The document is a lecture on materials science and crystallography given by Hari Prasad. It begins by outlining the learning objectives which include differences between crystalline and non-crystalline structures, crystal systems, atomic packing factors, and unit cells. It then defines key concepts such as space lattices, unit cells, crystal systems, coordination number, and lattice parameters. Examples are provided of different crystal structures including simple cubic, body centered cubic, face centered cubic, and hexagonal close packed. Miller indices and how to determine plane intercepts are also discussed.
2. Learning objectives
• After the chapter is completed, you will be able
to answer:
• Difference between crystalline and
noncrystalline structures
• Different crystal systems and crystal structures
• Atomic packing factors of different cubic crystal
systems
• Difference between unit cell and primitive cell
• Difference between single crystals and poly
crystals
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3. What is space lattice?
• Space lattice is the distribution of points in
3D in such a way that every point has
identical surroundings, i.e., it is an infinite
array of points in three dimensions in
which every point has surroundings
identical to every other point in the array.
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4. Common materials: with various ‘viewpoints’
Glass: amorphous
Ceramics
Crystal
Graphite
Metals Polymers
5. Common materials: examples
Metals and alloys Cu, Ni, Fe, NiAl (intermetallic compound), Brass (Cu-Zn alloys)
Ceramics (usually oxides, nitrides, carbides) Alumina (Al2O3), Zirconia (Zr2O3)
Polymers (thermoplasts, thermosets) (Elastomers) Polythene, Polyvinyl chloride, Polypropylene
Based on Electrical Conduction
Conductors Cu, Al, NiAl
Semiconductors Ge, Si, GaAs
Insulators Alumina, Polythene*
Based on Ductility
Ductile Metals, Alloys
Brittle Ceramics, Inorganic Glasses, Ge, Si
* some special polymers could be conducting
6. The broad scientific and technological segments of Materials Science are shown
MATERIALS SCIENCE & ENGINEERING
PHYSICAL MECHANICAL ELECTRO-CHEMICAL
TECHNOLOGICAL
• Extractive
• Casting
• Metal Forming
•Welding
• Powder Metallurgy
• Machining
• Structure
• Physical
Properties
Science of Metallurgy
• Deformation
Behaviour
• Thermodynamics
• Chemistry
• Corrosion
in the diagram below.
To gain a comprehensive understanding of materials science, all these aspects
have to be studied.
7. Crystal = Lattice +
Motif
Motif or Basis:
typically an atom or a group of atoms associated with each lattice point
Lattice the underlying periodicity of the crystal
Basis Entity associated with each lattice points
Lattice how to repeat
Motif what to repeat
Definition 1
Crystal
Translationally periodic
arrangement of motifs
Lattice
Translationally periodic
arrangement of points
9. Space Lattice
A lattice is also called a Space Lattice
An array of points such that every point has
identical surroundings
In Euclidean space infinite array
We can have 1D, 2D or 3D arrays (lattices)
or
Translationally periodic arrangement of points in space is called a lattice
10. Unit cell: A unit cell is the sub-division of the
space lattice that still retains the overall
characteristics of the space lattice.
Primitive cell: the smallest possible unit cell of a
lattice, having lattice points at each of its eight
vertices only.
A primitive cell is a minimum volume cell
corresponding to a single lattice point of a structure
with translational symmetry in 2 dimensions, 3
dimensions, or other dimensions.
A lattice can be characterized by the geometry of its
primitive cell.
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11. Materials and Packing
Crystalline materials...
• atoms pack in periodic, 3D arrays
-metals
-many ceramics
-some polymers
Non-crystalline materials...
• atoms have no periodic packing
-complex structures
-rapid cooling
crystalline SiO2 (Quartz)
"Amorphous" = Noncrystalline
Si Oxygen
• typical of:
• occurs for:
noncrystalline SiO2 (Glass)
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12. Crystal Systems
Unit cell: smallest repetitive volume which contains
the complete lattice pattern of a crystal.
7 crystal systems
14 crystal lattices
a, b, and c are the lattice constants
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13. The Unite Cell is the smallest group of atom showing
the characteristic lattice structure of a particular metal.
It is the building block of a single crystal.
A single crystal can have many unit cells.
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14. Crystal systems
Cubic Three equal axes, mutually perpendicular
a=b=c ===90˚
Tetragonal Three perpendicular axes, only two equal
a=b≠c ===90˚
Hexagonal Three equal coplanar axes at 120˚ and a fourth unequal
axis perpendicular to their plane
a=b≠c == 90˚ =120˚
Rhombohedral Three equal axes, not at right angles
a=b=c ==≠90˚
Orthorhombic Three unequal axes, all perpendicular
a≠b≠c ===90˚
Monoclinic Three unequal axes, one of which is perpendicular to the
other two
a≠b≠c ==90˚≠
Triclinic Three unequal axes, no two of which are perpendicular
a≠b≠c ≠ ≠≠90˚
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15. Some engineering applications require single crystals:
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--diamond single
crystals for abrasives
--turbine blades
16. What is coordination number?
• The coordination number of a central atom in a
crystal is the number of its nearest neighbours.
