CRYSTAL STRUCTURE - zamirkhansite.files.wordpress.com · repeated three dimensional pattern having a definite geometrical shape. Crystalline solids may occur in single crystal form

Department Of Applied Physics ACET (2016-17) BE-First Semester - Engineering Physics Page 1

UNIT -III

CRYSTAL STRUCTURE

Introduction:

Matter is normally divided into three states namely solid, liquid & gas; plasma is considered as the

fourth state of matter. Some of the substances exhibit an additional state intermediate to solid and liquid states.

It is known as liquid crystal state. Each of the state consists of some characteristics distribution of atoms (or

molecules) in it. Finally matter is classified as follows

Solid State :- A solid consist of atoms or cluster of atoms arranged in close proximity . The physical structure

of a solid and its properties are closely related to the scheme of arrangement of atoms within the solid.

Solids can be classified into two categories namely crystalline and amorphous solids.

Crystalline solids:

In crystalline solids, the atoms, ions or molecules of which they are composed of are arranged in regular

repeated three dimensional pattern having a definite geometrical shape.

Crystalline solids may occur in single crystal form or polycrystalline form.

Single Crystal:

Single crystals are polyhedrons that have a distinctive shape for each material. They have smooth faces

and straight edges.

A large single crystal of regular shape is called monocrystal. Quartz, alum, diamond and rock salt are

examples of solids that occurs as large size single crystal.

Matter

Solid

Crystalline

Single Crystalline

Polycrystalline

Non Crystalline

Amorphous

Liquid Gas


Polycrystalline Solids: Polycrystalline solids consists of fine grains having size of 103 to 10

4 A

o, separated by

well defined boundaries and oriented in different directions. Each such grain is a single crystal of an irregular

shape. Majority of natural solids have polycrystalline structure. Metals are example of polycrystalline solids.

Non-Crystalline (Amorphous) Solid: Some material maintains a fixed volume, shape and resemble in their

external features, but internally they do not have the ordered crystalline state. Such materials are known as

amorphous solids. The atomic arrangement in these materials exhibit only short range order as in liquids. They

are in fact considered as super cooled liquids having very high viscosity in which atoms have become fixed in a

random arrangement. Glass, rubber and polymers are examples of amorphous substance.


Crystalline Material Amorphous Material

i) The atoms or molecules constituting a

crystalline solid are arranged in a definite

and regular way throughout the body of the

crystal and possesses a definite external

geometrical shape.

ii)The orderly arrangement of atoms and

molecules in a crystal extends over a large

volume of the crystal i.e. crystal exhibits a

long range order of the atoms and

molecules.

iii) Crystalline materials are anisotropic i.e.

physical properties like thermal

conductivity, electrical conductivity,

compressibility etc. have different values in

different directions.

iv) Crystalline materials possesses uniform

chemical composition i.e. bonds between

all the atoms , ions or molecules are of

equal strength.

v) A crystalline material has sharp melting

point. It is because of the bond between its

constituents are of equal strength and on

heating all the bonds get ruptured suddenly

at a fixed temperature. Due to this the

change from solid state to liquid state

occurs suddenly.

Eg. Sugar, sodium chloride, Quartz.

i)The amorphous materials are those in

which atoms or molecules are arranged in

an irregular manner and do not have

definite external geometrical shape.

ii)The crystal exhibits a short range order

of atoms and molecules.

iii) The amorphous materials are isotropic

i.e. they have same physical properties in

all directions.

iv) The amorphous materials do not

possess uniform chemical composition.

Due to this bonds between all the atoms

and molecules are not of equal strength.

It does not have sharp melting points. It is

because bond between all the atoms and

molecules are not of equal strength. On

heating the weak bonds get ruptured first,

and stronger one later on. Due to this

change of state does not occur at fixed

temperature.

Eg. Glass, Rubber , Sulphur

Space Lattice:

A space lattice is defined as array of points arranged in a regular periodic fashion in three

dimensional space such that each point in the lattice has identical surroundings.

The points are called lattice points or lattice sites.

By using network of lines , the lattice points may be defined as shown in figure. The lines are called

lattice lines or lattice directions. A plane containing lattice points is known as a lattice plane or atomic plane.


A space lattice can be generated by successive translations of an initial point.

