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SOLID STATE PHYSICS
Crystal Structure
Dr. Maneesh B. Matte
HOD and Assistant Professor
Department of Physics
Rashtrapita Mahatma Gandhi Art’s and Science College, Nagbhid, Dist:-Chandrapur (M.S.) 441205
Email: maneesh.matte@gmail.com
Contact No:- 7743990371/9518981764
October 16, 2021
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 1 / 74
Outline
1 Basic Of Crystal Structure
Lattice
Basis
2 Crystal Structure
Single Crystal or Crystalline Solids
Polycrystalline
Amorphous Solids
3 Unit Cell
4 Cubic Crystal System
Simple Cubic(SC)
Body Centered Cubic(BCC)
Face Centered Cubic(FCC)
5 Hexagonal Closely Packed Structure(HCP)
6 Bravais Lattice
Bravais Lattice in Two dimension-Space lattice
Bravais Lattice in Three dimension-Space lattice
7 Lattice Translation Vector
8 Reciprocal Lattice
9 Brillouin Zones
10 Diffraction of Crystal
Bragg’s Law
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 2 / 74
Basic Of Crystal Structure
Basic Of Crystal Structure
Crystal structure = Lattice + Basis
Crystal structure is the manner in which atoms, ions, or molecules are spatially arranged.
In a crystal, atoms are arranged in straight rows in a three-dimensional periodic pattern.
It can be constructed by the infinite repetition of these identical structural units in space.
A basic concept in crystal structures is the Unit Cell. It is the smallest unit of volume that permits
identical cells to be stacked together to fill all space.
An ideal crystal is a periodic array of structural units, such as atoms or molecules.
Figure: Some common crystal structures
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 3 / 74
Basic Of Crystal Structure Lattice
Lattice
"An infinite periodic array of points in a space "
The arrangement of points defines the lattice symmetry
A lattice may be one, two or three dimensional.
(a) 1D (b) 2D (c) 3D
Figure: Lattice arrangement
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 4 / 74
Basic Of Crystal Structure Basis
Basis
A group of one or more atoms, located in a particular way with respect to each
other and associated with each point, is known as the Motif or Basis.
A basis of atoms is attached to every lattice points, with every basis is identical in
composition, arrangement and orientation.
Figure: Motif or Basis
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 5 / 74
Basic Of Crystal Structure Basis
Crystal structure = Lattice + Basis
(a) Latice (b) Basis (c) Crystal structure
Figure: Lattice + Basis = Crystal structure
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 6 / 74
Crystal Structure
Crystal Structure
Crystal structure: The manner in which atoms, ions, or molecules are spatially arranged.
It having three type as follows.
Crystalline Solids
Polycrystalline Solids
Amorphous Solids
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 7 / 74
Crystal Structure Single Crystal or Crystalline Solids
Single Crystal or Crystalline Solids
It is mostly common type of solid.
It having a definite and fixed shape, are rigid and incompressible.
The arrangement of particles in a crystalline solid is in a very orderly fashion.
The spaces between the atoms are very less due to high intermolecular forces. This
results in crystals having high melting and boiling points.
The intermolecular force is also uniform throughout the structure. Crystals have a
long-range order, which means the arrangement of atoms is repeated over a great
distance.
In this type of crystal periodicity is maintained throughout the body. It is having
Break points.
Figure: Single Crystal or Crystalline Solids
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 8 / 74
Crystal Structure Polycrystalline
Polycrystalline
Polycrystalline materials is polycrystals solids.
It consists of small crystals, also known as crystallites of varying size and orientation.
Polycrystalline cells can be recognized by a visible grain, a "metal flake effect".
Most inorganic solids are polycrystalline, including all common metals, many
ceramics, rocks, and ice.
Figure: Polycrystalline
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 9 / 74
Crystal Structure Amorphous Solids
Amorphous Solids
It is Randomly organised.
It having no specific shape and Melting and boiling points have range.
It having short order arrangement of particles and no break points hence break
unevenly.
Another name for Amorphous Solids is Super-Cooled Liquids.
The most common example of an amorphous solid is Glass. Gels, plastics, various
polymers, wax, thin films are also good examples of amorphous solids.
Figure: Amorphous Solids
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 10 / 74
Crystal Structure Amorphous Solids
Difference Between Crystalline and Amorphous Solids
Crystals have an orderly arrangement of their constituent particles. In comparison,
amorphous solids have no such arrangement. Their particles are randomly organised.
Crystals have a specific geometric shape with definite edges. Amorphous solids have
no geometry in their shapes
Crystalline solids have a sharp melting point on which they will definitely melt. An
amorphous solid will have a range of temperature over which it will melt, but no
definite temperature as such
Crystals have a long order arrangement of their particles. This means the particles
will show the same arrangement indefinitely. Amorphous solids have a short order
arrangement. Their particles show a lot of variety in their arrangement
Crystalline solids cleavage (break) along particular points and directions. Amorphous
solids cleavage into uneven parts with ragged edges.
Crystals are also known as True Solids, whereas another name for Amorphous Solids
is Super-Cooled Liquids.
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 11 / 74
Unit Cell
Unit Cell
Atoms or group of atoms forming a building block of the smallest acceptable size of
the whole volume of a crystal is defined as a unit cell.
The smallest repeating structure in a crystal.
The basic building block of the crystal structure.
It defines the entire crystal structure with the atom positions within.
Lattice points are located at the corner of the unit cell and in some cases, at either
faces or the centre of the unit cell.
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Unit Cell
Primitive and Non-Primitive Unit Cell
Primitive Unit Cell :- Primitive Cell is a minimum volume unit cell and has only
one lattice point in it. Simple cube is a primitive cell. Number of atoms per unit cell
is one for it.
Non-Primitive Unit Cell :- Non-primitive unit cells contain additional lattice points,
either on a face of the unit cell or within the unit cell, and so have more than one
lattice point per unit cell. BCC and FCC are non primitive. Number of atoms per
unit cell is 2 and 4 respectively.
Figure: Diagram shows primitive and non-primitive unit cell
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Unit Cell
Characteristics of the Unit Cell
No of atoms per unit cell :The number of atoms possessed by a unit cell is known as number of atoms
per unit cell.The distribution of atoms is different for different lattice structure. This can be determined
if the arrangement of atoms inside the cell is known.
Coordination Number: The coordination number of an atom in a crystal is the number atoms directly
surrounding with that atom. If the coordination number is high, then the structure will be more closely
packed. It signifies the tightness of packing of atoms in the crystal.
Atomic radius: Atomic radius is defined as half of the distance between any two nearest neighboring
atoms, which have direct contact with each other, in a crystal of a pure element. It is usually expressed
in terms of cube edge ‘a’(Lattice paramenter).
Packing factor (or) packing density: The packing density is the ratio between the total volume occupied
by the atoms or the molecule in an unit cell and the volume of the unit cell.
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 14 / 74
Cubic Crystal System
Cubic Crystal System
The three Bravais lattices in the cubic crystal system are:
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 15 / 74
Cubic Crystal System Simple Cubic(SC)
Simple Cubic(SC)
Location of lattice point at corner.
Volume of unit cell = a3
. (a is lattice parameter)
Number of atom in a unit cell = 8(1/8) = 1 atom, atomic radius(r) = a/2
Packing efficiency = 52 % (Packing Fraction = Π/6 =0.524)
Number of nearest neighbours or coordination number = 6
Nearest neighbour distance = a
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 16 / 74
Cubic Crystal System Body Centered Cubic(BCC)
Body Centered Cubic(BCC)
Location of lattice point at Corners + Body centre.
