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Brief introduction to Crystallography
Crystallography Structure Symmetry d-spacing Defines results one achieves with XRD, TEM, SEM, etc.
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Crystalline vs. Amorphous Materials
Crystalline material atoms are situated in aperiodic array over large distances.
Amorphous (or non-crystalline) material where
long range order is absent. Therefore no realsymmetry.
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Close packed directions are cube edges.
SIMPLE CUBIC STRUCTURE (SCC)
Coordination # = 6
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Coordination # = 8
Adapted from Fig. 3.2,
Callister 6e.
(Courtesy P.M. Anderson)
Close packed directions are cube diagonals.--Note: All atoms are identical; the center atom is shaded
differently only for ease of viewing.
BODY CENTERED CUBIC STRUCTURE (BCC)
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(Courtesy P.M. Anderson)
Close packed directions are face diagonals.--Note: All atoms are identical; the face-centered atoms are shaded
differently only for ease of viewing.
FACE CENTERED CUBIC STRUCTURE (FCC)
Adapted from Fig. 3.1(a),
Callister 6e.
Coordination Number = 12 Atomic packing factor
(APF) = 0.74 (ideal)
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ABCABC... Stacking Sequence along cube diagonal 2D Projection:
A sites
B sites
C sites
B B
B
BB
B BC C
C
A
A
AB
C
FCC STACKING SEQUENCE
Coordination #= 12
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Coordination # = 12
ABAB... Stacking Sequence
Only difference from FCC is in the stacking (and the
orientation we generally consider when describing/drawing
the stacking):
- ABAB along z-axis, instead of ABCABC along cube
diagonal
3D Projection 2D Projection
A sites
B sites
A sites Bottom layer
Middle layer
Top layer
Adapted from Fig. 3.3,
Callister 6e.
HEXAGONAL CLOSE-PACKED STRUCTURE (HCP)
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Lattices and Unit Cells
What if we want to consider more complicated crystals?
Take advantage of symmetry and well known structures
A lattice is a repeating array of special points in space that haveidentical surroundings (nearest neighbors, etc).
A unit cell defines a repeating unit within the lattice strictly the unit cell is the smallest repeating cell (though this is
sometimes totally impractical and thus more convenient, but larger,
unit cells are employed).
A lattice does not define the crystal, only its symmetry.
Atoms (or ions, molecules, etc.)will be placed at any number of
basis points at given distances/directions (i.e. vectors) from thelattice points. ONLY THEN do we have a crystal.
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2-d Lattices
A lattice repeats in all directions,
and defines the symmetry.
The unit cell is the simplest
repeating structure.
Square
unit cell
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Basis
BUT, a lattice only identifies repeating points and
their symmetry.
Whats the lattice for the wall or floor in the
classroom?
Must be combined with a basis to know where
atoms are in the crystal
What kind of bricks are hung at each lattice point?
Are there smaller stones patterned regularly in
between the bricks?
Only by knowing a lattice AND a basis can a crystal
structure (or anything else symmetric) truly be built.
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2-d Lattice and Basis
Anions at lattice positions (0,0) Cations at (,0)
Cations at (0,)
squareLattice:
Basis:
tetragonal
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Lattice:
A minimized pattern that will be followed
Basis:
a) A vector that defines the position of hooks onwhich something (atoms, molecules, bricks) will behung with relation to the underlying pattern(lattice)
and b) What exactly will be hung on each hook(atom, molecule)
Rules:
At least one hook (basis) per lattice point (but can be more than 1)
Basis vectors must put atoms within the unit cell(not in an adjacent cell)
Each basis vector applies to every single latticepoint
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Seven Crystal Systems
Cubic a=b=c ===90
Tetragonal a=bc ===90
Orthorhombic abc ===90
Hexagonal a=bc ==90 =120
Monoclinic abc ==90
Triclinic abc
Rhombohedral a=b=c ==90
In 3 dimensions, there are only 7 basic crystal systems.Easily defined by: 3 edge lengths (a, b and c) and 3
interaxial angles ( and).
This
class
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Cubic Bravais Lattices
Why no base centered cubic?
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/
Tetragonal
a=bc
/
Why no base centered tetragonal?
Tetragonal Bravais Lattices
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Orthorhombic Bravais Lattices
/ /
/ // /
b
/ /
/ /
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Other Bravais Lattices
a=bc
90
=120
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Fourteen Bravais Lattices (13-14)
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Defining Basis Positions Using
Fractional Co-ordinates
Note that positions are indicated as ( , , ).To locate atom positions in cubic unit cells we use
rectangular x, y, and zaxes.
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The FCC Structure
Note: All atoms are not always
shown because technically thereare only 4/unit cell (for FCC)
Lattice=Face Centered Cubic
Basis=red atoms at (0,0,0)
-of course it looks like there are more atoms, but these simply result
from the repeating lattice and symmetry; ie they are not distinct
Each position might be provided, but probably the fewest
number of positions will be giventhe rest fill in as long asyou know the Bravais lattice (ie the basic symmetry).