1) A repeated application of a translation of a given length and direction to an initial point generates a sequence

of periodically spaced points . This row of points s called linear lattice or 1-D lattice.

●— t1 →●— t1 →●— t1 →●— t1 →●— t1 →●— t1 →●— t1 →●

2) The repeated application of some other translation which is not along the same line (i.e. non-collinear) to the

above row generates a planer array of points. Called as 2-dimensional lattice or net.

3) A third translation which is not in the same plane(i.e. non-coplanar) applied to a 2-dimensional lattice

generates a three dimensional array of points called space lattice or simply lattice which is defined earlier as

shown in figure.

Environment : The arrangement of points around a given lattice point is called its environment.

Order : The arrangement of the atoms in specific relation to each other is called order.

In crystals the order exist in the immediate neighborhood of a given atom as well as over a large distances

corresponding to several layers of atoms. Therefore crystal possesses both short range order and long range

order.


BASIS:-

An atom or a group of atoms that is associated with every lattice points is identical in

composition, arrangement and orientations is called basis

Crystal structure : Crystal lattice or crystal structure is defined as a space lattice in which lattice sites are

occupied by atoms or clusters of atoms (basis) so each lattice point is associated with the same unit of

group of atoms.

Crystal structure is obtained when a basis is added at each point in the lattice. Thus crystal

structure is therefore the result of two quantities; namely space lattice and a basis. Figure shows the basis

consisting of a group of two atoms.

Space Lattice + Basis = Crystal Structure

Unit Cell:-

Unit Cell it cell may be defined as the smallest geometrical unit ,which when repeated in space

indefinitely , generates the space lattice and carries a full description of entire lattice.

Fig. represents a unit cell in general. A unit cell can be chosen in a number of ways. Usually, it is

chosen to be a parallelogram.


Crystallographic axes:-

These are the lines drawn parallel to the lines of intersection of any three faces of the unit cell which do

not lie in the same plane.

In fig. OX, OY, Oz are the crystallographic axes.

Lattice constants or Primitives:- The three sides of the unit cell are called primitives or the intercepts. a, b, c

defines the dimensions of the unit cell and are known as primitives or lattice constants.

Interaxial Angles:

The angles between three crystallographic axes of the unit cell are called interaxial angles.

The angle γ represents the angle between a and b axes, t he angle α represents the angle between b and

c axes The angle β represents the angle between c and a axes .

Lattice parameters:-The axial lengths a, b, c and the three inter-axial or interfacial angles α, β, γ are the basic

lattice parameters.

Primitive Unit Cell:-

It is the minimum volume unit cell that can be defined for a given lattice. All the lattice points belonging

to a primitive unit cell lie at its corners. Therefore, the effective number of lattice points in a primitive unit cell

is one.

Non- Primitive Unit Cell:-

Non- Primitive Unit Cell may have the lattice points at the corner as well as at the other locations both

inside and on the surface of the cell. Therefore the effective number of lattice points in Non- Primitive Unit Cell

is greater than one.

The unit cells marked a, b and c have a point at each corner and nowhere else. The cells marked d and e

contain additional points and are known as non primitive cells.


Fig : Different ways of choosing unit cell in a net

BRAVAIS LATTICES :

Bravais introduced the concept of space lattice in the study of crystal structures. One would expect that many

lattices can be generated in three dimensions with different primitives and non-primitive cells. However in 1848

, Aauguste Bravais,the French Crystallographer proved that there are only 14 different ways of arranging

identical points in three dimensional space ,satisfying the condition of periodicity so that they are in every way

equivalent in their surroundings. These fourteen types of arrangement are called Bravais Lattices.

Crystal System : The sets of the values that the six lattice parameters a,b,c ,α,β and γ can take are limited to 7

only and accordingly all the crystal structures can be classified into 7 crystal system. So the total 14 lattices

form 7 crystal system. They are Cubic , tetragonal, ,orthorhombic, monoclinic, Triclinic, Hexagonal and

rhombohedral.

