10-Aug-151

Chapter 1

Crystallography

Session 1

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Classification of matter

Matter

GASES LIQUIDS SOLIDS

CrystalsQusicrystalsAmorphous

Single

CrystalsPolycrystals

Crystalline Solids

• Regular arrangement of

atoms along the 3 D.

• Have long range order.

• Sharp melting point.

• High and fixed heat of

fusion.

• They are anisotropic.

• They show all

characteristic of solids.

• They have regular cut.

• No regular arrangement of

atoms.

• Have short range order.

• No sharp melting point.

• No fixed heat of fusion.

• They are isotropic.

• They don’t show all

characteristics of solids.

• They give irregular cut.

Amorphous Solids

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Two dimensional representation of Single, poly

crystalline and amorphous materials

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Basic Definitions

Lattice

Linear Lattice

Plane Lattice

Space Lattice

a

Translation vector

Ox

y

a

b

B

2a

b

OB 2 a b

• An infinite array of points

• Each point will have

Identical surrounding.

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A group of atoms (or ions) which when attached

to every lattice points produces the crystal structure.

Basis

Basis

Lattice + Basis = Crystal structure10-Aug-15 7

The building block of the crystal consisting of atoms, ions or

molecules which when translated along the three axes gives

the actual crystal structure. It should occupy a minimum

volume.

There are two distinct types of unit cell:

• Primitive unit cell

• Non-primitive unit cell

• Smallest repeatable unit in a point Lattice.

• Choice of unit cell is not unique

Unit Cell:

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UNIT CELL

Primitive Non-primitive

• Single lattice point per cell

• Smallest area in 2D, or

• Smallest volume in 3D

• More than one lattice point

per cell

• Integral multibles of the are

of primitive cell

Body centered cubic(bcc)

Conventional ≠ Primitive cell

Simple cubic(sc)

Conventional = Primitive cell

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Crystallographic axes

Axes resulted by the intersection of the three non-coplanar

faces of the unit cell.

The angles between these faces or crystallographic axes

are known as interfacial or inter-axial angles.

Lattice parameters

The translational vectors (or primitives) a, b, c of a unit cell

along X, Y, Z axes and inter-axial angles α, β, γ

: Angle between the axes Y and Z

: Angle between the axes Z and X

: Angle between the axes X and Y

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1. Cubic Crystal System

a = b = c = = = 90°

Crystal systems

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2.Tetragonal system

a = b c = = = 90°

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3. Orthorhombic system

= = = 90°a b c

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4. Monoclinic system

a b c = = 90°, 90°

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5. Triclinic system

90°a b c

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= = 90°

6. Rhombohedral (Trigonal) system

a = b = c

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7. Hexagonal system

a = b c = = 90°, = 120°

Summary of crystal systems

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Click here Crystal systems10-Aug-15 19

Session 2

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Bravais Lattice

An infinite array of discrete points generated by a set of discrete

translation operations described by

where ni are integers and ai are primitive vectors

14 Bravais lattices are possible in 3-dimentional space.

These are obtained by combination of lattice system with lattice

types

Lattice system (crystal system): gives the information regarding

the lattice parameters

Lattice types: Gives the information about lattice points present

in the unit cell.

1 1 2 2 3 3R n a n a n a

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Lattice types

Primitive

(P)

Body

centered (I)

Face

centered (F)

Base

centered (C)

18 1

8n

18 1 2

8n

1 18 6 4

8 2n

1 18 2 2

8 2n

Lattice types (animation)10-Aug-15 22

Crystal systemPrimitive

(P)

Body

centered

(I)

Face

centered

(F)

Base

centered

(C)

Cubic

Tetragonal Orthorhombic

Monoclinic

Triclinic

Rhombohedral

Hexagonal

14 Bravais lattices

Click here for animation for 14 Bravais lattices10-Aug-15 23

It is the ratio of volume occupied by the atoms in a unit cell to the

total volume of the unit cell.

Atomic Packing Factor or Packing Fraction

3

3

Volume of packed ions =

Volume of primitive unitcell

Volume of packed ions = ×Z

4

3

Packing fraction (%)

Vo

=

lume of unitc

e l

l

Z

r

a

Now, we are going to apply this for three crystal structures

having cubic symmetry

• Simple cubic

• Body centered cubic

• Face centered cubic

where, r = radius of atoms

a = unit cell edge

Z = no. of lattice points per unit cell

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Relation between atomic radius and edge length

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Simple cubic Body centered

cubic

Face centered

cubic

3

3

4

3PF = Z

r

a

3

3

4

3PF = Z

r

a

3

3

4

3PF = Z

r

a

/ 2r a Z = 1 3 / 4r a Z = 2 / 2 2r a Z = 4

PF = 52% PF = 68% PF = 74%

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Miller Indices

A notation of planes in Bravais lattices

Miller Indices are the reciprocals of the fractional intercepts

(with fractions cleared) which the plane makes with the

crystallographic x, y, z axes of the three nonparallel edges of the

cubic unit cell.