Volume of unit cell = a3
. (a is lattice parameter)
Number of atom in a unit cell = 8(1/8) + 1(1) = 2 atoms, atomic radius(r) = a
√
3
/4
Packing efficiency = 68 % (Packing Fraction =
√
3Π/8 =0.680)
Number of nearest neighbours or coordination number = 8
Nearest neighbour distance = a
√
3/2 = 0.866a
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 17 / 74
Cubic Crystal System Face Centered Cubic(FCC)
Face Centered Cubic(FCC)
Location of lattice point at Corners + Faces
Volume of unit cell = a3
. (a is lattice parameter)
Number of atom in a unit cell = 8(1/8)+6(1/2) = 4 atoms, atomic radius(r) =
a/2
√
2
Packing efficiency = 74 % (Packing Fraction =
√
2Π/8 =0.740)
Number of nearest neighbours or coordination number = 12
Nearest neighbour distance = a/
√
2 = 0.707a
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Cubic Crystal System Face Centered Cubic(FCC)
Characteristics of Cubic Lattice
Cubic Crystal System SC BCC FCC
Volume of Unit Cell a3
a3
a3
Number of atom per unit cell 1 2 4
Volume of primitive cell a3
a3
/2 a3
/4
lattice point per unit volume 1/a3
2/a3
4/a3
Number of nearest neighbours 6 8 12
Nearest neighbours Distances a
√
3a/2 a/
√
2
Number of second neighbours 12 6 6
Second neighbours Distances
√
2a a a
Packing factor/Packing density 0.52 0.68 0.74
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 19 / 74
Hexagonal Closely Packed Structure(HCP)
Hexagonal Closely Packed Structure(HCP)
Location of lattice point at Corners + Centre layer + top and bottom faces
Volume of unit cell = 3
√
3a2
c/2. (c is height and a is base side, c=1.633a)
Number of atom in a unit cell = 12(1/6) + 2(1/2) + 3 = 6 atoms, atomic radius(r) = a/2
Packing efficiency = 74.04 % (Packing Fraction = Π/3
√
2 =0.7404)
Number of nearest neighbours or coordination number = 12
Figure: The hexagonal layers form a close-packed structure with unit cell
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Bravais Lattice
Bravais Lattice
The smallest group of symmetrically aligned atoms which can be repeated in an
array to make up the entire crystal is called a unit cell.
There are several ways to describe a lattice. The most fundamental description is
known as the Bravais lattice
a Bravais lattice is an array of discrete points with an arrangement and orientation
that look exactly the same from any of the discrete points, that is the lattice points
are indistinguishable from one another.
These lattices are named after the French physicist Auguste Bravais.
In two-dimensional space, there are 5 Bravais lattices, grouped into four crystal
families.
The 14 Bravais lattices are grouped into seven lattice systems in three-dimensional
space.
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Bravais Lattice Bravais Lattice in Two dimension-Space lattice
Oblique (Monoclinic) a 6= b, θ 6= 900
(a) Oblique (Monoclinic)
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Bravais Lattice Bravais Lattice in Two dimension-Space lattice
Rectangular (Orthorhombic) a 6= b, θ = 900 , c=d, φ 6= 900
(b) Rectangular (Orthorhombic)
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Bravais Lattice Bravais Lattice in Two dimension-Space lattice
Centered Rectangular (orthorhombic), a 6= b, θ = 900 , c=d, φ 6= 900
(c) Centered Rectangular (Orthorhombic)
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Bravais Lattice Bravais Lattice in Two dimension-Space lattice
Hexagonal a = b, θ = 1200
(d) Hexagonal
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Bravais Lattice Bravais Lattice in Two dimension-Space lattice
Square (Tetragonal) a = b, θ = 900
(e) Square (Tetragonal)
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Bravais Lattice Bravais Lattice in Two dimension-Space lattice
Bravais Lattice in Two dimension-Space lattice
Bravais Cell Axes Primitive(P) Centred(I) Number of
Lattices Angles Primitive(P) Centred(I) Lattices
Oblique (Monoclinic) a 6= b X 1
θ 6= 900
Orthorhombic a 6= b,c=d X X 2
θ = 900
,φ = 900
Hexagonal a = b X 1
θ = 1200
Tetragonal a=b X 1
θ = 900
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Bravais Lattice Bravais Lattice in Three dimension-Space lattice
1. Cubic Crystals:- a=b=c and α = β = γ = 900
Figure: Structure in Cubic Crystals
(a) SC (b) BCC (c) FCC
Figure: Name of Cubic Crystals
(a) Fluorite
Octahedron
(b) Pyrite Cube
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Bravais Lattice Bravais Lattice in Three dimension-Space lattice
2. Tetragonal Crystals:- a=b6=c and α = β = γ = 900
Figure: Structure in Tetragonal Crystals
(a) SC (b) BCC
Figure: Name of Tetragonal Crystals
(a) Zircon
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Bravais Lattice Bravais Lattice in Three dimension-Space lattice
3. Orthorhombic Crystals:- a6=b6=c and α = β = γ = 900
Figure: Structure in Orthorhombic Crystals
(a) SC (b) BCC (c) FCC (d) Base Centre
Figure: Name of Orthorhombic Crystals
(a) Topaz
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Bravais Lattice Bravais Lattice in Three dimension-Space lattice
4. Rhombohedral Crystals:- a=b=c and α = β = γ 6= 900
Figure: Structure in Rhombohedral Crystals
(a) SC
Figure: Name of Rhombohedral Crystals
(a) Tourmaline
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Bravais Lattice Bravais Lattice in Three dimension-Space lattice
5. Hexagonal Crystals:- a=b6=c and α = β = 900, γ = 1200
Figure: Structure in Hexagonal Crystals
(a) SC
Figure: Name of Hexagonal Crystals
(a) Corundum
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Bravais Lattice Bravais Lattice in Three dimension-Space lattice
6. Monoclinic Crystals:- a6=b6=c and α = β = 900, γ = 1200
Figure: Structure in Monoclinic Crystals
(a) SC (b) BCC
Figure: Name of Monoclinic Crystals
(a) Kunzite
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Bravais Lattice Bravais Lattice in Three dimension-Space lattice
7. Triclinic Crystals:- a6=b6=c and α, β, γ 6= 900
Figure: Structure in Triclinic Crystals
(a) SC
Figure: Name of Triclinic Crystals
(a) Amazonite
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Bravais Lattice Bravais Lattice in Three dimension-Space lattice
THE 7 CRYSTAL SYSTEMS
Bravais Cell Axes Simple Body Face Base Number of
Lattices Angles Primitive(P) Centred(I) Centred(F) Centred(C) lattices
Cubic a=b=c X X X 3
α = β = γ = 900
Tetragonal a=b6=c X X 2
α = β = γ = 900
Orthorhombic a6=b6=c X X X X 4
α = β = γ = 900
Rhombohedral a=b=c X 1
α = β = γ 6= 900
Hexagonal a=b6=c X 1
α = β = 900
, γ = 1200
Monoclinic a6=b6=c X X 2
α = β = 900
, γ = 1200
Triclinic a6=b6=c X 1
α, β, γ 6= 900
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Lattice Translation Vector
Lattice Translation Vector
Translation vectors define the distance vectors that all atoms in the cluster are translated through to
form another cluster in the solid.
