Crystalline Solids :-In Crystalline Solids the atoms are arranged in some regular periodic geometrical pattern in three dimensions- long range order
Eg :- NaCl, CuSo4 ,CsCl, ZnS, etcAmorphous Solids :- In an amorphous solids the
atoms are not arranged in regular periodic geometrical pattern –short range order
Eg :- Boran trioxide (B2O3),Lamp Soot, glass etcUnit Cell :- A Unit cell is the volume of a solid
from which the entire crystal can be constructed by a translation repetition in three directions in sp
Crystal lattice :- An array of points in the space (2D,3D) in which every point has the same environment with respect
to all other points is called space lattice
Basis :- Atoms or molecules attached to each
lattice point in the crystal system is called basis.
Lattice+ Basis = Crystal
CsCl structure
Lattice
n-dimensional, infinite, periodic array of points,
each of which has identical surroundings.
Lattice
n-dimensional, infinite, periodic array of points,
each of which has identical surroundings.
use this as test for lattice points
lattice points
Crystallographic axes
Crystallographic axes:- The lines drawn parallel to the line of intersection of any three faces of the unit Cell which do not lie in the same plane are called Crystallographic axes X,Yand Z
Lattice ParametersInterfacial angles :- The angles between
three Crystallographic axes represented by
α,β,Ƴ are called interfacial angles * Primitives :- The intercepts a,b and c,
which define the dimensions of the unit cell on the respective Crystallographic axes are called as primitives of the Unit Cell
An array of points such that every point has identical surroundings
In Euclidean space infinite array
We can have 1D, 2D or 3D arrays (lattices)
Space Lattice
Translationally periodic arrangement of points in space is called a lattice
or
A lattice is also called a Space Lattice
Note: points are drawn with finite size for clarity in reality they are 0D (zero dimensional)
Crystals are grouped into seven crystal systems, according to characteristic symmetry of their unit cell.
The characteristic symmetry of a crystal is a combination of one or more rotations and inversions.
Crystal SystemsThere are Seven Basic Crystal SystemsCubicTetragonalOrthorhombicMonoclinicTriclinicRhombohedral (Trigonal)Hexagonal
Shape of UC Used as UC for crystal: Lattice Parameters
Cube Cubic (a = b = c, = = = 90)
Square Prism Tetragonal (a = b c, = = = 90)
Rectangular Prism Orthorhombic (a b c, = = = 90)
Parallelogram Prism Monoclinic (a b c, = = 90 )
Parallelepiped (general) Triclinic (a b c, )
Parallelepiped (Equilateral, Equiangular)
Rhombohedral (Trigonal)
(a = b = c, = = 90)
120 Rhombic Prism Hexagonal (a = b c, = = 90, = 120)
Latti
In 1848, Auguste Bravais demonstrated that in a 3-dimensional system there are fourteen possible lattices
A Bravais lattice is an infinite array of discrete points with identical environment
seven crystal systems + four lattice centering types = 14 Bravais lattices
Lattices are characterized by translation symmetry
Auguste Bravais (1811-1863)
Bravais showed that there are only 14 independent ways of arranging points in space so that the environment looks the same from each point. These lattices are called Bravais lattices
Cubic space lattices
Summary: Fourteen Bravais Lattices in Three Dimensions
Fourteen Bravais Lattices …
1 Cubic Cube P I F C
Lattice point
PPII
FF
a b c 90
4 23
m m
Symmetry of Cubic latticesSymmetry of Cubic lattices
P I F C
2 Tetragonal Square Prism (general height)
II
PP
a b c
90
Symmetry of Tetragonal latticesSymmetry of Tetragonal lattices
4 2 2
m m m
P I F C
3 Orthorhombic Rectangular Prism (general height)
PPII
FFCC
a b c
90
Symmetry of Orthorhombic latticesSymmetry of Orthorhombic lattices
2 2 2
m m m
Note the position of ‘a’ and ‘b’
a b c One convention
Why is Orthorhombic called Ortho-’Rhombic’?Why is Orthorhombic called Ortho-’Rhombic’?Is there a alternate possible set of unit cells for OR?Is there a alternate possible set of unit cells for OR?
