crystal structure and x – rays diffraction

1/23/2017 1Dr A K Mishra, Academic Coordinator, JIT Jahangirabad

Engineering Physics II

Dr. A K Mishra

Associate Professor

Applied Science Department

Jahangirabad Institute of Technology, Barabanki

• CRYSTAL STRUCTURE AND X – RAYS DIFFRACTION

• DIELECTRIC AND MAGNETIC PROPERTIES OF MATERIALS• DIELECTRIC PROPERTIES

• MAGNETIC PROPERTIES

• ELECTROMAGNETIC THEORY

• PHYSICS OF SOME TECHNOLOGICAL IMPORTANT MATERIALS• SEMICONDUCTOR

• SUPERCONDUCTOR

• NANO-MATERIALS

1/23/2017 2Dr A K Mishra, Academic Coordinator, JIT Jahangirabad

Crystal Structure: Introduction

Matter is classified into three kinds

Solids :– Atoms or molecules are arranged in a fixed manner– Solids have definite shape and size– On basis of arrangement of atoms or molecules, solids are classified into

two categories, they are crystalline solids and amorphous solids.Liquids :

– Atoms or molecules are not arranged in a fixed manner– Solids have not definite shape and size

Gases: – Atoms or molecules are not arranged in a fixed manner– Solids have not definite shape and size

1/23/2017 Dr A K Mishra, Academic Coordinator, JIT Jahangirabad 3

CRYSTALLINE SOLIDS AMORPHOUS SOLIDS

In crystalline solids, the atoms or molecules are arranged in a regular and orderly manner in 3-D pattern, called lattice.

In amorphous solids, the atoms or molecules are arranged in an irregular manner, otherwise there is no lattice structure.

These solids passed internal spatial symmetry of atomic or molecular orientation.

These solids do not posses any internal spatial symmetry.

If a crystal breaks, the broken pieces also have regular shape. Eg: M.C : Au, Ag,Al, N.M.C: Si, Nacl, Dia.

If an amorphous solid breaks, the broken pieces are irregular in shape.Eg : Glass, Plastic, Rubber.


Crystal Structure

LATTICE POINTS : Lattice points denote the position of atoms or molecules in the crystals.SPACE LATTICE :The angular arrangement of the space positions of the atoms or molecules ina crystals is called space lattice or lattice array.Three Dimensional- Space Lattice:It is defined as an infinite array of points in 3D-Space in which every pointhas the same environment w.r.t. all other points.In this case the resultant vector can be expressed as


lyrespective axis z, y, x,along vectorsonal translatiare ,, and

integeresarbitrary are ,, 321321

cba

nnnwherecnbnanT

BASIS :• The unit assembly of atoms (molecules or ions) identical in

composition ,arrangement and orientation is called basis. In elemental crystals like, Aluminum , barium, copper , silver , sodium etc. The basis is a single atom.

• In NaCl, KCl,AgI etc. basis has two atom or basis is diatomic .

• In case of CaF2,Sno2,Sio2 etc. the basis has three atoms or basis is Triatomic.

• The basis more than two atom is called multi-atomic

• Space lattice + Basis = CRYSTAL STRUCTURE.


• The arrangement within it, when repeated in three dimension gives the total structure of the crystal.

• An arbitrary arrangement of crystallographic axismarked X,Y,&Z.

• The angles between theThree crystallographic axes• are known as interfacial angles or interaxial angles.• The angle between the axes X and Y= α• The angle between the axes Z and X = β• The angle between the axes Z and Y = γThe intercepts a,b&c define the dimensions of an unit celland are known as its primitive. The three quantities a,b&c arealso called the fundamental translational vectors.


b

c

α

βϒ a

UNIT CELL AND LATTICE PARAMETERS

Primitive Unit Cell

• A primitive cell or primitive unit cell is a volume of space that when translated through all the vectors in a Bravaislattice just fills all of space without either overlapping itself or leaving voids.