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. 28

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Different lattice planes in a crystal

d

29

Determine the coordinates of the intercepts made by the plane

along the three crystallographic axes.

Express the intercepts as multiples of unit cell dimensions or

lattice parameters along the axes.

Determine the reciprocals of these numbers.

Reduce these reciprocals into the smallest whole numbers by

multiplying each with their L.C.M to get the smallest whole

number.

Enclose the smallest whole numbers in ( )

This give the Miller indices (h k l) of the plane.

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Procedure for finding Miller Indices

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Crystallographic Planes

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z

x

ya b

c

4. Miller Indices (110)

Example-2: a b cz

x

ya b

c

4. Miller Indices (100)

1. Intercepts 1 1

2. Reciprocals 1/1 1/1 1/

1 1 03. Reduction 1 1 0

1. Intercepts 1/2

2. Reciprocals 1/½ 1/ 1/

2 0 03. Reduction 2 0 0

Example-1: a b c

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Crystallographic Planes

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z

x

ya b

c

4. Miller Indices (634)

Example-3:1. Intercepts 1/2 1 3/4

a b c

2. Reciprocals 1/½ 1/1 1/¾

2 1 4/3

3. Reduction 6 3 4

(001)(010),

Family of Planes {hkl}

(100), (010),(001),Ex: {100} = (100),

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Crystal planes

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Important features of Miller indices

If a plane is parallel to any of the coordinate axis, then its

intercepts will be infinity.

All the parallel equidistant planes have the same Miller Indices.

Miller Indices define a set of equidistant parallel planes.

If the Miller Indices of two planes have the same ratio then the

planes are parallel to each other.

If (hkl) are the Miller Indices of a plane, then the plane cuts the

axis into h, k and l equal parts.

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SUMMARY OF MEANINGS OF PARENTHESES

(q,r,s) represents a point – note the exclusive use of

commas

(hkl) represents a plane

{hkl} represents a family of planes

[hkl] represents a direction

<hkl> represents a family of directions

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The inter-planar spacing between any crystallographic planes

belonging to the same (h,k,l) family is denoted by (dhkl).

Distances between planes defined by the same set of Miller

indices are unique for each material.

Inter-planar spacing in

Crystals

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Inter-planar spacing in different crystal systems

Inter-planar spacing in other crystal system is more complex

in nature

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Session 3

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Q1:

Q2:

Problems on Miller indices

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Q3: Determine the miller indices for the planes shown in

the following unit cell

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Q4: What are Miller Indices? Draw (111) and (110) planes in a

cubic lattice.

Q5: Sketch the following planes of a cubic unit cell (001), (120),

(211)

Q6: Obtain the Miller indices of a plane which intercepts at a, b/2

and 3c in simple cubic unit cell. Draw a neat diagram showing the

plane

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Problems on inter-planar

spacing

1. Explain how the X-ray diffraction can be employed to determine

the crystal structure. Give the ratio of inter-planar distances of

(100), (110) and (111) planes for a simple cubic structure.

2. The distance between (110) planes in a body centered cubic

structure is 0.203 nm. What is the size of the unit cell? What is

the radius of the atom?

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Diffraction:

The light rays when pass through a sharp edge of an object can

form some bright regions inside the geometrical shadow of the object.

This is due to the bending nature of light called diffraction.

Main criteria for diffraction:

The thickness of obstacle or slit is of the order of incident light

wavelength.

Diffraction in Crystals:

In the year 1912, a German physicist Laue suggested that the three-

dimensional arrangement of atoms in a crystal can produce diffraction.

(But with which radiation)

X-ray diffraction in Crystals:

The wavelength of X-rays is of the order of an angstrom which is in the

range of spacing between the atoms in crystals. Hence, produce

Diffraction. (Friedrich and Knipping)

Diffraction through Crystals

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Importance of X-rays

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Production of X-Rays

• A typical X-ray tube.

• X-rays are produced whenever high-

speed electrons collide with a metal

target.

