1.
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Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad
Engineering Physics II
Dr. A K Mishra
Associate Professor
Applied Science Department
Jahangirabad Institute of Technology, Barabanki

2.
• 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
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Dr A K Mishra, Academic Coordinator,
JIT Jahangirabad

3.
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
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4.
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.
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Crystal Structure

5.
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 in
a 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 point
has the same environment w.r.t. all other points.
In this case the resultant vector can be expressed as
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JIT Jahangirabad
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lyrespectiveaxisz,y,x,alongvectorsonaltranslatiare,,and
integeresarbitraryare,, 321321
cba
nnnwherecnbnanT 

6.
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.
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7.
• The arrangement within it, when repeated in three
dimension gives the total structure of the crystal.
• An arbitrary arrangement of crystallographic axis
marked 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 = γ
The intercepts a,b&c define the dimensions of an unit cell
and are known as its primitive. The three quantities a,b&c are
also called the fundamental translational vectors.
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b
c
α
β
ϒ a
UNIT CELL AND LATTICE PARAMETERS

8.
Primitive Unit Cell
• A primitive cell or primitive unit cell is a volume of space
that when translated through all the vectors in a Bravais
lattice 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.
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9.
SEVEN CRYSTAL SYSTEM AND FOURTEEN BRAVAIS
LATTICES
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Crystal system Type of lattice
CUBIC SIMPLE,FACE CENTERED,BODY CENTERED
TETRAGONAL SIMPLE,BODY CENTERED
ORTHOROMBIC SIMPLE,FACE CENTERED,BODY CENTERED
END CENTERED
MONOCLINIC SIMPLE,END CENTERED
RHOMBOHEDRAL SIMPLE
TRICLINIC SIMPLE
HEXAGONAL SIMPLE

10.
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, ϒǂ900
Calcite, 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
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11.
Space lattice of cubic system
SIMPLE 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.
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SCC

12.
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
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8
1
BCC

13.
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 face
shared 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 one
cell is ( x8) + ( x 6 )=4
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8
1
2
1
FCC

14.
Coordination Number and Atom positions in cubic unit cell
•The number of nearest equidistant neighbors around any lattice point in
the crystal lattice.
•Coordination number decides whether the structure is loosely or closely
packed.
• 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)
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(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,alongvectorsare,,z,, kjiwherecbjia 

15.
Coordination Number in BCC
• The distance between any two nearest neighbour can be
determine as follows
In 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/
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a
a
a
A B
C
D
O
a
2
3
2
a
2
2
3
2
3
2
aa
22
2 

16.
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 =
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a
a
A B
2
a
SIMPLE CUBIC CELL
Atomic radius

17.
FOR BCC CUBIC CELL
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4r
A
BC
F
a
r
2r
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 4
3 a

18.
For FCC
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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= 2
4
a

19.
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.
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itcellVolumeofun
celltomperunitVolumesofa
ingfactorAtomicPack 

20.
SIMPLE CUBIC CELL
No. of atom per unit cell = 01
Volume of one atom =
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3
4 πr3
Volume of one atom of SCC lattice = (a/2 )3π =
3
4
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)3
Atomic Packing factor = = = = 0.52
APF for SCC is 0.52 or 52% Hence SCC is loosely Packed structure.
a
6
3

a
6
3


21.
FOR BCC CUBIC CELL
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No. of atom per unit cell = 02
Volume of one atom =
Volume of one atom of BCC lattice = =
3
4
3
r
3
2)
4
3
(4 a
a
3
3
16

Volume of per unit cell = 02x
APF = 02x
=
= 0.68
APF of BCC is0.68 or 68% Hence BCC structure is closely packed structure.
a
a3
3
16
3
3
8

a
a3
3
16
3

22.
FOR FCC CUBIC CELL
Number of atom per unit cell = 04
Volume of one atom of FCC lattice = =
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3
4
3
r
3
4 _)2
4
(
3
a
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
a
3
212
4

23
3
3
a
a

23.
Crystal structure of sodium chloride
(NaCl) Or Rock Salt
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a
Cl
Na
8
1
2
1
X 8 + x 6 = 4 Cl
12 x + 1 = 4 Na
4
1

24.
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, , )
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2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1

25.
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, then
Volume 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.
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a
3
N
M
N
M
 a
3
N
M
a 3
N
nM

26.
 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%.
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itcellvolumeofun
itcelleatomperunvolumeofth
Atomic packing factor

27.
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).
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MILLER INDICES
X
Y
A
B
C
Z
pa
qb
rc
O

28.
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.
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p
1
q
1
r
1
p
1
q
1
r
1

29.
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.
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30.
Reciprocal lattice
• Real lattice
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31.
Continued………..
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32.
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.
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33.
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,lying
between (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.
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34.
X – Rays Production Coolidge tube
• Electron beam
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Heavy metal target
Electron beam
Filament
Vaccume
HT
X-Ray Beam
Copper Anode

35.
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.
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36.
Laue Experimental demonstration
• Determination of crystal structure:
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37.
Laue Experimental demonstration:
•In 1913,Laue,in collaboration with W Friedrich and P Knipping
was 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.
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38.
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.
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39.
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
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d
d
 

A
B
C

40.
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
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41.
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.
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42.
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.
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43.
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 = nλ
For n=1 λ = 2dsin 1
N=2 2λ = 2dsin 2
N=3 3λ = 2dsin 3
Hence sin1: sin2: sin3 = 1: 2 : 3
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44.
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.
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
2
Crystal
X-Ray