P I F C
4 Hexagonal 120 Rhombic Prism
What about the HCP?(Does it not have an additional atom somewhere in the middle?)
What about the HCP?(Does it not have an additional atom somewhere in the middle?)
A single unit cell (marked in blue) along with a 3-unit cells forming a
hexagonal prism
Note: there is only one type of hexagonal lattice (the primitive one)
a b c
90 , 120
Symmetry of Hexagonal latticesSymmetry of Hexagonal lattices
6 2 2
m m m
P I F C
5 Trigonal Parallelepiped (Equilateral, Equiangular)
90
a b c
Symmetry of Trigonal latticesSymmetry of Trigonal lattices
Rhombohedral
23
m
Note the position of the origin and of ‘a’, ‘b’ & ‘c’
P I F C
6 Monoclinic Parallogramic Prism
90
a b c
Symmetry of Monoclinic latticesSymmetry of Monoclinic lattices
2
m
a b c
Note the position of ‘a’, ‘b’ & ‘c’
One convention
P I F C
7 Triclinic Parallelepiped (general)
a b c
Symmetry of Triclinic latticesSymmetry of Triclinic lattices
1
Arrangement of lattice points in the Unit Cell
& No. of Lattice points / Cell
Position of lattice points Effective number of Lattice points / cell
1 P 8 Corners = [8 (1/8)] = 1
2 I8 Corners + 1 body centre
= [1 (for corners)] + [1 (BC)] = 2
3 F8 Corners +
6 face centres= [1 (for corners)] + [6 (1/2)] = 4
4
A/
B/
C
8 corners +2 centres of opposite faces
= [1 (for corners)] + [2 (1/2)] = 2
PH 0101 UNIT 4 LECTURE 2 29
MILLER INDICES
d
DIFFERENT LATTICE PLANES
PH 0101 UNIT 4 LECTURE 2 30
MILLER INDICES
The orientation of planes or faces in a crystal can be
described in terms of their intercepts on the three
axes.
Miller introduced a system to designate a plane in a
crystal.
He introduced a set of three numbers to specify a
plane in a crystal.
This set of three numbers is known as ‘Miller Indices’
of the concerned plane.
PH 0101 UNIT 4 LECTURE 2 31
MILLER INDICES
The orientation of planes or faces in a crystal can be
described in terms of their intercepts on the three
axes.
Miller introduced a system to designate a plane in a
crystal.
He introduced a set of three numbers to specify a
plane in a crystal.
This set of three numbers is known as ‘Miller Indices’
of the concerned plane.
PH 0101 UNIT 4 LECTURE 2 32
MILLER INDICES
Procedure for finding Miller Indices
Step 1: Determine the intercepts of the plane along the axes X,Y and Z in terms of the lattice constants a,b and c.
Step 2: Determine the reciprocals of these numbers.
PH 0101 UNIT 4 LECTURE 2 33
Step 3: Find the least common denominator (lcd) and multiply each by this lcd.
Step 4:The result is written in paranthesis.This is called the `Miller Indices’ of the plane in the form (h k l).
This is called the `Miller Indices’ of the plane in the form (h k l).
MILLER INDICES
PH 0101 UNIT 4 LECTURE 2 34
ILLUSTRATION
PLANES IN A CRYSTAL
Plane ABC has intercepts of 2 units along X-axis, 3
units along Y-axis and 2 units along Z-axis.
PH 0101 UNIT 4 LECTURE 2 35
DETERMINATION OF ‘MILLER INDICES’
Step 1:The intercepts are 2,3 and 2 on the three axes.
Step 2:The reciprocals are 1/2, 1/3 and 1/2.
Step 3:The least common denominator is ‘6’. Multiplying each reciprocal by lcd, we get, 3,2 and 3.