The translation vector is the distance through which an atom must be moved in order to be in the next
unit cell.
The lattice is defined by fundamental translation vectors, the position vector of any lattice site of the two
dimensional lattice as T = n1a1 + n2a2
where a1 and a2 are the two vectors and n1 and n2 is a pair of integers whose values depend on the
lattice site.
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Lattice Translation Vector
The choices of primitive lattice vectors are NOT unique.
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 37 / 74
Lattice Translation Vector
The choices of primitive lattice vectors are NOT unique.
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Lattice Translation Vector
Wigner–Seitz cell
The Wigner–Seitz cell around a lattice point is defined as the locus of points in
space that are closer to that lattice point than to any of the other lattice points.
A Wigner–Seitz cell is an example of a primitive cell, which is a unit cell containing
exactly one lattice point.
To create a Wigner-Seitz cell simply complete the following steps.
Choose any lattice site as the origin.
Starting at the origin draw vectors to all neighbouring lattice points.
Construct a plane perpendicular to and passing through the midpoint of each vector.
The area enclosed by these planes is the Wigner-Seitz cell.
Figure: Construction of a Wigner-Seitz Cell
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Lattice Translation Vector
Miller Indices
Miller Indices are used to represent the directions and the planes in a crystal.
Miller Indices is a group of smallest integers which represent a direction or a plane.
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Lattice Translation Vector
Miller Indices of a Direction
Based on interaction with cell boundaries.
Select any point on the direction line other than the origin.
Find out the coordinates of the point in terms of the unit vectors along different axes.
Specify negative coordinate with a bar on top
Divide them by the unit vector along the respective axis
Convert the result into the smallest integers by suitable multiplication or division and
express as [UVW].
Figure: Miller Indices of a Direction
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Lattice Translation Vector
Example: determine Miller indices of a direction
Figure: Miller Indices of a Direction
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Lattice Translation Vector
Miller Indices of planes
Based on reciprocal of interaction of the plane with cell axes, indicated with
parenthesis(h,k,l)
Find the intercepts of the plane with the three axes: (pa, qb, rc).
Take the reciprocals of the numbers (p, q, r)
Reduce to three smallest integers (h, k, l) by suitable multiplication or division.
Miller indices of the plane: (hkl)
Negative indices are indicated by a bar on top.
Same indices for parallel planes
A family of crystallographically equivalent planes (not necessarily parallel) is denoted by [hkl]
Figure: Miller Indices of planes
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Lattice Translation Vector
Example: determine Miller indices of a plane
Intercepts: (a/2, 2b/3, ∞)
p= 1/2, q= 2/3, r= ∞
Reciprocals: h = 2, k = 3/2, l = 0
Miller indices: (430)
Figure: Miller Indices of planes
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Lattice Translation Vector
Example: determine Miller indices of a plane
Figure: Crystal Planes
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Reciprocal Lattice
Reciprocal Lattice
The fundamental property of a crystal is its triple periodicity and a Crystal may be
generated by repeating a certain unit of pattern through the translations of a certain
lattice called the Direct lattice.
To describe the morphology of a crystal, the simplest way is to associate, with each
set of lattice planes parallel to a natural face, a vector drawn from a given origin and
normal to the corresponding lattice planes.
To complete the description it suffices to give to each vector a length directly related
to the spacing of the lattice planes.
A reciprocal lattice is regarded as a geometrical abstraction.It is essentially identical
to a "wave vector" k-space
Why do we need a reciprocal lattice?
Reciprocal lattice provides simple geometrical basis for understanding:
a)All things of "wave nature" (like behaviour of electron and lattice vibrations in crystals.
b)The geometry of x-ray and electron diffraction patterns.
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 46 / 74
Reciprocal Lattice
Mathematical Definition of Reciprocal Lattice
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Reciprocal Lattice
Note
Since Ḡ • R̄ = 2πn this implies that b̄i • ¯
aj = 2πδij where δij = 1 if i=j and δij = 0 if
i 6= j ( ¯
b1 • ¯
a1 = 2 π and ¯
b1 • ¯
a2 = 0 ).
for any Ḡ = k1
¯
b1 + k2
¯
b2 + k3
¯
b3 and R̄ = n1 ¯
a1 + n2 ¯
a2 + n3 ¯
a3 ,Ḡ • R̄ =
2π(k1n1 + k2n2 + k3n3)
The two lattices (reciprocal and direct)are related by the above definitions in 1.
Rotating a crystal means rotating both the direct and reciprocal lattices.
The direct crystal lattice has the dimension of [L] while the reciprocal lattice has the
dimension of [L−1
]
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 48 / 74
Reciprocal Lattice
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Reciprocal Lattice
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Reciprocal Lattice
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Reciprocal Lattice
The reciprocal lattice of a SC is also a SC
The BCC primitive lattice vectors in the reciprocal lattice are just the primitive
vectors of an FCC lattice.
The FCC primitive lattice vectors in the reciprocal lattice are just the primitive
vectors of an BCC lattice.
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Brillouin Zones
Brillouin Zones.
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Brillouin Zones
Brillouin Zones.
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Brillouin Zones
Brillouin Zones.
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Brillouin Zones
Construction of a Wigner-Seitz cell in the Reciprocal lattice (First Brillouin
zone)
To construct the first Brillouin zone, we need to find the link between the incident beam (likeelectron or
neutron or phonon beam) of wave vector k̄ and the reciprocal lattice vector Ḡ. The procedure to build up the
first Brillouin zone is as follows: -
Select a vector Ḡ from the origin to a reciprocal lattice point
Construct a plane normal to the vector Ḡ at itsmid point. This plane forms a part of the zone boundary.
The diffracted beam will be in the direction k̄ − Ḡ.
Thus the Brillouin construction exhibits all the wave vectors k̄ which can be Bragg-reflected by the
crystal.
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Brillouin Zones
First, Second and Third Brillouin Zones.
Wigner-Seitz cell: smallest possible primitive cell, which consist of one lattice point and all the surrounding space closer to it than
to any other point. The construction of the W-S cell in the reciprocal lattice delivers the first Brillouin zone (important for
diffraction)
The importance of Brillouin zone:
The Brillouin zones are used to describe and analyze the electron energy in the band
energy structure of crystals.
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 57 / 74
Diffraction of Crystal
Diffraction
It occurs significantly when the size of the aperture or obstacle is of similar linear dimensions to the
wavelength of the incident wave.
It happens when a part of the travelling wavefront is obscured.
It is defined as the bending of waves around the corners of an obstacle or through an aperture into the
region of geometrical shadow of the obstacle/aperture.
For very small aperture sizes, the vast majority of the wave is blocked. For large apertures the wave
passes by or through the obstacle without any significant diffraction, and that largely at the edges.
In an aperture with width smaller than the wavelength, the wave transmitted through the aperture
spreads all the way round and behaves like a point source of waves (they spread out below).
(a) d < λ (b) d > λ (c) waves moving
directly forward
(d) slit at an angle θ
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 58 / 74
Diffraction of Crystal
Scattering and Diffraction
A diffracted beam may be defined as a beam composed of a large number of
scattered rays mutually reinforcing each other.
Figure: Scattering and Diffraction
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 59 / 74
Diffraction of Crystal
X-Rays
X-Rays discovered by Wilhelm Roentgen.
The name as X-Rays because in the nature at first was unknown and also Roentgen
Rays.
The wavelength range 0.1 to 100 A0
It is short wavelength electromagnetic radiation produced by deceleration of high
energy electrons or by electronic transition of electrons in the inner orbital of atom.