• A primitive cell must contain precisely one lattice point.• Unit cell drawn with lattice point at each corner but lattice

point at the center of certain faces Unit cell with lattice point at corners only ,called primitive cells.

• Unit cell may be primitive cell but all primitive cell is not necessarily unit cell.


SEVEN CRYSTAL SYSTEM AND FOURTEEN BRAVAIS LATTICES


Crystal system Type of lattice

CUBIC SIMPLE,FACE CENTERED,BODY CENTERED

TETRAGONAL SIMPLE,BODY CENTERED

ORTHOROMBIC SIMPLE,FACE CENTERED,BODY CENTEREDEND CENTERED

MONOCLINIC SIMPLE,END CENTERED

RHOMBOHEDRAL SIMPLE

TRICLINIC SIMPLE

HEXAGONAL SIMPLE

SYSTEM LATTICE PARAMETER

INTERFACIAL ANGLE

EXAMPLE

CUBIC a=b=c α=β=ϒ=900 Nacl,Kcl,Diamond

TETRAGONAL a = b ǂ c α=β=ϒ=900 SiO2 , TiO2

ORTHOROMBIC a ǂ b ǂ c α=β=ϒ=900 KNO3

MONOCLINIC a ǂ b ǂ c α=β=900, ϒǂ900 Na2SO4 , 10H2O

RHOMBOHEDRAL a=b=c α=β=900, ϒǂ900Calcite, NaNO3

TRICLINIC a ǂ b ǂ c α ǂ β ǂ ϒ ǂ 900 CuSO4 , 5H2O

HEXAGONAL a = b ǂ c α=β=900, ϒ=1200 Graphite, ZnO


Space lattice of cubic systemSIMPLE CUBIC CELL (SCC):• More than half elements crystallized into the cubic system

(a=b=c, α=β=ϒ=900 ).• In SCC atoms are situated at the corners of the cell such that

they are touches each other.• Only one-eight part of atom remains in each cell rest

contributing others.• Therefore there is only one atoms per unit cell.• The cell containing only one atom is called primitive cell, SCC

is also known as cubic P cell.


SCC

BODY CENTERED CUBIC CELL (BCC)• Atoms are situated at each corner of the unit cell and also at the

intersection of the body diagonal also known as cubic 1 cell.

• Each cell has eight corners and eight cell meet at each corner.

• One-eight atom contribute to any one cell and one atom at center of each cell.

• BCC has two atom per unit cell.

• Total No. of atom in any one cell is( x8 +1)=2


81

BCC

FACE CENTERED CUBIC CELL (FCC)• Atoms are situated at all eight corners of the cell and also at

the center of all the six faces of the unit cell.

• It is also known as cubic F cell.

• One-eight of the atom belongs to any one cell, similarly an atom at the corner of the faceshared by two cell only i.e only one-half belongs to any one cell.

• FCC cell has four atom per unit cell.

• Total no. of atoms belongs to any onecell is ( x8) + ( x 6 )=4


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21

FCC

Coordination Number and Atom positions in cubic unit cell•The number of nearest equidistant neighbors around any lattice point inthe crystal lattice.•Coordination number decides whether the structure is loosely or closelypacked.

• The position of nearest atom can be term as follow in SCC

and a is the length of each edge.The coordination of nearest neighbour at O are;( a,0,0),( 0, a,0),( 0,0, a)


(0,a,0) (0,0,a)

O(0,0,0) (1)

(a,0,0)(-a,0,0)(2)

(0,0,-a)(0,-a,0) (6)

(4)

(5)(3)

axis. z, y, x,along vectors are ,, z,, kjiwherecbjia

Coordination Number in BCC• The distance between any two nearest neighbour can be

determine as followsIn triangle ABC( AC)2= (AB)2 + (BC)2

= a2 + a2

AC = a In triangle ADC

( AD)2= (AC)2 + (CD)2

=

AD = a Thus the distance between any two nearest neighbour is a

Similarly for FCC : The coordination number of FCC is twelve and the distance between any two nearest neighbour is a/


a

a

a

A B

C

D

O

a23

2a

2

23

23

2

aa 222

The atomic radius is half a distance between any nearest neighbors in crystal of pure element .it is necessary to calculate the atomic radius of atom of sc,bcc and fcc it should be remember that any two nearest neighbor touches each other.