• A source of electrons – hot W filament,

a high accelerating voltage between the

cathode (W) and the anode

• metal target, Cu, Al, Mo, Mg works as

anode.

n = 2dsin

English physicists Sir W.H. Bragg and his son Sir W.L. Bragg

developed a relationship in 1913 to explain why the cleavage

faces of crystals appear to reflect X-ray beams at certain

angles of incidence (theta, ).

The variable d is the distance between atomic layers in a

crystal, and the variable lambda is the wavelength of the

incident X-ray beam; n is an integer.

This observation is an example of X-ray wave interference,

commonly known as X-ray diffraction (XRD), and was direct

evidence for the periodic atomic structure of crystals

postulated for several centuries.

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Bragg’s Law in one dimension

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Deriving Bragg’s Law: n = 2dsin

Constructive interferenceoccurs only when

PATH DIFFERENCE = n

n = AB + BC

AB=BC

n = 2AB

Sin = AB/d

AB = dsin

n =2dsin

= 2dhkl sinhkl

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Session 4

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Problem on Bragg’s law

1. A beam of X-rays of wavelength 0.071 nm is diffracted by (110) plane of rock salt

with lattice constant of 0.28 nm. Find the glancing angle for the second-order

diffraction.

2. A beam of X-rays is incident on a NaCl crystal with lattice plane spacing 0.282

nm. Calculate the wavelength of X-rays if the first-order Bragg reflection takes

place at a glancing angle of 8°35′. Also calculate the maximum order of

diffraction possible.

3. Monochromatic X-rays of λ = 1.5 A.U are incident on a crystal face having an

inter-planar spacing of 1.6 A.U. Find the highest order for which Bragg’s

reflection maximum can be seen.

4. For BCC iron, compute (a) the inter-planar spacing, and (b) the diffraction angle

for the (220) set of planes. The lattice parameter for Fe is 0.2866 nm. Also,

assume that monochromatic radiation having a wavelength of 0.1790 nm is used,

and the order of reflection is 1.

5. The metal niobium has a BCC crystal structure. If the angle of diffraction for the

(211) set of planes occurs at 75.990 (first order reflection) when monochromatic X-

radiation having a wavelength 0.1659 nm is used. Compute (a) the inter-planar

spacing for this set of planes and (b) the atomic radius for the niobium atom.

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X-ray diffraction Methods

Laue Method

Rotating Crystal method

Powder Method

• Polychromatic Beam

• Diffraction angle is fixed

• Single Crystals

• Monochromatic Beam

• Variable diffraction angle

• Single Crystal

• Monochromatic Beam

• Variable diffraction angle

• Polycrystals (powder)

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Laue Method

First diffraction method ever used for structural studies.

It was demonstrated by von Laue.

We can use this method for identification of any orientation

present in single crystals.

Quality of the single crystals can also be checked with this

method.

In this method, single crystal is fixed at a particular position.

(Diffraction angle is fixed)

Continuous spectrum of radiation from X-ray tube will be

used in this method. (polychromatic beam)

Each set of planes diffract out particular wavelength which

satisfy Bragg’s law for a given value of d and .

Two categories

• Back-reflection method

• Transmission method

2tan(180 2 )r

D

Back-reflection method

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r1

2

X-rays

Collimator Single crystal

B

F

S

D

Transmission method

r2

(180 -2 )

X-rays

CollimatorSingle crystal

B

F

S

D

1tan 2r

D

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Positions of spots on the film depends on the orientation of

single crystal relative to incident beam.

Spots are distorted or smeared out if the crystal is bent or

twisted.

With these two, we can analyze the quality and orientation of

crystals.

Transmission

method

Back-reflection

method

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Rotating Crystal Method

A single crystal is used in this method.

It is mounted with one of its axis normal to a monochromatic

X-ray beam.

A cylindrical film is placed around it & the crystal is rotated

about the chosen axis.

As the crystal rotates, sets of lattice planes will at some point

make the correct Bragg angle for the monochromatic incident

beam, and at that point a diffracted beam will be formed.

The reflected beams are located on the surfaces of imaginary cones.

shape & size of unit cell as well as the arrangement of

atoms (Crystal structure of unknown materials) inside the cell

can be determined with the help of this method.

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Experimental setup of

Rotation Crystal method

Cylindrical film

Axis of Crystal

Single crystal

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Powder Method

Very fine powder sample (crystal) is used instead of a single

crystal.

This powder sample is an assembly of tiny crystals oriented in

random direction with respect to incident beam (monochromatic

X-rays).

Diffraction will come from all set of planes present in the powder

samples.Debye-Scherrer Camera

A small amount of powdered material is sealed into a fine

capillary tube made from glass that does not diffract X-Rays.

