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8/8/2019 UNIT I Crystallography
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UNIT-I Crystallography
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UNIT-I Crystallography
1.1 Review of Crystal Structure & Space Lattice:
1.1.1 Review-Types of solid
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1.1.2 Space Lattice & No. of Atoms/ Unit Cell-
It is infinite array of points in 3D arrangement. Each point
having identical surrounding as that of every other points in
the array.
X
Y
Z
ba
g
Centres of the atoms are considered points & all points are
joined togather by straight lines to form space or crystal
lattice.
The space lattice of a crystal is described by means of 3D
coordinate system in which co-ordinate axes coincide with
any three edges of the crystal that intersect at one point.
Here (a,b,c) and (E, F,G) are lattice parameters. Various
combinations of these parameters give 14 types of spacelattices called Bravais Lattices.
ab
c
Space Lattice Unit Cell
O
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UNIT-I Crystallography
1.1.3 Bravais Space Latice
The most common space lattices out of 14 are BCC, FCC &
HCP
Bravais
Lattices
Primitives Angles No. of
atoms/ Unit
cell
Examples
Simple
Cubic (SC)a = b = c E = F = K 8x1/8 = 1
Body
Centred
Cubic
(BCC)
a = b = c E = F = K 8x1/8+1 = 2 Mo, V, Mn
Face
Centred
Cubic
(FCC)
a = b = c E = F = K 8x1/8+6x1/2
= 4
Al, Pb, Ag
Hexagonal
ClosePacking
(HCP)
a = b = c E= F = 90
K = 120
12x1/6+2x
1/2+3 = 6
Mg, Ca, Zn
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1.1.4 Unit Cell : I is the smallest structural unit or buildingblock that can describe the crystal structure. Repetition of
the unit cell generates the entire crystal.
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UNIT-I Crystallography1.2 Co-ordination Numbers & Atomic Power Factor:
1.2.1 Coordination Number: No. of nearest atoms
to which given atom is boned.
1.2.2 Atomic Power Factor: Fraction of the volume
occupied by spherical atoms to that of volume of the unit
cell. i.e. APF =No. of Atoms per unit cell x (Volume of
Atoms/Volume of Unit Cell)
1.2.3 Effective No. of Atoms/Unit cell : Sum of the
total fraction of atoms in a unit cell that is shared with no. of
adjacent unit cells at their respective position (i.e. corners,
body or faces).
8 corner atoms shared with 8 cells = 8x1/8 = 16 faces atoms shared with 2 cells = 6x1/2 = 3
1 body atom shared with only 1 unit cell = 1x1 =1
Simple Cube
- Atoms are located at each corner of the unit cell
- Hard Spherical atoms touch each other across the length
=> a = 2r
- Coordination No. => CN = 6
- No. of Atoms per Unit Cell = 1
-Atomic Power Factor => APF= 0.52
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UNIT-I Crystallography
Face Centred Cubic
- Atoms are located at each corner & the centre of all the
faces of the cubic unit cell.
- Hard Spherical atoms touch each other across the length
=> a = 2 r
- Coordination No. =>CN= 12- No. of Atoms per Unit Cell = 4
- Atomic Power Factor => APF
= 0.74
Body Centred Cubic
- Atoms are located at each corner & the centre of the body
the cubic unit cell.
- Hard Spherical atoms touch each other across the length
=> a = 4r/
- Coordination No. =>CN= 8
- No. of Atoms per Unit Cell= 2
-Atomic Power Factor => APF
= 0.68
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UNIT-I Crystallography
Hexagonal Closed Packed Crystal Structure- Six atoms form regular hexagon, surrounding one atom in
centre. Another plane is situated halfway up unit cell (c-axis),
with 3 additional atoms situated at interstices of hexagonal
planes.
-Unit cell has two lattice parameters ³a´ & ³c´.-Ideal ratio c/a = 1.633
- Coordination No. =>CN= 12
- Effective No. of Atoms per Unit Cell = 6
- Atomic Power Factor => APF = 0.74
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- Hard Spherical atoms touch each other across the length=> a = 2r, c = 1.633a
- Coordination No. =>CN= 12
- No. of Atoms per Unit Cell= 6
-Atomic Power Factor => APF
Volume of primitive unit cell =3/2 a^2 x 1.633a
= 82 x r^3
Volume of atoms = 4pi/3 x r^3
APF = 6 x (4pi/3 x r^3)/ (3 x 82 x r^3)
APF = 0.74
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UNIT-I Crystallography
Co-ordination Numbers & Atomic Power Factor
Bravais
Lattices
No. of
atoms/
Unit Cell
Co-
ordin.
No. (CN)
Atomic
radius
Atomic
Power
Factor
Simple
Cubic (SC)1 6 r = a/2 0.52
Body
Centred
Cubic
(BCC)
2 8 r = a/4 0.68
Face
Centred
Cubic
(FCC)
4 12 r = a/2 0.74
Hexagonal
Close
Packing
(HCP)
6 12 c/a =
1.633
0.74
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UNIT-I Crystallography
1.3 Crystal Planes and Directions :The measurement of physical properties of a
material involves direction. Physical properties measured in
a direction will remain same after performing symmetry
operation and measuring in the same direction.
Thus the physical properties of crystal depend on thedirection of measurement, and so becomes necessary to
identify crystal directions.