THE UNIT CELL CHARACTERISTICS:-

(i) Unit cell Volume, V

The volume of a unit cell is given by

V= abc [1-cos2 α-cos

2 β-cos

2 γ + 2cos αcos βcos γ]

1/2

If α= β= γ = 90o, then V = abc

Further, when a=b=c, as in the case of cubic cells

V= a3

(ii) Effective number of atoms per unit cell (Z) / Rank Of Unit cell:-

The effective number of atoms per unit cell is equal to the product of the number of atoms per lattice

point and the number of lattice points per unit cell.

Unit cell is a part of an infinite scheme and is not an isolated entity. Therefore the Lattice points do not

exclusively belong to one unit cell but each one is shared by several adjacent cells. As a result an atom located

at a lattice site contributes only a fraction of its mass and volume to one unit cell.

(iii) Atomic Radius,(r):

The relationship between the apparent size of the atom and the edge of the unit cell can be determined

where one atom is attached to a lattice site.,


(iv) Nearest neighbour distance, (2r):

The nearest neighbour distance, in metallic structures is the distance between the centres of the atoms in

contact expressed in terms of the length of the edge of the unit cell

(v) Coordination number (CN):

The coordination number of an atom in a crystal is the number of nearest neighbour atoms which are

simultaneously in contact with that atom. It signifies the tightness of packing of atoms in the crystal.

(vi) Atomic packing fraction (APF):

The fraction of the space occupied by atoms in the unit cell is known as atomic packing fraction.

It is defined as the ratio of the volume of effective number of atoms in the unit cell to the total

volume of the unit cell. Thus,

𝐀𝐏𝐅 = No. of atoms /unit 𝑐𝑒𝑙𝑙 (Volume of each atom)

Volume of the unit cell

Therefore, 𝐀𝐏𝐅 =𝐙𝐯𝐕

The atomic packing fraction denotes the efficiency with which the available space in a unit cell is utilized.

(vii) Void space:

The void space in the unit cell is the vacant space left unutilized in the cell.

It is equal to (1- APF). It is often expressed as %. . It is commonly known as interstitial space.

Thus, %Void space= (1 – APF) x 100

(viii) Density:

The density of unit cell must be the same as that of bulk crystal.

Thus,

Density, ρ =𝑚𝑎𝑠𝑠

𝑣𝑜𝑙𝑢𝑚𝑒 ρ =

ZW

V

Where, W is the mass of each atom. W is given by w = M NA

Where, M →the molecular weight of the material &

NA→ the Avogadro number.

Therefore,

ρ =ZM

NA V

𝛒 =𝐙𝐌

𝐍𝐀𝐚𝟑


Cubic Lattice or Cubic Crystal System:

The lattice for which the unit cell is a cube is called a cubic lattice. In this case a=b=c and α=β= γ=90.There are three types of unit cells in the cubic structure. They are:

1) Simple Cubic Cell (SC)

A simple Cubic Cell is a primitive Cell with lattice points located at eight corners of the cube.

2) Body Centered Cubic Cell (BCC)

The Body Centered Cubic Cell is a non-primitive cell having a lattice point within the cell in addition to

the eight lattice points at eight corners of the cube.

3) Face Centered Cubic Cell (FCC)

The Face Centered Cubic Cell is a Non-primitive cell having six lattice points at the centers of its six

faces and eight lattice points at eight corners of the cube.

Simple Cubic Cell Body Centered Cubic Cell Face Centered Cubic Cell

1) Simple Cubic Cell (SC):

(i) Unit cell Volume:


If α= β= γ=90o, then V = abc


V= a3

(ii) Atomic radius:


In a SC cell the atoms would be in contact along the edges of the cube, as shown in fig. If ‘a’ is the edge of the

cubic cell & r be the radius of the atom.

a = 2 r or r = a / 2

(iii) Effective number of atoms per unit cell (Z) :

A simple cubic cell has eight lattice point at its eight corners ,which are occupied by eight atoms.