The sample is placed in the Debye Scherrer camera and is

accurately aligned to be in the center of the camera. X-Rays

enter the camera through a collimator.

The powder diffracts the X-Rays in accordance with Braggs Law

to produce cones of diffracted beams.

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These cones intersect a strip of photographic film located in

the cylindrical camera to produce a characteristic set of arcs

on the film.

When the film is removed from the camera, flattened and

processed, it shows the diffraction lines and the holes for the

incident and transmitted beams.

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For a sample of several randomly orientated single crystals, the

diffracted beams will lie on the surface of several cones. The

cones may emerge in all directions, forwards and backwards.

In case of powdered sample, the diffracted beams form

continuous cones. A circular film is used to record the diffraction

pattern. Each cone intersects the film giving diffraction lines.

These lines are seen as arcs on the film.

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1

2

S

W

212

S

W

2

2 2 2 2

2sin

4h k l

a

Determination of Lattice parameter

By knowing the h, k, l values one can easily calculate lattice parameter

of cubic crystal using the equation

Applications of XRD

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X-Ray Diffraction technique is used to

Distinguishing between crystalline & amorphous

materials.

Determination of the structure of crystalline materials.

Determination of electron distribution within the atoms, &

throughout the unit cell.

Determination of the orientation of single crystals.

Determination of the texture of polygrained materials.

Measurement of strain and small grain size…..etc.

Advantages

XRD is a nondestructive technique

X-Rays are the least expensive, the most convenient &the most widely used method to determine crystal structures.

X-Rays are not absorbed very much by air, so the sampleneed not be in an evacuated chamber.

Disadvantages X-Rays do not interact very strongly with lighter elements.

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Advantages & Disadvantages of

X-Ray Diffraction

1. Chromium has BCC structure. Its atomic radius is 0.1249 nm .

Calculate the free volume of unit cell.

2. Lithium crystallizes in BCC structure. Calculate the lattice

constant, given that the atomic weight and density for lithium

are 6.94 and 530 kg/m3 respectively.

3. Iron crystallizes in BCC structure. Calculate the lattice constant,

given that the atomic weight and density of iron are 55.85 and

7860 kg/m3 respectively.

4. If the edge of the unit cell of a cube in the diamond structure is

0.356 nm, calculate the number of atoms/m3.

5. A metal in BCC structure has a lattice constant 3.5Ao. Calculate

the number of atoms per sq. mm area in the (200) plane.

6. Germanium crystallizes in diamond (from) structures with 8

atoms per unit cell. If the lattice constant is 5.62 Ao, calculate its

density.10-Aug-15 Confidential 63

Problems

5. A beam of X-rays of wavelength 0.071 nm is diffracted by (100)

plane of rock salt with lattice constant of 0.28nm. Find the

glancing angle for the second – order diffraction.

6. A beam of X-rays is incident on a NaCl crystal with lattice plane

spacing 0.282 nm. Calculate the wavelength of X-rays if the first-

order Bragg reflection takes place at a glancing angle of 8o 35’.

Also calculate the maximum order of diffraction possible.

7. The fraction of vacant sites in a metal is 1 X 10-10 at 500oC. What

will be the fraction of vacancy sites at 1000o C?

8. Calculate the ratios of d100 : d110 : d111 for a simple cubic structure

9. The Bragg’s angle in the first order for (220) reflection from

nickel (FCC) is 38.2o. When X-rays of wavelength 1.54 Ao are

employed in a diffraction experiment, determine the lattice

parameter of nickel.

10.Monochromatic X-rays of = 1.5 A.U are incident on a crystal

face having an inter-planar spacing of 1.6 A.U. Find the highest

order for which Bragg’s reflection maximum can be seen.10-Aug-15 Confidential 64

11. The distance between (100) planes in a BCC structure is 0.203

nm. What is the size of the unit cell? What is the radius of the

atom?

12. Monochromatic X-rays of = 1.5 A.U are incident on a crystal

face having an inter-planar spacing of 1.6 A.U. Find the highest

order for which Bragg’s reflection maximum can be seen.

13. Copper has FCC structure with lattice constant 0.36nm.

Calculate the inter-planar spacing for (111) and (321) planes.

14. The first order diffraction occurs when a X-ray beam of

wavelength 0.675 Ao incident at a glancing angle 5o 25’ on a

crystal. What is the glancing angle for third-order diffraction to

occur?

15. The Bragg’s angle in the first order for (220) reflection from

nickel ( FCC) is 38.2o . When X-rays of wavelength 1.54 Ao are

employed in a diffraction experiment. Determine the lattice

parameter of nickel.

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