Crystals may be thought of being made up of large
number of atomic planes oriented in various directions. So to
study crystal structure it is necessary to specify various
atomic planes in the crystal.
1.3.1 Crystal Directions: Indices of crystal direction
called Miller indices of direction are defined as follows:
Let the direction vector of given
direction is r = n1X + n2Y + n3Z where n1, n2 & n3 are the
intercepts in terms of unit cell parameters a, b & c resp.Then the Miller¶s direction indices (denoted as [h k l] in
respective X, Y & Z directions) can be found out as follows.
The family of the directions are denoted by <h k l>.
Miller indices of -ve intercept is shown with overhead( -) bar
as [h k l]
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UNIT-I Crystallography
Direction
Vectors
OA OB OC OH
Interceptsin terms of
unit cell
parameters
n1 = 0n2 = b
n3 = 0
n1 = an2 = b
n3 =0
n1 = an2 = b
n3 = c
n1 = ½ an2 = ½ b
n3 = c
Unit Cells
Parameters
a, b, c a, b, c a, b, c a, b, c
Ratio 0, 1, 0 1, 1, 0 1, 1, 1 ½, ½, 1
Smallest
Integer
0, 1, 0 1, 1, 0 1, 1, 1 1, 1, 2
Miller
Indices of directions
[0 1 0] [1 1 0] [1 1 1] [1 1 2]
Direction indices of shown
vectors in the fig. are found
out & given in the table.
H
A
GF
E
D C
B
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UNIT-I Crystallography
Direction
Vectors
Intercepts in
terms of unit
cell parametersUnit Cells
Parameters
Ratio
Smallest
Integer Miller Indices of
directions
More Direction indices (including negative intercepts)
O
K
N
M
L
P
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UNIT-I Crystallography
1.3.2 Crystal Planes: The planes are identified bytheir orientations w.r.t. crystallographic directions. The
orientation is specified by three parameters known as miller
indices of planes and is denoted by (h k l).
The family of the planes are denoted by {h k l}
Miller indices of -ve intercept is shown with overhead( -) bar
as (h k l)
The Miller indices of the planes are obtained as follows:
1. Find the intercepts of the plane in terms of lattice
parameters a, b, c
2. Find the ratio of the intercepts to that of latticeparameters.
3. Take reciprocals of the three ratios
4. Reduce it to smallest integer
5. Arrange the set in (h k l) form.
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1.3.3 Salient Features of Miller Indices:i. Miller indices of equally spaced parallel planes
are the same.
ii. A plane parallel to one of the coordinate axes has
an intercept at infinity.
iii. The plane passing through origin is defined byshifting the origin.
iv. Any two planes having Miller indices (h1 k1 l1) &
(h2 k2 l2) will be perpendicular, if
h1h2+k1k2+l1l2 = 0
v. When Miller indices contains integer of morethan one digit, the indices are separated by comma as
(3 4, 12) or (4, 12,15)
vi. For cubic crystals, the direction [h k l] is normal
to the plane having Miller index (h k l)
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Planes A B C D
Intercepts in
terms of unit
cell parameters
Unit CellsParameters
Ratio
Reciprocal
Smallest
Integer Miller Indices of
planes
Y
Z
ZZ
Y
Y
Y
Z
A
DC
B
Tutorial -1
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Tutorial -1
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UNIT-I Crystallography
Unit
Cell
Crystal
Direction
No. of
effectiveatoms (Ne)
Area
(A)
Planer
density Vp
(atoms/sq.mm)
SC
(1 0 0) 4x1/4 = 1 a x a 1/(a^2)
(1 1 0) 4x1/4 = 1 2 a x a 1/(2 a^2)
(1 1 1) 3x1/6 = 0.5 ½ (2 a x
2 a sin60)0.5/(3/2a^2)
1.4 Planer Density: Planer density is the no. of effective atoms per unit area of a crystal plane.
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UNIT-I Crystallography
Unit
Cell
Crystal
Direction
No. of
effective
atoms (Ne)
Area
(A)
Planer
density Vp
(atoms/sq.mm)
FCC
(1 0 0) 4x1/4+1=2 a x a 2/(a^2)
(1 1 0) 4x1/4 +
2x1/2= 22 a x a 2/(2 a^2)
(1 1 1) 3x1/6 +1/2x3= 2
½ (2 a x2 a sin60)
2/(3/2a^2)
BCC
(1 0 0) 4x1/4 = 1 a x a 1/(a^2)
(1 1 0) 4x1/4+1 =
2
2 a x a 2/(2 a^2)
(1 1 1) 3x1/6 +1 =
1.5
½ (2 a x
2 a sin60)1.5/(3/2a^2)
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UNIT-I Crystallography
Q.1 Explain the term crystal lattice and crystal structure.Distinguish between single crystal & polycrystalline structure.
Q.2 Explain with example three most common lattices observe
in metal.
Q.3 What do you understand by Co ordination number,
Effective no. of atoms per unit cell and atomic packing factor ?Q4. Calculate the APF for BCC, FCC & HCP crystal. What is
the significance of atomic packing Factor ?
Q.5 Show the following in a unit cell
(1 1 1), (1 0 1), (1 1 0), (1 1 2) & [1 1 0], [1 1 1]
Q.6 Lead has a BCC structure and an atomic radius of 3.499 A . Calculate the number of atoms per square mm of
(1 0 0) planes.
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