In the three dimensional array, each corner atom is linked to eight surrounding cell, hence in effect, each

corner atom contributes 1/8th

of its content to a unit cell. The total contribution (Z) from all the corner atoms to

the unit cell is given by,

𝐙 =𝟏

𝟖 𝐱

𝐚𝐭𝐨𝐦𝐬

𝐜𝐨𝐫𝐧𝐞𝐫 𝐱

𝐜𝐨𝐫𝐧𝐞𝐫

𝐜𝐞𝐥𝐥

Z =1 atom / cell

(iv) Coordination Number (CN) :

Around an atom in a SC cell, there would be six equally spaced nearest neighbour atoms each at a

distance ‘a’ from that atom. Four atoms lie in the plane of the atom while one is vertically above it and one

vertically below. Therefore, coordination number, CN=6

(v) Atomic Packing Fraction (APF):

APF = (Z )/V

For SC cell, Z = 1

= Volume of spherical atom

= 4/3r3

V = Volume of the unit cell

= a3 = (2r)

3 = 8r

3

APF = [1 x (4/3r3)] / (8r

3)

APF = /6 = 0.52

(vi) Percentage void space:

a

a

a

rr


% Void space= (1 – APF) x 100

= (1-0.52) x100

= 48%

(vii) Density:

𝛒 =𝐙𝐌

𝐍𝐀𝐚𝟑 In case of SC cell, Z= 1 Hence, 𝛒 =

𝟏𝐌

𝐍𝐀𝐚𝟑

2) Body Centered Cubic Cell (BCC) :

i) Unit cell Volume:


V= abc [1-cos2 α-cos

2 β-cos


1/2

If α = β = γ = 90o, then V = abc

Further, when a = b = c, as in the case of cubic cells

V = a3

ii) Atomic Radius : In a BCC cell atom are in contact along the body diagonal. From

iii) Effective Number of atom per unit cell (Z):


A BCC cell has eight atoms at corners of the cube and one atom within the volume of the cell.. An atom

at the lattice point within the cell belongs completely to the cell and the atoms at the corners of the cell

contributes 1/8th

share. The effective number of atoms per BCC cell is then,

𝐙 = 𝟏 𝐱𝐁𝐨𝐝𝐲 𝐜𝐞𝐧𝐭𝐞𝐫𝐞𝐝 𝐚𝐭𝐨𝐦

𝐜𝐞𝐥𝐥+

𝟏

𝟖 𝐱

𝐚𝐭𝐨𝐦𝐬

𝐜𝐨𝐫𝐧𝐞𝐫 𝐱 𝟖 𝐱

𝐜𝐨𝐫𝐧𝐞𝐫

𝐜𝐞𝐥𝐥

Z = 1+ 1 =2 atoms/cell

iv) Coordination Number (CN) :

In the BCC cell each corner atom is in contact with the body centre atom. As there are eight unit cells

around each corner, the atom located at each corner would be simultaneously touching the eight body centre

atom around it. Therefore coordination number is eight.

CN = 8

v) Atomic Packing Fraction :

APF = (Z )/V

For BCC cell, Z = 2

= volume of spherical atom

= 4/3r3

V = volume of the unit cell

= a3 = (4r/3)

3 = 64r

3 /33

APF = [2 x (4/3r3)] / [64r

3/ 33]

APF = 3 /8 = 0.68

vi) % Void space:

Void Space = [1- APF] x 100

Void Space = [1-0.68] x 100 = 32%

(vii) Density:

ZM

ρ=

NA a3

In case of BCC cell, Z= 2


Hence, 2M

ρ =

NA a3

3) FACE CENTRED CUBIC CELL (FCC)

i) Unit cell Volume:


V= abc[1-cos2 α-cos

2 β-cos


1/2

If α= β= γ=90o, then V = abc


V= a3

ii) Atomic Radius :

In a FCC cell, atoms are in contact along the face diagonal of the cube, as shown in fig

iii) Effective Number of atom per unit cell (Z):


In a FCC cell ,eight atoms are located at the eight corners and six atoms are at the centres of the six

faces.Therefore each face centered atom contributes only half of its mass & volume to one cell, while each

corner atom contributes 1/8th

of its content. Thus, the effective number of atoms per FCC cell

1 atoms faces 1 atom corners

Z= ( )x 6( )+ ( )x 8( )

2 face cell 8 corners cell

= ( ½)x 6 + 8 x (1/8)

= 3 + 1

= 4 atom/cell

iv) Coordination number : In FCC cell, each corner atom is in contact with face centre atom. It would be simultaneously

touching 4 atoms in the xy-plane,4 atoms in the yz-plane,4 atoms in zx-plane , making up a total of

12 atoms.