For SCC: If r be the atomic radius and a be the length of the edge of the cube, then

AB = 2r = a

or r =


a

a

A B

2a

SIMPLE CUBIC CELL

Atomic radius

FOR BCC CUBIC CELL


4r

A

BC

F

a

r2r

r

C

F

( AC)2= (AB)2 + (BC)2

= a2 + a2

=2 a2

( FC)2= (AC)2 + (AF)2

= 3 a2

= 2 a2 + a2

FC = 4r

r=4

3 a

The radius of each atom of a BCC cell is 43 a

For FCC


A B

CD(a)

FCC has six atoms at the center of each six faces and eight

at each of the eight corner of the cube.

Hence the radius of each atom of a FCC cell is

B

CD a

4r

(b)

( BD)2= (DC)2 + (BC)2

(BD)2= (a)2 + (a)2

(4r)2 = 2(a)2

r= 24a

ATOMIC PACKING FACTOR (APF)•Space occupied by all atoms in a unit cell.

OR•Ratio of the volume of the atoms per unit cell to the volume of the unit cell.


itcellVolumeofuncelltomperunitVolumesofaingfactorAtomicPack

SIMPLE CUBIC CELLNo. of atom per unit cell = 01

Volume of one atom =


34 πr3

Volume of one atom of SCC lattice = (a/2 )3π = 34 a

6

3

SCC has one atom per unit cell Volume of atom per unit cell = No. of atoms per unit cell x volume of unit cell

= 1x =Volume of SCC cell = (a)3Atomic Packing factor = = = = 0.52

APF for SCC is 0.52 or 52% Hence SCC is loosely Packed structure.

a6

3 a6

3

FOR BCC CUBIC CELL


No. of atom per unit cell = 02Volume of one atom = Volume of one atom of BCC lattice = =

34 3r

3

2)43(4 a

a3316

Volume of per unit cell = 02x

APF = 02x

== 0.68

APF of BCC is0.68 or 68% Hence BCC structure is closely packed structure.

aa

3

3

163

38

aa

3

3

163

FOR FCC CUBIC CELLNumber of atom per unit cell = 04

Volume of one atom of FCC lattice = =


34 3r

3

4 _)24

(3a

Volume of atom per unit cell= No. of atoms per unit cell x volume of one atom

=

APF = = 0.74

APF for FCC is 0.74 or 74% Hence FCC is closely packed structure

a3

2124

233

3

aa

Crystal structure of sodium chloride (NaCl) Or Rock Salt


a

Cl

Na

81

21

X 8 + x 6 = 4 Cl

12 x + 1 = 4 Na41

Crystal structure of sodium chloride (NaCl) Or Rock Salt• NaCl ions is arranged in cubic pattern.• Cl ions are situated at each corner (8) and the center of each

face (6) of cell.• Cl ions lie on FCC lattice.• The Na ions are also arranged in the face centered cubic

lattice.• One Na ions is at center and others are located at the mid

point of the twelve edges.• The position of ions in unit cell are :

Na;( , , ),(0,0, ),(0, ,0), ( ,0,0 )

Cl; (0,0,0) ( , ,0),( ,0, ),( 0, , )


21

21

21

21

21

21

21

21

21

21

21

21

Unit cell and lattice constant of a space lattice

• Consider a cubic crystal (a=b=c) lattice of lattice constant a and be the density of crystal material, thenVolume of unit cell = mass of the unit cell = ……………….(1)Let n be the No. of atom per unit cell,M the molecular weight and N the Avogadro number then,Mass of each molecule = Mass of unit cell,m = n ………………(2)Comparing equation (1) and (2) ,we get

= n =

Thus lattice constant can be determined.


a 3

NM

NM

a 3

NM

a 3N

nM

Atomic packing factor is the ratio of volume occupied by the atoms in an unit cell to the total volume of the unit cell. It is also called packing fraction.

