalline Materials
o Crystals of different
• sizes
• orientations
• shapes
o Grain Boundaries
• mismatch between
two neighboring
crystals
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14
Polycrystalline Materials
o Most crystalline materials are composed of many
small crystals called grains
o Crystallographic directions of adjacent grains are
usually random
o There is usually atomic mismatch where two grains
meet - this is called a grain boundary
o Most powdered materials have a many randomly
oriented grains
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Reading Assignment
Shackelford 2001(5th Ed)
– Read Chapter 3, pp 59-64
Read ahead to page 88, 101-110
Check class web site:
www.public.iastate.edu\~bastaw\courses\Mate271.html
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