Coordination Number = 4+ 4 + 4 = 12

v) Atomic Packing Fraction :

APF = (Z ) /V

For FCC cell, Z = 4

= Volume of spherical atom

= 4/3r3

V = volume of the unit cell

= a3 = (22r)

3

APF = [4 x (4/3r3)] / [162r

3]

APF = /32 = 0.74

vi) % Void space:


% Void space= (1 – APF) x 100

= (1- 0.74) x100

=26%

vii) Density :

For FCC cell, Z =4

Hence, ρ = 4M / NAa3

Miller Indices: Miller introduced a system to designate a plane in a crystal. He introduced a set of three numbers that

specify a plane in a crystal. Indices are the numbers which describe the different planes in the crystal.

The method of designating a plane in the crystal by this set of three numbers (hkl) is known as Miller

Indices.

Definition: Miller Indices are the reciprocals of the intercepts, made by the plane on the crystallographic

axes, when reduced to smallest integers.

The Procedure For Finding The Miller Indices:

Step 1 : Identify the intercepts of planes along three crystallographic axes (The coordinates of the points of

interception are expressed as integral multiples of the axial lengths in the respective directions)

Step 2 : Determine the reciprocals of the integers

Step 3 : Reduce the reciprocals to the smallest set of integers h, k, l by taking LCM

Step 4 : The integers are written as (hkl) by enclosing in a parenthesis.

Characteristics SC BCC FCC

Z 1 2 4

R a/2 (3/4) a a/22

2r a (3/2) a a/2

CN 6 8 12

APF 0.52 0.68 0.74

Void Space 48% 32% 26%

Density M/(NAa

3

) 2M / (NAa

3

) 4M / (NAa

3

)


Example : Find the Miller indices for the given plane ABC

The plane ABC has intercepts of 2a on x-axis, 2b on y-axis and 1c on z axis.

Step 1 ; The intercepts are 2 , 2 , 1

Step 2 : The reciprocals of the intercepts are ½ , ½ , 1

Step 3 : Taking LCM we get 1, 1, 2

Step 4 : The miller Indices of the planes are (1 1 2)

Important features of Miller indices :

Parallel planes spaced equally have the same miller indices.

A plane parallel to one coordinate axis has miller index zero for that direction

When the intercepts of a plane is negative on an axis , a bar is placed on the corresponding miller index

MILLER INDICES

1. What are Miller indices? Determine the Miller indices of a plane that makes an intercept of 1 on a-axis, 2 on

b-axis, and is parallel to c-axis.

Sol. Miller Indices : Miller indices are the reciprocal of the intercepts made by the plane on three

crystallographic axes when reduced to the smallest integers.

Intercepts on X,Y, Z axes are a, 2b and c respectively

Step 1 : The intercepts are a, 2a, c

Step 2 : The reciprocals of integrals are 0,2

1,1

Step 3 : Taking L.C.M. m we get 2,1 , 0

Step 4 : The Miller indices of the set of planes are (2 1 0)

2. What are the Miller indices? Obtain Miller indices for the plane making intercepts 3, 2 and 1 on the crystal

axis a, b and c respectively.

Sol. For Defination refer solution of Que. (1)

Intercepts on X,Y, Z axes are 3a, 2b and 1c respectively

Step 1 : The intercepts are 3a, 2b, 1c


Step 2 : The reciprocals of integrals are 1

1,

2

1,

3

1

Step 3 : Taking L.C.M. we get 2,3,6


3. Find Miller indices of a set of planes with intercepts a, 2a and 3a on X, Y and Z-axes respectively for a

cubic crystal. Draw (1 2 0) plane.

Sol. Intercepts on X,Y, Z axes are a, 2a and 3 a respectively

Step 1 : The intercepts are a, 2a,3a

Step 2 : The reciprocals of integrals are 3

1,

2

1,1

Step 3 : Taking L.C.M. m we get 6 , 3 , 2


4. Find MI of set of parallel plane which makes intercept in the ratio 3a : 4b on X & Y axis & are parallel to

Z – axis.