What is lattice parameter?
• The lattice constant, or lattice parameter, refers
to the physical dimension of unit cells in a crystal
lattice.
• Lattices in three dimensions generally have
three lattice constants, referred to as a, b, and
c.
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17. Simple Cubic Structure (SC)
• Rare due to low packing density (only Po has this structure)
• Close-packed directions are cube edges.
• Coordination # = 6
(# nearest neighbors)
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21. Body Centered Cubic Structure (BCC)
• Atoms touch each other along cube diagonals.
--Note: All atoms are identical; the center atom is shaded
differently only for ease of viewing.
ex: Cr, W, Fe (), Tantalum, Molybdenum
• Coordination # = 8
2 atoms/unit cell: 1 center + 8 corners x 1/8
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24. Atomic Packing Factor: BCC
a
APF =
4
3
Close-packed directions:
p ( 3 a/4 ) 3
2
atoms
length = 4R =
volume
unit cell atom
a 3
volume
unit cell
3 a
• APF for a body-centered cubic structure = 0.68
a
R
2 a
3 a
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25. Face Centered Cubic Structure (FCC)
• Atoms touch each other along face diagonals.
--Note: All atoms are identical; the face-centered atoms are shaded
differently only for ease of viewing.
ex: Al, Cu, Au, Pb, Ni, Pt, Ag
• Coordination # = 12
4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8
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27. Atomic Packing Factor: FCC
• APF for a face-centered cubic structure = 0.74
maximum achievable APF
Close-packed directions:
length = 4R = 2 a
atoms
unit cell atom
APF =
4
3
4 p ( 2a/4)3
volume
a3
volume
unit cell
Unit cell contains:
6 x 1/2 + 8 x 1/8
= 4 atoms/unit cell
2 a
a
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28. FCC Stacking Sequence
• ABCABC... Stacking Sequence
• 2D Projection
A sites
C
B B
B
B B
B B
C sites
C C
A
B sites
• FCC Unit Cell
A
B
C
29. Putting atoms in the B position in the II layer and in C positions in the III layer we get
a stacking sequence ABC ABC ABC…. The CCP (FCC) crystal
+ +
A B
FCC
=
A
B
C
A
B
C
C
30. Hexagonal Close-Packed Structure
• ABAB... Stacking Sequence
• 3D Projection • 2D Projection
c
a
• Coordination # = 12
• APF = 0.74
(HCP)
Top layer
Middle layer
6 atoms/unit cell
ex: Cd, Mg, Ti, Zn
• c/a = 1.633
A sites
B sites
A sites Bottom layer
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31. APF for HCP
a Area of triangle =
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c
a
A sites
B sites
A sites
C=1.633a
Number of atoms in HCP unit cell=
(12*1/6)+(2*1/2)+3=6atoms
Vol.of HCP unit cell=
area of the hexagonal face X height of the hexagonal
Area of the hexagonal face=area of each triangle X6
h
a
풃풉
ퟐ
=
풂풉
ퟐ
=
ퟏ
ퟐ
풂.
풂 ퟑ
ퟐ
Area of hexagon = ퟔ.
풂ퟐ ퟑ
ퟒ
Volume of HCP= ퟔ.
풂ퟐ ퟑ
ퟒ
. 퐂 = ퟔ.
풂ퟐ ퟑ
ퟒ
. ퟏ. ퟔퟑퟑ퐚
APF= 6∗
ퟒ흅풓ퟑ
ퟑ
/(
ퟑ
ퟒ
∗ ퟔ ∗ ퟏ. ퟔퟑퟑ ∗ 퐚ퟑ)
a=2r
APF =0.74
39. Theoretical Density, r
Density = r =
Massof AtomsinUnitCell
TotalVolumeofUnitCell
n A
VC NA
r =
where n = number of atoms/unit cell
A = atomic weight
VC = Volume of unit cell = a3 for cubic
NA = Avogadro’s number
= 6.023 x 1023 atoms/mol
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40. Theoretical Density, r
• Ex: Cr (BCC)
A = 52.00 g/mol
R = 0.125 nm
n = 2
a = 4R/ 3 = 0.2887 nm
rtheoretical
ractual
a
R
r =
a 3
2 52.00
atoms
unit cell
g
mol
volume atoms
unit cell
mol
6.023 x 1023
= 7.18 g/cm3
= 7.19 g/cm3
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41. Polymorphism
• Two or more distinct crystal structures for the same
material (allotropy/polymorphism)
titanium
, -Ti
carbon
diamond, graphite
iron system
BCC
FCC
BCC
1538ºC
-Fe
1394ºC
-Fe
912ºC
-Fe
liquid
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42. Miller indices
Miller indices: defined as the reciprocals of
the intercepts made by the plane on the three
axes.