• The arrangement of atoms in different layers and the way of stacking of different layers result in different crystal manner.

• Atomic packing factor =

• Metallic crystals have closest packing in two forms • (i) hexagonal close packed • (ii) face- centered cubic with packing factor 74%.• The packing factor of simple cubic structure is 52%.• The packing factor of body centered cubic structure is 68%.


itcellvolumeofunitcelleatomperunvolumeofth

Atomic packing factor

In a crystal orientation of planes or faces can be described in terms of their intercepts on the three crystallographic axes. Miller suggested a method of indicating the orientation of a plane by reducing the reciprocal of the intercepts into smallest whole numbers. These indices are called Miller indices generally represented by (h k l).


MILLER INDICES

X

Y

A

B

C

Z

paqb

rc

O

Procedure to determine Miller Indices• Choose the plane that does not pass through the origin at

(0,0,0).• Find the intercept OX,OY,OZ.let it be pa, qb&rc,where a,b,c

are primitives and p,q,r may be integer or fraction.• Take reciprocal of these intercepts as follows

( , , )Reduce the to three smallest hole number having the same

ratio.the smallest possible integers h,k,l are:

h:k:l = , ,

It is done by multiplying the reciprocal by LCM. these numbers are known as Miller indices.


p1

q1

r1

p1

q1

r1

Reciprocal Lattice• P P Ewald devised reciprocal lattice.• Each set of parallel plane can be represented by normal to

these planes having equal length equal to the reciprocal of the interplanar spacing of the corresponding set.

• The normal's are drawn from a common origin and points marked at the end of the normal.

• The points at the end of the normal form a lattice array.• Since the distance in the array are reciprocal to distance in

the crystal, the array of points called reciprocal lattice of the crystal.

• The point in the reciprocal lattice are called reciprocal lattice points.


Reciprocal lattice• Real lattice


Continued………..


X – Rays Diffraction• When Wilhelm Rőntgen first discovered X-rays in 1895, no one had

any about phenomenon was - hence the name X-rays (although they were referred to as Rőntgen rays for a while).

• X- rays are electromagnetic radiation of exactly the same nature as light but of very much shorter wavelength between(0.01 –10)A0.

• When electrons with high value of kinetic energy collide with the target metal of high melting point and large atomic weight, x-rays are produced. In 1913, Dr Coolidge introduced new type of tube Which now most commonly used to produced x – rays.

• in this tube filament is heated to produced electrons by thermionic emission and the incident on the target, after striking the target, X – rays produced. Since most KE of the electrons is absorbed by the target metal in the form of heat, a cooling system is required to maintain the temperature.


Important Properties of X-rays: Properties of X-Rays • X – rays are electromagnetic radiations like

light.thus they posses all the properties of electromagnetic radiations and shows a phenomena like reflection, refraction, diffraction and interference.

• X-rays have very short range of wavelength,lyingbetween (0.01 – 10)A0.

• X – rays are absorbed by the material.• X – rays are not deviated by either electric or

magnetic field.• X –rays ionize the gas through which they pass.• X – rays can penetrate the solid materials.• X – rays affect the photographic plate.


X – Rays Production Coolidge tube• Electron beam


Heavy metal targetElectron beam

Filament

VaccumeHT

X-Ray Beam

Copper Anode

Diffraction of X –rays• For testing of wave nature of X – rays, scientists tried to

obtain the diffraction pattern of X – rays, man-made transmission grating (ruled with 6000 lines per inch) could not produce appreciable amount of diffraction with X – rays. thus it was concluded that diffraction of X –rays by manmade grating is impossible. Therefore to obtained the diffraction pattern of a wave the transmission grating having 40 million lines per cm are required for diffracting the X – rays.