Step 4:Hence Miller indices for the plane ABC is (3 2 3)
ILLUSTRATION
PH 0101 UNIT 4 LECTURE 2 36
IMPORTANT FEATURES OF MILLER INDICES
For the cubic crystal especially, the important features of Miller indices are,
A plane which is parallel to any one of the co-ordinate axes has an intercept of infinity (). Therefore the Miller index for that axis is zero; i.e. for an intercept at infinity, the corresponding index is zero.
MILLER INDICES
PH 0101 UNIT 4 LECTURE 2 37
EXAMPLE
( 1 0 0 ) plane
Plane parallel to Y and Z axes
PH 0101 UNIT 4 LECTURE 2 38
EXAMPLE
In the above plane, the intercept along X axis is 1 unit.
The plane is parallel to Y and Z axes. So, the intercepts along Y and Z axes are ‘’.
Now the intercepts are 1, and .
The reciprocals of the intercepts are = 1/1, 1/ and 1/.
Therefore the Miller indices for the above plane is (1 0 0).
PH 0101 UNIT 4 LECTURE 2 39
MILLER INDICES
IMPORTANT FEATURES OF MILLER INDICES
A plane passing through the origin is defined in terms of a parallel plane having non zero intercepts.
All equally spaced parallel planes have same ‘Miller indices’ i.e. The Miller indices do not only define a particular plane but also a set of parallel planes. Thus the planes whose intercepts are 1, 1,1; 2,2,2; -3,-3,-3 etc., are all represented by the same set of Miller indices.
PH 0101 UNIT 4 LECTURE 2 40
MILLER INDICES
IMPORTANT FEATURES OF MILLER INDICES
It is only the ratio of the indices which is important in this notation. The (6 2 2) planes are the same as (3 1 1) planes.
If a plane cuts an axis on the negative side of the origin, corresponding index is negative. It is represented by a bar, like (1 0 0). i.e. Miller indices (1 0 0) indicates that the plane has an intercept in the –ve X –axis.
PH 0101 UNIT 4 LECTURE 2 41
MILLER INDICES OF SOME IMPORTANT PLANES
l + k + h
a = d
222hkl
Spacing between planes in a cubic crystal
where dhkl = inter-planar spacing between planes with Miller indices h,k,and l.a = lattice constant (edge of the cube)h, k, l = Miller indices of cubic planes being considered.
PH 0101 UNIT 4 LECTURE 2 43
PROBLEMS
Worked Example:Calculate the miller indices for the plane with
intercepts 2a, - 3b and 4c the along the crystallographic axes.
The intercepts are 2, - 3 and 4
Step 1: The intercepts are 2, -3 and 4 along the 3 axes
Step 2: The reciprocals are
Step 3: The least common denominator is 12.
Multiplying each reciprocal by lcd, we get 6 -4 and 3
Step 4: Hence the Miller indices for the plane is
1 1 1, and
2 3 4
6 4 3
PH 0101 UNIT 4 LECTURE 2 44
Worked Example
The lattice constant for a unit cell of aluminum is 4.031Å Calculate the interplanar space of (2 1 1) plane.
a = 4.031 Å(h k l) = (2 1 1)Interplanar spacing
d = 1.6456 Å
PROBLEMS
10
2 2 2 2 2 2
4.031 10ad
h k l 2 1 1
PH 0101 UNIT 4 LECTURE 2 45
PROBLEMS
Worked Example:Find the perpendicular distance between the two planes
indicated by the Miller indices (1 2 1) and (2 1 2) in a unit cell of a cubic lattice with a lattice constant parameter ‘a’.We know the perpendicular distance between the origin and the plane is (1 2 1)
and the perpendicular distance between the origin and the plane (2 1 2),
1 2 2 2 2 2 21 1 1
a a ad
6h k l 1 2 1
2 2 2 2 2 2 22 2 2
a a a ad
39h k l 2 1 2
PH 0101 UNIT 4 LECTURE 2 46
PROBLEMS
The perpendicular distance between the planes (1 2 1) and (2 1 2) are,
d = d1 – d2 =
(or) d = 0.0749 a.
3a 6a a(3 6)a a
36 3 6 3 6
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