Sol. Intercepts on X,Y, Z axes are 3a, 4b and c respectively

Step 1 : The intercepts are 3a, 4b, c

Step 2 : The reciprocals of integrals are

1,

4

1,

3

1

Step 3 : Taking L.C.M. m we get 4 , 3 , 0


5.* Find the Miller indices of atomic planes having intercepts a, 2a and ∞ on X,Y and Z axes respectively. Find

the distance from the origin. (W/06)

(3)

Sol. Intercepts on X,Y, Z axes are a, 2a and respectively

Step 1 : The intercepts are a, 2a,

Step 2 : The reciprocals of integrals are 0,2

1,1

Step 3 : Taking L.C.M. m we get 2,1 , 0



a

ad

lkh

ad hkl

4472.0

012 222210

222

6. Obtain the Miller indices for planes making intercepts of 1 Ao, 2 A

o and 4 A

o along the a, b, and c crystal

axes with 2Ao, 3 A

o and 6 A

o as unit cell parameters respectively.

Sol. Unit cell parameters /Primitives /lattice constants : a= 2 A ,b=2 A , c=3 A

X intercept , 2

11 pApa

Y intercept , 3

22 qAqb

Z intercept , 3

2

6

44 rArc

Taking reciprocals of intercepts 2

3,

2

3,2

Taking LCM 4, 3, 3

Miller Indices are ( 4 3 3 )

7. A certain crystal has lattice constants of 4.24 A ; 10 A ; 3.66 A on the X,Y, and Z axes respectively .

determine the miller indices of two lattice planes of this crystal having intercepts of :

i) 2.12 A ,10 A ,1.83 A for one plane

ii) 4.24 A , ,1.22 A for the other plane.

Sol. : Lattice constants : a= 4.24 A ,b=10 A , c=3.66 A

i) X intercept , 2

1

24.4

12.212.2

A

ApApa

Y intercept , 110

1010 qAqb

Z intercept , 2

1

66.3

83.183.1

A

ArArc

Taking reciprocals of intercepts 2 ,1,2

Taking LCM 2,1,2

Miller Indices are (2 1 2 )

ii) X intercept , 124.4

24.424.4

A

ApApa

Y intercept , qb= or q =


Z intercept , 3

1

66.3

22.122.1

A

ArArc

Taking reciprocals of intercepts 1/1, 1/ .3/1

Taking LCM 1,0,3

Miller Indices are (1 0 3 )

Draw the Crystal planes having Miller Indices:

1) ( 1 0 0 ) 9) (1 0 2)

2) ( 0 1 0 ) 10) (2 0 0)

3) ( 0 0 1 ) 11) (2 0 1)

4) (1 1 0 ) 12) (1 2 3)

5) (0 1 1 ) 13) (2 1 0)

6) (1 0 1 ) 14) (1 1 2)

7) (1 1 1 ) 15) (1 2 0)

8) (2 0 0 ) 16) (2 1 1)

Interplaner distance in a Cubic Crystal :

The distance between successive members of a series of a parallel planes is known as a interplaner distance.

Let dhkl represents the distance between two adjacent parallel planes having Miller Indices (hkl). Let the plane

ABC be one of the planes that has intercepts OA=a/h, OB=b/k and OC=c/l. The origin of the coordinates O is

in next plane of set parallel to ABC. The perpendicular OD in fig. from the origin to the plane is equal to dhkl.

Let the direction cosines of OD be cosα, cosβ, cosγ. The intercepts of the plane on the coordinate axes are

OA= a/h; OB=a/k; OC=a/l (As a=b=c= Unit length of edge a)

Where ‘a’ is the length of the cube edge.

Denoting OD=d

OD hd OD kd

cos α = = ; cos β= =

OA a OB a

OD ld

cos γ = =

OC a

Since cos2

α+ cos2

β+ cos2

γ= 1

h2d

2 + k

2d

2 + l

2d

2 =1

a2 a

2 a

2

d2

(h2

+ k2+ l

2) = 1

a2


a2

d2=

(h2

+ k2+ l

2)

a

d =

√ (h2

+ k2+ l

2)


Voids: All the available space in the crystal is not filled with atoms. Even in the close packed structure it is

seen that about 26% of the volume is left vacant .

These vacant spaces between the atoms in the crystal are called voids. They are also known as

interstices.

There are two kinds of void forms in close packed structure.

(I) Tetrahedral Void:

A tetrahedral void is formed by a sphere fitting into the valley formed between three adjacent

sphere of a closed packed layer, as shown in figure.