• In 1913, German physicist Max Von Laue suggested that atoms in a natural crystal are regularly arranged at equal distances, the separation between successive layers is of the order of the wavelength of X – rays so crystal act as space grating.


Laue Experimental demonstration• Determination of crystal structure:


Laue Experimental demonstration:•In 1913,Laue,in collaboration with W Friedrich and P Knippingwas successful in obtaining diffraction pattern of X –rays passing through the three dimensional crystal gratting.the experimental arrangement of laue demonstration is shown above.

•X – rays are produced from the Coolidge tube and are collimated into fine pencil beam by passing through pin holes, the beam is allowed to pass through the crystal of ZnS or NaCl which is acting as a grating.

•diffraction pattern is obtained at photographic plate. This pattern consist of a central spot, surrounded by a series of spot in a definite pattern. The symmetrical pattern of spot is called Laue's spots.


Continued………• Laue spot prove that X – rays are electromagnetic waves.• The proper explanation of laue is given by W.L Braggs he

said that these spots corresponds to the constructive interference between the rays reflected from the various set of parallel crystal planes, there by satisfying the equation,2dsin = n,where is the glancing angle, d is the inter-atomic space and λ is the wavelength of the x –rays.Laue,s established following two facts:

• X- Rays are electromagnetic waves of short wavelength.• In crystal, atoms are arranged in a three- dimensional

lattice.


Bragg reflection• The diagram below shows X-rays being reflected from a

crystal. Each layer of atoms acts like a mirror and reflects X-rays strongly at an angle of reflection that equals the angle of incidence. The diagram shows reflection from successive layers


d

d

A

B

C

Bragg reflection• If the path difference between the beams from

successive layers of atoms is a whole number of wavelengths, then there is constructive interference.

• The path difference is the distance (AB + BC )as above:

AB = BC =dsinθ(AB + BC) = 2dsinθ

• The reflected beam has maximum intensity when 2dsinθ = nλ


Braggs spectrometer• W H Braggs and his son W L Braggs devised a spectrometer in

which a crystal is used as a reflection grating instead of transmission grating.


Continued………• It consist of a source of X- rays ,slits and crystal mounted on the

prism table consisting of two venire scale to note down the angle, is attached with the prism table. An ionization chamber is attached with the prism table, along with a galvanometer.

• rays from the Coolidge tube are passed through slits to obtain pencil beam.

• This beam is allowed to fall on the crystal mounted on the prism table, capable of rotating about a vertical axis passing through the center of the turn table.

• the beam after reflection enter into the ionization chamber through slit the chamber is filled with ethyl bromide, and is capable of rotating with the prism table the venire scale gives the position of the ionization chamber the position of crystal and and ionization chamber is arranged in such a way that the rotation of angle in the position of crystal produces the rotation of 2 in the position of ionization chamber.


Continued………• Due to this arrangement the X- ray beam reflected from the

surface of the crystal is always received in the ionization chamber. the intensity of X- rays in terms of the ionization current is observed for different values of the glancing angle. the resulting ionization current is observed through a galvanometer.

• The graph plotted between the ionization current I and glancing angle for the sodium chloride crystal.

• It is clear from the graph for a certain values of the ionization current I increases abruptly, the peaks of the curve for glancing angle satisfy Bragg’s equation i.e.

2dsin = nλFor n=1 λ = 2dsin 1

N=2 2λ = 2dsin 2N=3 3λ = 2dsin 3

Hence sin1: sin2: sin3 = 1: 2 : 3


Continued………• Thus , λ, and d if two are known then the value of the third

one can be calculated. From the above experiment the following fact can be calculated.

• As the order of spectrum increases, the intensity of the reflected rays decreases.

• The ionization current does not falls to zero for any values of the glancing angle ,but it does not attain maximum values for certain glancing angles. it indicate that there is a continuous spectrum.

1/23/2017Dr A K Mishra, Academic Coordinator,

JIT Jahangirabad 44

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Crystal

X-Ray