The penetrating power of X-rays depends on energy, there are two types of x-rays
Soft X-rays which have less penetrating and low energy.
Hard X-rays which have more penetrating and more energy.
It is used for determine the structure of crystal by XRD.
XRD in crystal discovered by Max Von Laue.
(a) Wilhelm Roentgen (b) Max Von Laue
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 60 / 74
Diffraction of Crystal
X-Ray Diffraction
Principle:-
XRD based on constructive interference of monochromatic x-rays and crystalline sample. These X-rays are
generated by cathode rays tube , filtered to produce monochromatic radiation,collimated to concentrated and
directed towards sample. The interaction of incident rays with sample produces constructive interference when
condition satisfy the Bragg’s Law.
Every Crystalline substance gives pattern, the same substance gives same pattern and in the mixture of
substances each produce its pattern independently of the others.
XRD as finger print of the substance. It is based on scattering of x-rays by crystals.
The atomic planes of crystal cause as incident beam of x-rays to interfere with another as they leave the
crystals. This phenomenon is called XRD.
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 61 / 74
Diffraction of Crystal
Why XRD?
Measure the average spacing’s between layers or rows of atoms.
Determine the orientation of a single crystal or grain.
Find the crystal structure of an unknown material.
Measure the size, shape and internal stress of small crystalline regions.
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 62 / 74
Diffraction of Crystal
Essential Parts of the Diffractometer
X-ray Tube: the source of X Rays
Incident-beam optics: condition the X-ray beam before it hits the sample
The goniometer: the platform that holds and moves the sample, optics, detector,
or tube
The sample and sample holder
Receiving-side optics: condition the X-ray beam after it has encountered the
sample
Detector: count the number of X Rays scattered by the sample
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 63 / 74
Diffraction of Crystal
Basic components and Features of XRD
Production
Diffraction
Detection
Interpretation
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 64 / 74
Diffraction of Crystal Bragg’s Law
Bragg’s Law
In 1913, Sir William Henry Bragg and his son Sir Lawrence Bragg develop the relationship to explain why
the cleave faces of crystals appears to reflect x-rays beam at certain angle of incidence(θ).
The variable d is distance between atomic layers in a crystal and variable λ is wavelength of incident
beam n is integer.
Although Bragg’s law was used to explain interference pattern of x-rays scattered by crystals, diffraction
has been developed to study the structure of all state of matter with any beam.
Bragg’s carried out series of experiment and creates Bragg’s Equation
nλ = 2dsinθ
where , n = 1 for first order reflection
λ = wavelength of incident beam
θ = angle of incidence
d = distance between atomic layers.
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 65 / 74
Diffraction of Crystal Bragg’s Law
Derivation of Bragg’law
For Constructive interference ,path difference is integral multiple of wavelength.
So, nλ = path difference = AB + BC (from figure)
but AB = BC
therefore nλ = 2 × AB
From right angle ∆ ABZ , Sinθ = AB
d
i.e. AB = d Sinθ
so,
nλ = 2dhkl Sinθhkl
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 66 / 74
Diffraction of Crystal Bragg’s Law
Conditions are required for Diffraction in Bragg’s Law
For parallel planes of atoms, with a space dhkl between the planes, constructive
interference only occurs when Bragg’s law is satisfied.
In diffractometers, the X-ray wavelength λ is fixed.
Consequently, a family of planes produces a diffraction peak only at a specific angle θ
Additionally, the plane normal must be parallel to the diffraction vector.
Plane normal: the direction perpendicular to a plane of atoms
Diffraction vector: the vector that bisects the angle between the incident and diffracted beam
The space between diffracting planes of atoms determines peak positions.
The peak intensity is determined by what atoms are in the diffracting plane.
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 67 / 74
Diffraction of Crystal Bragg’s Law
XRD-Methods
Laue photographic method
Braggs X-Ray spectrometer
Rotating crystal method
Powder method
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 68 / 74
Diffraction of Crystal Bragg’s Law
Braggs X-Ray spectrometer
x-rays from x-rays tube
S1 and S2 adjustable slit
C - crystal
Ionization Chamber
One plate of ionization chamber is connected to the positive terminal of a H.T
Battery , while negative terminal is connected to quadrant electro-meter(measures
the strength of ionization current)
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 69 / 74
Diffraction of Crystal Bragg’s Law
Working of Bragg’s x-ray spectrometer
Crystal is mounted such that θ= 0 and ionization chamber is adjusted to receive x-rays.
Crystal and ionization chamber are allowed to move in small steps
The angle through which the chamber is moved is twice the angle through which the crystal is rotated
X-ray spectrum is obtained by plotting a graph between ionization current and the glancing angle θ
Peaks are obtained, corresponding to Bragg’s diffraction.
Different order glancing angles are obtained with known values of λ and n and from the observed value
of θ and d can be measured.
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 70 / 74
Diffraction of Crystal Bragg’s Law
DETERMINATION OF CRYSTAL STRUCTURE BY BRAGG’S LAW
X-Rays fall on crystal surface
The crystal is rotated and x-rays are made to reflect from various lattice planes
The intense reflections are measured by bragg’s spectrometer and the glancing
angles for each reflection is recorded
Then on applying bragg’s equation ratio of lattice spacing for various groups of
planes can be obtained.
Ratio’s will be different for different crystals
Experimentally observed ratio’s are compared with the calculated ratio’s ,particular
structure may be identified.
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 71 / 74
Diffraction of Crystal Bragg’s Law
XRD Applications
Phase Composition of a Sample.
Quantitative Phase Analysis: determine the relative amounts of phases in a mixture
by referencing the relative peak intensities
Unit cell lattice parameters and Bravais lattice symmetry
Index peak positions
Lattice parameters can vary as a function of, and therefore give you information about,
alloying, doping, solid solutions, strains, etc.
Crystal Structure
Crystallite Size and Micro-strain
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 72 / 74
Diffraction of Crystal Bragg’s Law
Numericals
Calculate the wavelength for the first-order spectrum if the angle of incidence is 30
degrees.
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.
What is the distance between the adjacent Miller planes if the first order reflection
from X-rays of wavelength 2.29 A0
occurs at 27o
80
Gold crystallizes into an FCC structure. The edge length of the FCC unit cell is 4.07
Calculate a) the closest distance between two gold atoms and b) the density of gold
if its atomic weight is 197.
X rays of λ = 0.1537 nm from a Cu target are diffracted from the (111) planes of an
FCC metal. The Bragg angle is 19.2o
, find lattice parameter.
X- rays of wavelength = 179 pm produce a reflection at 2 = 47.2o
from the 110
planes of BCC lattice. Calculate the edge length of the unit cell.
Plane 1 cuts the axes at a / 3, b / 2 and c /4 ,What are the Miller indices of these
planes ?
The spacing of one set of crystal planes in NaCl is 0.282 nm. A monochromatic
beam of X-rays produces a Bragg maximum when its glancing angle with these
planes is θ= 70
first order maximum, find the wavelength of the X-rays. What is the
minimum possible accelerating voltage V0 that produced the X-rays?