Thus an tetrahedral void is surrounded by four spheres.

If the centres of four spheres are joined, a regular tetrahedron is produced. The vacant site

enclosed by the tetrahedron is known as tetrahedral void.

In a three dimensional structure two tetrahedral voids form for every sphere.

They are smaller in size but larger in number compared to octahedral voids.


(II) Octahedral Void:

An octahedral void is produced when a void formed by three adjacent spheres in one layer

comes on the top of the void formed by three sphere in contact in the adjacent layer as shown

in figure.

Thus an octahedral void is surrounded by six spheres.

When the centres of the six spheres are joined an octahedron is obtained as shown in fig.

hence the space enclosed is known as octahedral void.

In a closed packed three dimensional structure one octahedral void forms per sphere.

Octahedral voids are bigger in size and smaller in number compared to the tetrahedral voids.

Tetrahedral Octahedral

Two voids per sphere One void per sphere

Surrounded by four atoms Surrounded by six atoms

Smaller in size and greater number

Bigger in size and smaller in number


Bragg’s Law :

William Henry Bragg and William Lawrence Bragg, the team of son & father of British Physicist derived

in1913, a simple equation relating the wavelength of X-ray to the angular position of the scattered beams

and the separation of the atomic plane in the crystal.

A crystal may be regarded as a stack of the parallel planes of atoms, the atomis planes are often

called Bragg planes. Every crystal has several sets of Bragg’s planes oriented in different directions and each

plane in a given set has the same distribution of atoms.

When an X-ray beam impinges on the crystal , each atom in a given Bragg plane scatters a portion of

the incident beam. The scattered waves proceed in all directions.

The condition for constructive interference of waves is obtained by treating the crystal as a set of

equally spaced parallel semitransparent planes and by regarding the complex phenomenon of diffraction of

X-ray as a simple reflection at the Bragg’s planes.

Let us consider a series of Bragg planes separated by a distance, d. let a beam of monochromatic X-rays

represented by the parallel lines AB and DE be incident on these planes. The scattered beam emerges along

BC and EF as shown in fig.

The rays BC and DE are coherent and reinforce each other, if they are in phase. In fig. the

contribution of two adjacent atoms in successive planes MN and PQ are considered. The path difference, Δ

between the rays reflected along BC and EF is,


Δ =GE + EH -------------(A)

The rays BC and EF will be in phase only when their path difference, Δ is an integral multiple of

wavelength. Thus,

Δ =GE + EH = mλ (m= 1, 2, 3,……..)

From the figure,

<ABG = θ + α =90-------------------(1)

BE is normal to the plane MN, therefore,

<MBE =α + <GBE = 90-----------------(2)

Comparing (1) & (2), we get,

<GBE =θ

In the ΔBGE, BE = d sinθ and sinθ = GE/ BE

∴ GE = BE. Sinθ = d sinθ

Similarly, EH = d sinθ

∴ the path difference,

Δ =GE + EH = 2d sinθ------------------(B)

Combining equations (A) & (B), we obtain the condition for reinforcement of scattered waves as,

2d sinθ =mλ (m=1, 2, 3,………….)

This equation is known as Bragg’s law or Bragg’s equation. The angle θ in the above equation

represents the angle between the Bragg plane and the direction of the incident beam. It is known as glancing

angle or Bragg angle.

The angle 2θ is called diffraction angle.

For a given set of atomic planes, d has a fixed value and the reinforcement of reflected waves depend

on the value of θ. As θ is increased gradually, a number of positions will be found at which the reflections

are intense. These positions corresponds to m=1, 2, 3,……. Values. Thus m denotes the order of reflection.

If the value of θ is determined experimentally, and knowing the wavelength λ , the interplaner

spacing ‘d’ can be determined with help of Bragg’s law.

Application of X-ray diffraction Techniques /Bragg’s Law :


The X-Ray diffraction techniques /Bragg’s law are very useful in determining the crystal structure

and identification of materials. The diffraction pattern gives information about the d values and making use

of d values the unit cell and the atomic positions in the cell can be obtained.

Documents

CRYSTAL STRUCTURE - zamirkhansite.files.wordpress.com · repeated three dimensional pattern having a definite geometrical shape. Crystalline solids may occur in single crystal form