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43. Procedure for finding Miller indices
Determine the intercepts
of the plane along the axes
X,Y and Z in terms of the
lattice constants a, b and c.
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Step 1
45. Find the least common
denominator (lcd)
and multiply each by
this lcd
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Step 3
46. The result is written in
parenthesis.
This is called the `Miller
Indices’ of the plane in
the form (h k l).
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Step 4
47. (2,0,0)
Miller Indices for planes
(0,3,0)
(0,0,1)
Find intercepts along axes → 2 3 1
Take reciprocal → 1/2 1/3 1
Convert to smallest integers in the same ratio → 3 2 6
Enclose in parenthesis → (326)
48. Plane ABC has intercepts of 2 units
along X-axis, 3 units along Y-axis and 2
units along Z-axis.
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X
Z
Y
A
C
B
49. DETERMINATION OF ‘MILLER INDICES’
Step 1: The intercepts are 2, 3 and 2 on the three
axes.
Step 2: The reciprocals are 1/2, 1/3 and 1/2.
Step 3: The least common denominator is ‘6’.
Multiplying each reciprocal by lcd,
we get, 3,2 and 3.
Step 4:Hence Miller indices for the plane ABC is (3
2 3)
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50. IMPORTANT FEATURES OF MILLER INDICES
For the cubic crystal especially, the important features of
Miller indices are,
A plane which is parallel to any one of the co-ordinate
axes has an intercept of infinity ().
Therefore the Miller index for that axis is zero; i.e. for an
intercept at infinity, the corresponding index is zero.
A plane passing through the origin is defined in terms
of a parallel plane having non zero intercepts.
All equally spaced parallel planes have same ‘Miller
indices’ i.e. The Miller indices do not only define a
particular plane but also a set of parallel planes.
Thus the planes whose intercepts are 1, 1,1; 2,2,2; -3,-3,-3
etc., are all represented by the same set of Miller
indices.
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51. Worked Example:
Calculate the miller indices for the plane with intercepts 2a, - 3b
and 4c the along the crystallographic axes.
The intercepts are 2, - 3 and 4
Step 1: The intercepts are 2, -3 and 4 along the 3 axes
Step 2: The reciprocals are
Step 3: The least common denominator is 12.
Multiplying each reciprocal by lcd, we get 6 -4 and 3
Step 4: Hence the Miller indices for the plane is 6 4 3
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61. Crystallographic Directions
The crystallographic directions are fictitious
lines linking nodes (atoms, ions or molecules) of
a crystal.
Similarly, the crystallographic planes are
fictitious planes linking nodes.
The length of the vector projection on each of the
three axes is determined; these are measured in
terms of the unit cell dimensions a, b, and c.
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62. To find the Miller indices of a direction,
Choose a perpendicular plane to that direction.
Find the Miller indices of that perpendicular
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plane.
The perpendicular plane and the direction have
the same Miller indices value.
Therefore, the Miller indices of the
perpendicular plane is written within a square
bracket to represent the Miller indices of the
direction like [ ].
63. Summary of notations
Symbol
Alternate
symbols
Directio
n
[ ] [uvw] → Particular direction
< > <uvw> [[ ]] → Family of directions
Plane
( ) (hkl) → Particular plane
{ } {hkl} (( )) → Family of planes
Point
. . .xyz. [[ ]] → Particular point
: : :xyz: → Family of point
*A family is also referred to as a symmetrical set
64. For each of the three axes, there will exist
both positive and negative coordinates.
Thus negative indices are also possible,
which are represented by a bar over
the appropriate index. For example, the 1
The above image shows [100], [110], and [111] directions within a
unit cell
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65. The vector, as drawn, passes through the origin of the coordinate system,
and therefore no translation is necessary. Projections of this vector onto the
x, y, and z axes are, respectively,1/2, b, and 0c, which become 1/2, 1, and 0 in
terms of the unit cell parameters (i.e., when the a, b, and c are dropped).
Reduction of these numbers to the lowest set of integers is accompanied by
multiplication of each by the factor 2.This yields the integers 1, 2, and 0,
which are then enclosed in brackets as [120].
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67. Worked Example
Find the angle between the directions [2 1 1] and [1 1 2] in a
u u v v w w
½ ½
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cubic crystal.
The two directions are [2 1 1] and [1 1 2]
We know that the angle between the two directions,
1 2 1 2 1 2
2 2 2 2 2 2
1 1 1 2 2 2
cos
(u v w ) (u v w )
68. In this case, u1 = 2, v1 = 1, w1 = 1, Type equation here.u2 = 1,
(2 1) (1 1) (1
2) 5
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v2 = 1, w2 = 2
(or) cos = 0.833
= 35° 3530.
2 2 2 2 2 2
cos
2 1 l 1 1
2 6