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 73 / 74
Diffraction of Crystal Bragg’s Law
Thank You
Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 74 / 74

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  • 1. SOLID STATE PHYSICS Crystal Structure Dr. Maneesh B. Matte HOD and Assistant Professor Department of Physics Rashtrapita Mahatma Gandhi Art’s and Science College, Nagbhid, Dist:-Chandrapur (M.S.) 441205 Email: maneesh.matte@gmail.com Contact No:- 7743990371/9518981764 October 16, 2021 Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 1 / 74
  • 2. Outline 1 Basic Of Crystal Structure Lattice Basis 2 Crystal Structure Single Crystal or Crystalline Solids Polycrystalline Amorphous Solids 3 Unit Cell 4 Cubic Crystal System Simple Cubic(SC) Body Centered Cubic(BCC) Face Centered Cubic(FCC) 5 Hexagonal Closely Packed Structure(HCP) 6 Bravais Lattice Bravais Lattice in Two dimension-Space lattice Bravais Lattice in Three dimension-Space lattice 7 Lattice Translation Vector 8 Reciprocal Lattice 9 Brillouin Zones 10 Diffraction of Crystal Bragg’s Law Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 2 / 74
  • 3. Basic Of Crystal Structure Basic Of Crystal Structure Crystal structure = Lattice + Basis Crystal structure is the manner in which atoms, ions, or molecules are spatially arranged. In a crystal, atoms are arranged in straight rows in a three-dimensional periodic pattern. It can be constructed by the infinite repetition of these identical structural units in space. A basic concept in crystal structures is the Unit Cell. It is the smallest unit of volume that permits identical cells to be stacked together to fill all space. An ideal crystal is a periodic array of structural units, such as atoms or molecules. Figure: Some common crystal structures Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 3 / 74
  • 4. Basic Of Crystal Structure Lattice Lattice "An infinite periodic array of points in a space " The arrangement of points defines the lattice symmetry A lattice may be one, two or three dimensional. (a) 1D (b) 2D (c) 3D Figure: Lattice arrangement Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 4 / 74
  • 5. Basic Of Crystal Structure Basis Basis A group of one or more atoms, located in a particular way with respect to each other and associated with each point, is known as the Motif or Basis. A basis of atoms is attached to every lattice points, with every basis is identical in composition, arrangement and orientation. Figure: Motif or Basis Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 5 / 74
  • 6. Basic Of Crystal Structure Basis Crystal structure = Lattice + Basis (a) Latice (b) Basis (c) Crystal structure Figure: Lattice + Basis = Crystal structure Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 6 / 74
  • 7. Crystal Structure Crystal Structure Crystal structure: The manner in which atoms, ions, or molecules are spatially arranged. It having three type as follows. Crystalline Solids Polycrystalline Solids Amorphous Solids Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 7 / 74
  • 8. Crystal Structure Single Crystal or Crystalline Solids Single Crystal or Crystalline Solids It is mostly common type of solid. It having a definite and fixed shape, are rigid and incompressible. The arrangement of particles in a crystalline solid is in a very orderly fashion. The spaces between the atoms are very less due to high intermolecular forces. This results in crystals having high melting and boiling points. The intermolecular force is also uniform throughout the structure. Crystals have a long-range order, which means the arrangement of atoms is repeated over a great distance. In this type of crystal periodicity is maintained throughout the body. It is having Break points. Figure: Single Crystal or Crystalline Solids Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 8 / 74
  • 9. Crystal Structure Polycrystalline Polycrystalline Polycrystalline materials is polycrystals solids. It consists of small crystals, also known as crystallites of varying size and orientation. Polycrystalline cells can be recognized by a visible grain, a "metal flake effect". Most inorganic solids are polycrystalline, including all common metals, many ceramics, rocks, and ice. Figure: Polycrystalline Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 9 / 74
  • 10. Crystal Structure Amorphous Solids Amorphous Solids It is Randomly organised. It having no specific shape and Melting and boiling points have range. It having short order arrangement of particles and no break points hence break unevenly. Another name for Amorphous Solids is Super-Cooled Liquids. The most common example of an amorphous solid is Glass. Gels, plastics, various polymers, wax, thin films are also good examples of amorphous solids. Figure: Amorphous Solids Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 10 / 74
  • 11. Crystal Structure Amorphous Solids Difference Between Crystalline and Amorphous Solids Crystals have an orderly arrangement of their constituent particles. In comparison, amorphous solids have no such arrangement. Their particles are randomly organised. Crystals have a specific geometric shape with definite edges. Amorphous solids have no geometry in their shapes Crystalline solids have a sharp melting point on which they will definitely melt. An amorphous solid will have a range of temperature over which it will melt, but no definite temperature as such Crystals have a long order arrangement of their particles. This means the particles will show the same arrangement indefinitely. Amorphous solids have a short order arrangement. Their particles show a lot of variety in their arrangement Crystalline solids cleavage (break) along particular points and directions. Amorphous solids cleavage into uneven parts with ragged edges. Crystals are also known as True Solids, whereas another name for Amorphous Solids is Super-Cooled Liquids. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 11 / 74
  • 12. Unit Cell Unit Cell Atoms or group of atoms forming a building block of the smallest acceptable size of the whole volume of a crystal is defined as a unit cell. The smallest repeating structure in a crystal. The basic building block of the crystal structure. It defines the entire crystal structure with the atom positions within. Lattice points are located at the corner of the unit cell and in some cases, at either faces or the centre of the unit cell. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 12 / 74
  • 13. Unit Cell Primitive and Non-Primitive Unit Cell Primitive Unit Cell :- Primitive Cell is a minimum volume unit cell and has only one lattice point in it. Simple cube is a primitive cell. Number of atoms per unit cell is one for it. Non-Primitive Unit Cell :- Non-primitive unit cells contain additional lattice points, either on a face of the unit cell or within the unit cell, and so have more than one lattice point per unit cell. BCC and FCC are non primitive. Number of atoms per unit cell is 2 and 4 respectively. Figure: Diagram shows primitive and non-primitive unit cell Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 13 / 74
  • 14. Unit Cell Characteristics of the Unit Cell No of atoms per unit cell :The number of atoms possessed by a unit cell is known as number of atoms per unit cell.The distribution of atoms is different for different lattice structure. This can be determined if the arrangement of atoms inside the cell is known. Coordination Number: The coordination number of an atom in a crystal is the number atoms directly surrounding with that atom. If the coordination number is high, then the structure will be more closely packed. It signifies the tightness of packing of atoms in the crystal. Atomic radius: Atomic radius is defined as half of the distance between any two nearest neighboring atoms, which have direct contact with each other, in a crystal of a pure element. It is usually expressed in terms of cube edge ‘a’(Lattice paramenter). Packing factor (or) packing density: The packing density is the ratio between the total volume occupied by the atoms or the molecule in an unit cell and the volume of the unit cell. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 14 / 74
  • 15. Cubic Crystal System Cubic Crystal System The three Bravais lattices in the cubic crystal system are: Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 15 / 74
