Crystal Structure PPT

1. Material Science

2. Unit I
Structure of Metals and Materials

3. Contents
Basic concepts of Crystal structures
Types of crystal systems
Crystal structure of metals( BCC, FCC and HCP systems)
ceramics & molecular arrangement of polymers
Miller indices
Indexing of lattice planes & directions
Lattice parameters (coordination number, no. of atoms per unit cell,
atomic packing factor, density)

4. Basic Concepts of Crystal Structure
Study of Properties of Metals by Structure
Types of Solids
Crystalline Solids
Built up of number of crystals (Metals or Non-metals)
E.g. Iron, Copper, Aluminum
Crystalline structures may be single crystal or aggregate of many
crystals known as Polycrystalline
Separated by well defined boundary
Non Crystalline Solids
Not crystalline in structure
Amorphous materials
E.g. Glass, wood, Plastics

5. Basic Concepts of Crystal Structure
Crystal Structure
Many unit cells repeat in 3 Dimensional space
Space Lattice
Infinite array of points in 3 Dimensional space
Every point is located symmetrically with respect to the other
Unit Cell
Basic structural part in the composition of materials

6. Crystal Systems
Sr.
No.
System Axial Lengths and Angles Unit Cell
01 Cubic a = b = c,
α = β = γ = 90O
02 Tetragonal a = b ≠ c,
α = β = γ = 90O
03 Orthohombic a ≠ b ≠ c,
α = β = γ = 90O
04 Rhombohedral a = b = c,
α = β = γ ≠ 90O

7. Crystal Systems
Sr.
No.
System Axial Lengths and Angles Unit Cell
05 Hexagonal a = b ≠ c,
α = β = 90O
, γ = 120O
06 Monoclinic a ≠ b ≠ c,
α = γ = 90O
≠β
07 Triclinic a ≠ b ≠ c,
α ≠ β ≠ γ ≠ 90O

8. Types of Crystal Structure
1. Simple Cubic Crystal Structure (SC)
2. Body Centered Crystal Structure (BCC)
3. Face centered Crystal Structure (FCC)
4. Hexagonal Closed Packed Structure (HCP)

9. Types of Crystal Structure
1. Simple Cubic Crystal Structure (SC)
Nc = No of corner atoms = 8
Nf = No of face atoms = 0
Ni = No of interior atoms = 0
Navg = Average no of atoms per unit cell
= Nc + Nf + Ni
8 2 1
= 8 + 0 + 0
8 2 1
Navg = 1

10. Types of Crystal Structure
2. Body centered Crystal Structure (BCC)
Nc = No of corner atoms = 8
Nf = No of face atoms = 0
Ni = No of interior atoms = 1
Navg = Average no of atoms per unit cell
= Nc + Nf + Ni
8 2 1
= 8 + 0 + 1
8 2 1
Navg = 2

11. Types of Crystal Structure
3. Face centered Crystal Structure (FCC)
Nc = No of corner atoms = 8
Nf = No of face atoms = 6
Ni = No of interior atoms = 0
Navg = Average no of atoms per unit cell
= Nc + Nf + Ni
8 2 1
= 8 + 6 + 0
8 2 1
Navg = 4

12. Types of Crystal Structure
4. Hexagonal Closed Packed Structure (HCP)
Nc = No of corner atoms = 12
Nf = No of face atoms = 2
Ni = No of interior atoms = 3
Navg = Average no of atoms per unit cell
= Nc + Nf + Ni
6 2 1
= 12 + 2 + 3
6 2 1
Navg = 6

13. Significance of Cubic Unit Cell*
Highest level of Geometrical Symmetry
Same symmetry as that of crystal structure

14. Lattice Parameters
Coordination Number (CN)
Number of Atoms per unit Cell
Atomic Packing Factor (APF)
Density

15. Lattice Parameters
Atomic Packing Factor (APF) and APF for Simple Cubic Structure
Hence APF for Simple Cubic Structure = 0.52

16. Lattice Parameters
Atomic Packing Factor for Body centered Cubic Structure
Hence APF for Body centered Cubic Structure = 0.68

17. Lattice Parameters
Atomic Packing Factor for Face centered Cubic Structure
Hence APF for Face centered Cubic Structure = 0.74

18. Lattice Parameters
Atomic Packing Factor for Hexagonal Close Packed Structure
No of atoms in HCP = 06, Atomic Radius r = a/2

19. Sr No Crystal
Structure
Average
No of
atoms per
unit cell
Co
ordination
No.
APF Materials
1. SCC 1 6 0.52 Do not
exist
2 BCC 2 8 0.68 Cr, Mo,
Alpha Fe,
Na
3 FCC 4 12 0.74 Al, Cu, Ag,
Pb, Au
4 HCP 6 12 0.74 Mg, Zn,
Cd, Ti

20. Lattice Parameters
Density = Mass of atoms in unit cell
Volume of unit cell
A. Linear Density (ρL
)
No of effective atoms NeL
per unit length on specific length L along any
direction in unit cell
ρL
= NeL
L
B. Planer Density (ρP
)
No of atoms per unit area of crystal plane
ρP
= Ne
A
Where, Ne
= Effective no of atoms on the plane with area A

21. Ceramics
Barium Titanate: Ceramic used in capacitor
Piezoelectric Material
BaTiO3
Ba at cubic corner
O at center of 6 faces
Ti at Body center
Metal give up electrons
Metallic ions – cataions – Positively charged
Non-metal gain electrons
Non metallic ions – anions – Negatively charged

22. Molecular Arrangement of Polymers
Polymer Molecular Arrangements
Polymer molecules are very large
Long and flexible chains with string of C- Atoms as a backbone
Side bonding of C Atoms to H Atoms
E.g. Ethylene
Polymer means repeated monomers
E.g.

23. Molecular Arrangement of Polymers
Applications of ethylene

24. Problems on volume density

25. Miller Indices for planes and Directions
Mathematical Notation to represent atomic planes and direction in crystrals
Use
Dislocation in crystals
Optical properties
Adsorptions (Adhesion of atoms) and Reactivity
Surface Tension

26. INTRODUCTION
NEED OF DIRECTIONS AND PLANES
GENERAL RULES AND CONVENTION
MILLER INDICES FOR PLANES
MILLER INDICES FOR DIRECTIONS
IMPORTANT FEATURES OF MILLER INDICES
Contents

27. The crystal lattice may be regarded as made up of
an infinite set of parallel equidistant planes passing
through the lattice points which are known as lattice
planes. In simple terms, the planes passing through
lattice points are called ‘lattice planes’. For a given
lattice, the lattice planes can be chosen in a different
number of ways.

28. The orientation of planes or faces in a crystal can
be described in terms of their intercepts on the three
axes.
Miller introduced a system to designate a plane in a
crystal.
He introduced a set of three numbers to specify a
plane in a crystal.
This set of three numbers is known as ‘Miller
Indices’ of the concerned plane.

29. Deformation under loading (slip) occurs on certain
crystalline planes and in certain crystallographic
directions.
Before we can predict how materials fail, we need
to know what modes of failure are more likely to
occur
Other properties of materials (electrical
conductivity, thermal conductivity, elastic modulus)
can vary in a crystal with orientation
NEED OF DIRECTIONS AND
PLANES

30. Procedure for finding miller indices of planes
Find intercepts X, Y and Z of the plane along with three axes.
Express the intercepts in terms of axial units.
Find the ratio of their reciprocals (i.e. 1/p)
Covert reciprocals into whole numbers by multiplying each of them by their
LCM
Enclose these nos. in round bracket which represents miller indices of the given
plane

31. Problems

37. Example-