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Simple unit cells
Based on Simple Cubic
Lattice with a basis of:
1) anions at the SC points
2) cations at the center
[offset by (, , )]
Based on FCC Lattice with a basis of:
1) anions at the FCC points
2) cations at the center
[offset by (, 0, 0)]
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Consider CaF2 : based on the CsCl structure.
BUT: there are half as many Ca2+ as F- ionshow can
we accommodate this?
CsCl structure w/only half of cation sites occupied.
The initial unit cell is no longer valid since every lattice
site doesnt have a basis attached to it (some are void).
Instead, we use a supercell. 2 possibilities here:
AmXp STRUCTURES
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AmXp STRUCTURES Consider CaF
2
: Based on the CsCl structure,
BUT: there are half as many Ca2+ as F- ions
This is accommodate with empty cubes.
i.e. CsCl based structure w/only half of cation sites
occupied. Thus, the proposed CsCl unit cell is no longer valid
since every Simple Cubic lattice site doesnt have a
basis attached to it (half are void!). Instead, use
a supercell.
Other
possibilities:
m is not equal to p
Adapted from Fig. 12.5, Callister 6e.
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Crystallographic Directions and Planes
Terminology
Positions (,,,)
Directions [] note no commas
Planes () note no commas
Families of directions note no
commas
Families of planes {} note no commas
100)=bar one zero zero and NOT one bar zero zero
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Directions in Cubic Unit Cells
Position co-ordinates of vector OM = 1, 0.5, 0 (but these arent
integers)
Position co-ordinates must be multiplied by 2 to obtain integers.
Direction indices of OM = 2 (1, 0.5, 0) = [2, 1, 0]
A negative index direction (e.g. see vector ON) is written with a
bar over the index
For cubic crystals, the crystallographic direction indicesare the vector components of the direction resolved along
each of the co-ordinate axes (x,y,z)
and reduced to the smallest integers.
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[Directions]
[100] [110]
[111]
[021] [011]
[200]
[210]
x
y
z
1. Draw cell + origin
2. Draw vector
3. reduce to
smallest integer
values
4. [xyz]
Group exercise
[Di ti ]
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more [Directions]
[100]
[011]
[011]
xy
z
For negative directions:
a. Add more unit cells.
OR
b. Shift the origin.
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Directions in Cubic Unit Cellsz
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
[001 ] [01 1 ][11 1]
[112 ][2 1 2]
[22 1 ]
The origin may be shifted. Thus:
a) If all positive, origin is at(0,0,0) b) If z is negative, origin will be at (x,y,1)
c) If y is negative, origin will be at (x,1,z) d) If x is negative, origin will be at (1,y,z)
Group exercise
f il f di ti
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=family of directions
[100]
[010]
[001]
[100]
[010]
[001]
xy
z
A includes all possible
directions with the same basic coordinates.
1 D h i i ll
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(Planes)
(xyz)
(100)
(110)
(111) (100)
(020)
(040) (04/30)
()
x
y
z1. Draw the origin, cell,
and normal vector.
2. Draw the plane at adistance from the origin
of 1/sqrt(a2+b2+c2).
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{Planes} =family of planes
{xyz}
{100}
{110}
{111}
{100}
x
y
z
How many
equivalentplanes are there
in each family
(multiplicity)?
multiplici
ty
3
6
?
?
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For a simple cubic lattice and (0,0,0) basis:
Along (xyz) plane
(100)
(110)
xy
z
Should be able to draw atoms in, and above/below,
a given plane in a crystal.
For CsCl (simple cubic lattice
Group exercise
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For CsCl (simple cubic lattice,
basis of Cl-(0,0,0) and Cs-
(,)) Along (xyz)plane
(100)
(110)
xy
z
Group exercise
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Crystallographic Planes Miller Indices
Intercept; 1, ,
Reciprocal; 1/1, 1/, 1/
Simplify; 1, 0, 0
Miller indices; (1 0 0)
1,1,
1/1,1/1,1/
1,1,0
(1 1 0)
1,1,1
1/1, 1/1, 1/1
1,1,1
(1 1 1 )
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Crystallographic Planes Miller
Indices
When planes are more complicated, follow these
rules:
1. Choose a plane that does not
pass through the origin at
(0,0,0)
2. Determine the intercepts of the
plane in terms of the x, y, and z
axes of the cube. 1/3,2/3,1
3. Form the reciprocals of these
intercepts 3, 3/2, 1
4. Bring to smallest integers by
multiplying or dividing through
by a common factor. 6, 3, 2
5. Enclose integers in parentheses
(6 3 2)
Group exercise
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Planes in Cubic Unit Cellsz
x
y
z
x
y
z
x
y
z
x
y
z
x
y
z
x
y
(010)(11 0)
(111 )
(012 ) (112) (342 )
Group exercise
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