  • 16. Cubic Crystal System Simple Cubic(SC) Simple Cubic(SC) Location of lattice point at corner. Volume of unit cell = a3 . (a is lattice parameter) Number of atom in a unit cell = 8(1/8) = 1 atom, atomic radius(r) = a/2 Packing efficiency = 52 % (Packing Fraction = Π/6 =0.524) Number of nearest neighbours or coordination number = 6 Nearest neighbour distance = a Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 16 / 74
  • 17. Cubic Crystal System Body Centered Cubic(BCC) Body Centered Cubic(BCC) Location of lattice point at Corners + Body centre. Volume of unit cell = a3 . (a is lattice parameter) Number of atom in a unit cell = 8(1/8) + 1(1) = 2 atoms, atomic radius(r) = a √ 3 /4 Packing efficiency = 68 % (Packing Fraction = √ 3Π/8 =0.680) Number of nearest neighbours or coordination number = 8 Nearest neighbour distance = a √ 3/2 = 0.866a Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 17 / 74
  • 18. Cubic Crystal System Face Centered Cubic(FCC) Face Centered Cubic(FCC) Location of lattice point at Corners + Faces Volume of unit cell = a3 . (a is lattice parameter) Number of atom in a unit cell = 8(1/8)+6(1/2) = 4 atoms, atomic radius(r) = a/2 √ 2 Packing efficiency = 74 % (Packing Fraction = √ 2Π/8 =0.740) Number of nearest neighbours or coordination number = 12 Nearest neighbour distance = a/ √ 2 = 0.707a Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 18 / 74
  • 19. Cubic Crystal System Face Centered Cubic(FCC) Characteristics of Cubic Lattice Cubic Crystal System SC BCC FCC Volume of Unit Cell a3 a3 a3 Number of atom per unit cell 1 2 4 Volume of primitive cell a3 a3 /2 a3 /4 lattice point per unit volume 1/a3 2/a3 4/a3 Number of nearest neighbours 6 8 12 Nearest neighbours Distances a √ 3a/2 a/ √ 2 Number of second neighbours 12 6 6 Second neighbours Distances √ 2a a a Packing factor/Packing density 0.52 0.68 0.74 Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 19 / 74
  • 20. Hexagonal Closely Packed Structure(HCP) Hexagonal Closely Packed Structure(HCP) Location of lattice point at Corners + Centre layer + top and bottom faces Volume of unit cell = 3 √ 3a2 c/2. (c is height and a is base side, c=1.633a) Number of atom in a unit cell = 12(1/6) + 2(1/2) + 3 = 6 atoms, atomic radius(r) = a/2 Packing efficiency = 74.04 % (Packing Fraction = Π/3 √ 2 =0.7404) Number of nearest neighbours or coordination number = 12 Figure: The hexagonal layers form a close-packed structure with unit cell Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 20 / 74
  • 21. Bravais Lattice Bravais Lattice The smallest group of symmetrically aligned atoms which can be repeated in an array to make up the entire crystal is called a unit cell. There are several ways to describe a lattice. The most fundamental description is known as the Bravais lattice a Bravais lattice is an array of discrete points with an arrangement and orientation that look exactly the same from any of the discrete points, that is the lattice points are indistinguishable from one another. These lattices are named after the French physicist Auguste Bravais. In two-dimensional space, there are 5 Bravais lattices, grouped into four crystal families. The 14 Bravais lattices are grouped into seven lattice systems in three-dimensional space. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 21 / 74
  • 22. Bravais Lattice Bravais Lattice in Two dimension-Space lattice Oblique (Monoclinic) a 6= b, θ 6= 900 (a) Oblique (Monoclinic) Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 22 / 74
  • 23. Bravais Lattice Bravais Lattice in Two dimension-Space lattice Rectangular (Orthorhombic) a 6= b, θ = 900 , c=d, φ 6= 900 (b) Rectangular (Orthorhombic) Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 23 / 74
  • 24. Bravais Lattice Bravais Lattice in Two dimension-Space lattice Centered Rectangular (orthorhombic), a 6= b, θ = 900 , c=d, φ 6= 900 (c) Centered Rectangular (Orthorhombic) Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 24 / 74
  • 25. Bravais Lattice Bravais Lattice in Two dimension-Space lattice Hexagonal a = b, θ = 1200 (d) Hexagonal Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 25 / 74
  • 26. Bravais Lattice Bravais Lattice in Two dimension-Space lattice Square (Tetragonal) a = b, θ = 900 (e) Square (Tetragonal) Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 26 / 74
  • 27. Bravais Lattice Bravais Lattice in Two dimension-Space lattice Bravais Lattice in Two dimension-Space lattice Bravais Cell Axes Primitive(P) Centred(I) Number of Lattices Angles Primitive(P) Centred(I) Lattices Oblique (Monoclinic) a 6= b X 1 θ 6= 900 Orthorhombic a 6= b,c=d X X 2 θ = 900 ,φ = 900 Hexagonal a = b X 1 θ = 1200 Tetragonal a=b X 1 θ = 900 Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 27 / 74
  • 28. Bravais Lattice Bravais Lattice in Three dimension-Space lattice 1. Cubic Crystals:- a=b=c and α = β = γ = 900 Figure: Structure in Cubic Crystals (a) SC (b) BCC (c) FCC Figure: Name of Cubic Crystals (a) Fluorite Octahedron (b) Pyrite Cube Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 28 / 74
  • 29. Bravais Lattice Bravais Lattice in Three dimension-Space lattice 2. Tetragonal Crystals:- a=b6=c and α = β = γ = 900 Figure: Structure in Tetragonal Crystals (a) SC (b) BCC Figure: Name of Tetragonal Crystals (a) Zircon Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 29 / 74
  • 30. Bravais Lattice Bravais Lattice in Three dimension-Space lattice 3. Orthorhombic Crystals:- a6=b6=c and α = β = γ = 900 Figure: Structure in Orthorhombic Crystals (a) SC (b) BCC (c) FCC (d) Base Centre Figure: Name of Orthorhombic Crystals (a) Topaz Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 30 / 74
  • 31. Bravais Lattice Bravais Lattice in Three dimension-Space lattice 4. Rhombohedral Crystals:- a=b=c and α = β = γ 6= 900 Figure: Structure in Rhombohedral Crystals (a) SC Figure: Name of Rhombohedral Crystals (a) Tourmaline Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 31 / 74
  • 32. Bravais Lattice Bravais Lattice in Three dimension-Space lattice 5. Hexagonal Crystals:- a=b6=c and α = β = 900, γ = 1200 Figure: Structure in Hexagonal Crystals (a) SC Figure: Name of Hexagonal Crystals (a) Corundum Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 32 / 74
  • 33. Bravais Lattice Bravais Lattice in Three dimension-Space lattice 6. Monoclinic Crystals:- a6=b6=c and α = β = 900, γ = 1200 Figure: Structure in Monoclinic Crystals (a) SC (b) BCC Figure: Name of Monoclinic Crystals (a) Kunzite Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 33 / 74
  • 34. Bravais Lattice Bravais Lattice in Three dimension-Space lattice 7. Triclinic Crystals:- a6=b6=c and α, β, γ 6= 900 Figure: Structure in Triclinic Crystals (a) SC Figure: Name of Triclinic Crystals (a) Amazonite Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 34 / 74
  • 35. Bravais Lattice Bravais Lattice in Three dimension-Space lattice THE 7 CRYSTAL SYSTEMS Bravais Cell Axes Simple Body Face Base Number of Lattices Angles Primitive(P) Centred(I) Centred(F) Centred(C) lattices Cubic a=b=c X X X 3 α = β = γ = 900 Tetragonal a=b6=c X X 2 α = β = γ = 900 Orthorhombic a6=b6=c X X X X 4 α = β = γ = 900 Rhombohedral a=b=c X 1 α = β = γ 6= 900 Hexagonal a=b6=c X 1 α = β = 900 , γ = 1200 Monoclinic a6=b6=c X X 2 α = β = 900 , γ = 1200 Triclinic a6=b6=c X 1 α, β, γ 6= 900 Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 35 / 74
  • 36. Lattice Translation Vector Lattice Translation Vector Translation vectors define the distance vectors that all atoms in the cluster are translated through to form another cluster in the solid. The translation vector is the distance through which an atom must be moved in order to be in the next unit cell. The lattice is defined by fundamental translation vectors, the position vector of any lattice site of the two dimensional lattice as T = n1a1 + n2a2 where a1 and a2 are the two vectors and n1 and n2 is a pair of integers whose values depend on the lattice site. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 36 / 74
  • 37. Lattice Translation Vector The choices of primitive lattice vectors are NOT unique. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 37 / 74
  • 38. Lattice Translation Vector The choices of primitive lattice vectors are NOT unique. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 38 / 74
