CRYSTAL STRUCTURE
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Transcript CRYSTAL STRUCTURE
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, etc
Amorphous 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 etc
Unit 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
Lattice
n-dimensional, infinite, periodic array of points,
each of which has identical surroundings.
use this as test for lattice points
CsCl structure
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 Parameters
Interfacial 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
Space Lattice
A lattice is also called a Space Lattice
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)
or
Translationally periodic arrangement of points in space is called a 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 Systems
There are Seven Basic Crystal Systems
Cubic
Tetragonal
Orthorhombic
Monoclinic
Triclinic
Rhombohedral (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)
Auguste Bravais
(1811-1863)
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
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
P
Cube
I
F
C
I
P
abc
90
Symmetry of Cubic lattices
4 2
3
m m
Lattice point
F
2
Tetragonal
Square Prism (general height)
P
I
F
C
I
P
Symmetry of Tetragonal lattices
4 2 2
mmm
abc
90
3
Orthorhombic Rectangular Prism (general height)
P
I
F
C
One convention
abc
I
P
Note the position of
‘a’ and ‘b’
abc
90
F
Symmetry of Orthorhombic lattices
C
2 2 2
mmm
Is there a alternate possible set of unit cells for OR?
Why is Orthorhombic called Ortho-’Rhombic’?
P
4
Hexagonal
I
F
C
120 Rhombic Prism
abc
90, 120
A single unit cell (marked in blue)
along with a 3-unit cells forming a
hexagonal prism
Symmetry of Hexagonal lattices
6 2 2
mmm
Note: there is only one type of hexagonal
lattice (the primitive one)
What about the HCP?
(Does it not have an additional atom somewhere in the middle?)
P
5
Rhombohedra
lTrigonal
Parallelepiped (Equilateral, Equiangular)
I
abc
90
Note the position of the origin
and of ‘a’, ‘b’ & ‘c’
Symmetry of Trigonal lattices
3
2
m
F
C
P
6
Monoclinic
Parallogramic Prism
One convention
abc
abc
90
Note the position of
‘a’, ‘b’ & ‘c’
Symmetry of Monoclinic lattices
2
m
I
F
C
P
7
Triclinic
Parallelepiped (general)
abc
Symmetry of Triclinic lattices
1
I
F
C
Arrangement of lattice points in the Unit Cell
& No. of Lattice points / Cell
Position of lattice points
Effective number of Lattice points / cell
P
8 Corners
= [8 (1/8)] = 1
I
8 Corners
+
1 body centre
= [1 (for corners)] + [1 (BC)] = 2
3
F
8 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
1
2
MILLER INDICES
d
DIFFERENT LATTICE PLANES
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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.
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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.
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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.
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MILLER INDICES
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).
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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.
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ILLUSTRATION
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)
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MILLER INDICES
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.
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EXAMPLE
( 1 0 0 ) plane
Plane parallel to Y and Z axes
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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).
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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.
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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.
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MILLER INDICES OF SOME IMPORTANT PLANES
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Spacing between planes in a cubic crystal
d hkl =
a
2
2
h + k +l
2
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.
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
1 1
1
,
and
2 3
4
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
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4 3
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PROBLEMS
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
a
h 2 k 2 l2
4.031 1010
22 12 12
d = 1.6456 Å
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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)
a
d1
a
h12 k12 l12
12 22 12
a
6
and the perpendicular distance between the origin and the plane (2 1 2),
d2
a
h
2
2
k
2
2
l
2
2
a
2
2
1
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2
2
a
a
3
9
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PROBLEMS
The perpendicular distance between the planes (1 2 1) and (2 1 2) are,
d = d1 – d2 =
a
6
(or)
a(3 6)
a 3a 6a
3
3 6
3 6
d = 0.0749 a.
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