UNIT I Crystallography

8/8/2019 UNIT I Crystallography

http://slidepdf.com/reader/full/unit-i-crystallography 1/23

UNIT-I Crystallography




1.1 Review of Crystal Structure & Space Lattice:

1.1.1 Review-Types of solid










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




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




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.



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




Face Centred Cubic

- Atoms are located at each corner & the centre of all the

faces of the cubic unit cell.


=> 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.


=> a = 4r/

- Coordination No. =>CN= 8

- No. of Atoms per Unit Cell= 2

-Atomic Power Factor => APF

= 0.68




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


- Effective No. of Atoms per Unit Cell = 6

- Atomic Power Factor => APF = 0.74




- Hard Spherical atoms touch each other across the length=> a = 2r, c = 1.633a


- 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




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




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]




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




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




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.




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)




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




Tutorial -1




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.




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)




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.




END

Documents

UNIT I Crystallography