  • 39. Lattice Translation Vector Wigner–Seitz cell The Wigner–Seitz cell around a lattice point is defined as the locus of points in space that are closer to that lattice point than to any of the other lattice points. A Wigner–Seitz cell is an example of a primitive cell, which is a unit cell containing exactly one lattice point. To create a Wigner-Seitz cell simply complete the following steps. Choose any lattice site as the origin. Starting at the origin draw vectors to all neighbouring lattice points. Construct a plane perpendicular to and passing through the midpoint of each vector. The area enclosed by these planes is the Wigner-Seitz cell. Figure: Construction of a Wigner-Seitz Cell Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 39 / 74
  • 40. Lattice Translation Vector Miller Indices Miller Indices are used to represent the directions and the planes in a crystal. Miller Indices is a group of smallest integers which represent a direction or a plane. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 40 / 74
  • 41. Lattice Translation Vector Miller Indices of a Direction Based on interaction with cell boundaries. Select any point on the direction line other than the origin. Find out the coordinates of the point in terms of the unit vectors along different axes. Specify negative coordinate with a bar on top Divide them by the unit vector along the respective axis Convert the result into the smallest integers by suitable multiplication or division and express as [UVW]. Figure: Miller Indices of a Direction Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 41 / 74
  • 42. Lattice Translation Vector Example: determine Miller indices of a direction Figure: Miller Indices of a Direction Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 42 / 74
  • 43. Lattice Translation Vector Miller Indices of planes Based on reciprocal of interaction of the plane with cell axes, indicated with parenthesis(h,k,l) Find the intercepts of the plane with the three axes: (pa, qb, rc). Take the reciprocals of the numbers (p, q, r) Reduce to three smallest integers (h, k, l) by suitable multiplication or division. Miller indices of the plane: (hkl) Negative indices are indicated by a bar on top. Same indices for parallel planes A family of crystallographically equivalent planes (not necessarily parallel) is denoted by [hkl] Figure: Miller Indices of planes Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 43 / 74
  • 44. Lattice Translation Vector Example: determine Miller indices of a plane Intercepts: (a/2, 2b/3, ∞) p= 1/2, q= 2/3, r= ∞ Reciprocals: h = 2, k = 3/2, l = 0 Miller indices: (430) Figure: Miller Indices of planes Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 44 / 74
  • 45. Lattice Translation Vector Example: determine Miller indices of a plane Figure: Crystal Planes Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 45 / 74
  • 46. Reciprocal Lattice Reciprocal Lattice The fundamental property of a crystal is its triple periodicity and a Crystal may be generated by repeating a certain unit of pattern through the translations of a certain lattice called the Direct lattice. To describe the morphology of a crystal, the simplest way is to associate, with each set of lattice planes parallel to a natural face, a vector drawn from a given origin and normal to the corresponding lattice planes. To complete the description it suffices to give to each vector a length directly related to the spacing of the lattice planes. A reciprocal lattice is regarded as a geometrical abstraction.It is essentially identical to a "wave vector" k-space Why do we need a reciprocal lattice? Reciprocal lattice provides simple geometrical basis for understanding: a)All things of "wave nature" (like behaviour of electron and lattice vibrations in crystals. b)The geometry of x-ray and electron diffraction patterns. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 46 / 74
  • 47. Reciprocal Lattice Mathematical Definition of Reciprocal Lattice Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 47 / 74
  • 48. Reciprocal Lattice Note Since Ḡ • R̄ = 2πn this implies that b̄i • ¯ aj = 2πδij where δij = 1 if i=j and δij = 0 if i 6= j ( ¯ b1 • ¯ a1 = 2 π and ¯ b1 • ¯ a2 = 0 ). for any Ḡ = k1 ¯ b1 + k2 ¯ b2 + k3 ¯ b3 and R̄ = n1 ¯ a1 + n2 ¯ a2 + n3 ¯ a3 ,Ḡ • R̄ = 2π(k1n1 + k2n2 + k3n3) The two lattices (reciprocal and direct)are related by the above definitions in 1. Rotating a crystal means rotating both the direct and reciprocal lattices. The direct crystal lattice has the dimension of [L] while the reciprocal lattice has the dimension of [L−1 ] Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 48 / 74
  • 49. Reciprocal Lattice Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 49 / 74
  • 50. Reciprocal Lattice Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 50 / 74
  • 51. Reciprocal Lattice Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 51 / 74
  • 52. Reciprocal Lattice The reciprocal lattice of a SC is also a SC The BCC primitive lattice vectors in the reciprocal lattice are just the primitive vectors of an FCC lattice. The FCC primitive lattice vectors in the reciprocal lattice are just the primitive vectors of an BCC lattice. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 52 / 74
  • 53. Brillouin Zones Brillouin Zones. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 53 / 74
  • 54. Brillouin Zones Brillouin Zones. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 54 / 74
  • 55. Brillouin Zones Brillouin Zones. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 55 / 74
  • 56. Brillouin Zones Construction of a Wigner-Seitz cell in the Reciprocal lattice (First Brillouin zone) To construct the first Brillouin zone, we need to find the link between the incident beam (likeelectron or neutron or phonon beam) of wave vector k̄ and the reciprocal lattice vector Ḡ. The procedure to build up the first Brillouin zone is as follows: - Select a vector Ḡ from the origin to a reciprocal lattice point Construct a plane normal to the vector Ḡ at itsmid point. This plane forms a part of the zone boundary. The diffracted beam will be in the direction k̄ − Ḡ. Thus the Brillouin construction exhibits all the wave vectors k̄ which can be Bragg-reflected by the crystal. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 56 / 74
  • 57. Brillouin Zones First, Second and Third Brillouin Zones. Wigner-Seitz cell: smallest possible primitive cell, which consist of one lattice point and all the surrounding space closer to it than to any other point. The construction of the W-S cell in the reciprocal lattice delivers the first Brillouin zone (important for diffraction) The importance of Brillouin zone: The Brillouin zones are used to describe and analyze the electron energy in the band energy structure of crystals. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 57 / 74
  • 58. Diffraction of Crystal Diffraction It occurs significantly when the size of the aperture or obstacle is of similar linear dimensions to the wavelength of the incident wave. It happens when a part of the travelling wavefront is obscured. It is defined as the bending of waves around the corners of an obstacle or through an aperture into the region of geometrical shadow of the obstacle/aperture. For very small aperture sizes, the vast majority of the wave is blocked. For large apertures the wave passes by or through the obstacle without any significant diffraction, and that largely at the edges. In an aperture with width smaller than the wavelength, the wave transmitted through the aperture spreads all the way round and behaves like a point source of waves (they spread out below). (a) d < λ (b) d > λ (c) waves moving directly forward (d) slit at an angle θ Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 58 / 74
  • 59. Diffraction of Crystal Scattering and Diffraction A diffracted beam may be defined as a beam composed of a large number of scattered rays mutually reinforcing each other. Figure: Scattering and Diffraction Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 59 / 74
  • 60. Diffraction of Crystal X-Rays X-Rays discovered by Wilhelm Roentgen. The name as X-Rays because in the nature at first was unknown and also Roentgen Rays. The wavelength range 0.1 to 100 A0 It is short wavelength electromagnetic radiation produced by deceleration of high energy electrons or by electronic transition of electrons in the inner orbital of atom. The penetrating power of X-rays depends on energy, there are two types of x-rays Soft X-rays which have less penetrating and low energy. Hard X-rays which have more penetrating and more energy. It is used for determine the structure of crystal by XRD. XRD in crystal discovered by Max Von Laue. (a) Wilhelm Roentgen (b) Max Von Laue Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 60 / 74
  • 61. Diffraction of Crystal X-Ray Diffraction Principle:- XRD based on constructive interference of monochromatic x-rays and crystalline sample. These X-rays are generated by cathode rays tube , filtered to produce monochromatic radiation,collimated to concentrated and directed towards sample. The interaction of incident rays with sample produces constructive interference when condition satisfy the Bragg’s Law. Every Crystalline substance gives pattern, the same substance gives same pattern and in the mixture of substances each produce its pattern independently of the others. XRD as finger print of the substance. It is based on scattering of x-rays by crystals. The atomic planes of crystal cause as incident beam of x-rays to interfere with another as they leave the crystals. This phenomenon is called XRD. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 61 / 74
  • 62. Diffraction of Crystal Why XRD? Measure the average spacing’s between layers or rows of atoms. Determine the orientation of a single crystal or grain. Find the crystal structure of an unknown material. Measure the size, shape and internal stress of small crystalline regions. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 62 / 74
  • 63. Diffraction of Crystal Essential Parts of the Diffractometer X-ray Tube: the source of X Rays Incident-beam optics: condition the X-ray beam before it hits the sample The goniometer: the platform that holds and moves the sample, optics, detector, or tube The sample and sample holder Receiving-side optics: condition the X-ray beam after it has encountered the sample Detector: count the number of X Rays scattered by the sample Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 63 / 74
  • 64. Diffraction of Crystal Basic components and Features of XRD Production Diffraction Detection Interpretation Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 64 / 74
  • 65. Diffraction of Crystal Bragg’s Law Bragg’s Law In 1913, Sir William Henry Bragg and his son Sir Lawrence Bragg develop the relationship to explain why the cleave faces of crystals appears to reflect x-rays beam at certain angle of incidence(θ). The variable d is distance between atomic layers in a crystal and variable λ is wavelength of incident beam n is integer. Although Bragg’s law was used to explain interference pattern of x-rays scattered by crystals, diffraction has been developed to study the structure of all state of matter with any beam. Bragg’s carried out series of experiment and creates Bragg’s Equation nλ = 2dsinθ where , n = 1 for first order reflection λ = wavelength of incident beam θ = angle of incidence d = distance between atomic layers. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 65 / 74
  • 66. Diffraction of Crystal Bragg’s Law Derivation of Bragg’law For Constructive interference ,path difference is integral multiple of wavelength. So, nλ = path difference = AB + BC (from figure) but AB = BC therefore nλ = 2 × AB From right angle ∆ ABZ , Sinθ = AB d i.e. AB = d Sinθ so, nλ = 2dhkl Sinθhkl Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 66 / 74
  • 67. Diffraction of Crystal Bragg’s Law Conditions are required for Diffraction in Bragg’s Law For parallel planes of atoms, with a space dhkl between the planes, constructive interference only occurs when Bragg’s law is satisfied. In diffractometers, the X-ray wavelength λ is fixed. Consequently, a family of planes produces a diffraction peak only at a specific angle θ Additionally, the plane normal must be parallel to the diffraction vector. Plane normal: the direction perpendicular to a plane of atoms Diffraction vector: the vector that bisects the angle between the incident and diffracted beam The space between diffracting planes of atoms determines peak positions. The peak intensity is determined by what atoms are in the diffracting plane. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 67 / 74
  • 68. Diffraction of Crystal Bragg’s Law XRD-Methods Laue photographic method Braggs X-Ray spectrometer Rotating crystal method Powder method Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 68 / 74
  • 69. Diffraction of Crystal Bragg’s Law Braggs X-Ray spectrometer x-rays from x-rays tube S1 and S2 adjustable slit C - crystal Ionization Chamber One plate of ionization chamber is connected to the positive terminal of a H.T Battery , while negative terminal is connected to quadrant electro-meter(measures the strength of ionization current) Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 69 / 74
  • 70. Diffraction of Crystal Bragg’s Law Working of Bragg’s x-ray spectrometer Crystal is mounted such that θ= 0 and ionization chamber is adjusted to receive x-rays. Crystal and ionization chamber are allowed to move in small steps The angle through which the chamber is moved is twice the angle through which the crystal is rotated X-ray spectrum is obtained by plotting a graph between ionization current and the glancing angle θ Peaks are obtained, corresponding to Bragg’s diffraction. Different order glancing angles are obtained with known values of λ and n and from the observed value of θ and d can be measured. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 70 / 74
  • 71. Diffraction of Crystal Bragg’s Law DETERMINATION OF CRYSTAL STRUCTURE BY BRAGG’S LAW X-Rays fall on crystal surface The crystal is rotated and x-rays are made to reflect from various lattice planes The intense reflections are measured by bragg’s spectrometer and the glancing angles for each reflection is recorded Then on applying bragg’s equation ratio of lattice spacing for various groups of planes can be obtained. Ratio’s will be different for different crystals Experimentally observed ratio’s are compared with the calculated ratio’s ,particular structure may be identified. Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 71 / 74
  • 72. Diffraction of Crystal Bragg’s Law XRD Applications Phase Composition of a Sample. Quantitative Phase Analysis: determine the relative amounts of phases in a mixture by referencing the relative peak intensities Unit cell lattice parameters and Bravais lattice symmetry Index peak positions Lattice parameters can vary as a function of, and therefore give you information about, alloying, doping, solid solutions, strains, etc. Crystal Structure Crystallite Size and Micro-strain Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 72 / 74
  • 73. Diffraction of Crystal Bragg’s Law Numericals Calculate the wavelength for the first-order spectrum if the angle of incidence is 30 degrees. 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. What is the distance between the adjacent Miller planes if the first order reflection from X-rays of wavelength 2.29 A0 occurs at 27o 80 Gold crystallizes into an FCC structure. The edge length of the FCC unit cell is 4.07 Calculate a) the closest distance between two gold atoms and b) the density of gold if its atomic weight is 197. X rays of λ = 0.1537 nm from a Cu target are diffracted from the (111) planes of an FCC metal. The Bragg angle is 19.2o , find lattice parameter. X- rays of wavelength = 179 pm produce a reflection at 2 = 47.2o from the 110 planes of BCC lattice. Calculate the edge length of the unit cell. Plane 1 cuts the axes at a / 3, b / 2 and c /4 ,What are the Miller indices of these planes ? The spacing of one set of crystal planes in NaCl is 0.282 nm. A monochromatic beam of X-rays produces a Bragg maximum when its glancing angle with these planes is θ= 70 first order maximum, find the wavelength of the X-rays. What is the minimum possible accelerating voltage V0 that produced the X-rays? Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 73 / 74
  • 74. Diffraction of Crystal Bragg’s Law Thank You Dr. M. B. Matte ⇒ B.SC. 3rd Year(SEM-V) Unit-I (Physics Paper-II) October 16, 2021 74 / 74