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8/9/2019 Crystal Structures by Mr.charis israel ancha
1/7
.
PREPARED BY
Mr. A.CHARIS ISRAEL. M.Sc., B.Ed., (Ph.D.)
Asst. Professor of PHYSICS
Mobile No: +91-9866934653
SECUNDERABAD
8/9/2019 Crystal Structures by Mr.charis israel ancha
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UNIT I (PART-B) CRYSTAL STRUCTURES & X-RAY DIFFRACTION 2008-2009
For Downloads visit: www.charis-ancha.blogspot.com +91-9866934653
Comment at: charisisrael@gmail.com
APPLIED PHYSIC
Page 2
Lattice &Lattice Points
Basis
Crystal Structure
Primitive Unit
Unit & Primitive Cells
CRYSTAL STRUCTURES
Basic Definitions:
1. Lattice and Lattice points:A lattice is a regular periodic arrangement ofimaginary points in
space that looks a net-like structure in which the environment about any
particular point is in every way the same as that about any other point.
These imaginary points are called lattice points. A lattice is a
mathematical concept.
OR
A lattice is defined as an infinite array of points in threedimensions in which every point has surroundings identical to that of
every other point in the array.
2. Basis:A basis or pattern is a structural or a building unit associatedidentically with every lattice point. In an ideal crystal, this structural unit
may be an atom or group of atoms or molecules identical in composition,arrangement and orientation.
OR
A group of atoms or molecules identical in composition is called
thebasis.
3. Crystal Structure:A crystal structure is formed by associating identically with
every lattice point a basis. Logically it can be defined as
Lattice + Basis = Crystal Structure4. Unit Cell1:
The Unit Cellis the smallest2
unitwhich, when repeated in space
indefinitely, generates the space lattice. We can suitably compare it with
the building block of a wall.
A unit cell may contain more than one effective lattice point3,
hence it isnot necessary that the unit cell should be a primitive cell . The
cells of this type are called as thenon-primitiveunit cells.
5. Primitive cell:Aprimitive cellis a smallestunit cell which consists ofonly one
effective lattice point.
1To know the properties of a crystal, it is only sufficient to know the properties of a unit cell of that crystal.
2The choice of a conventional unit cell is a matter of convenience. Usually, a cell with shortest possible size is chosen as a convenient unit cell.
3For example, a square obtained by joining four neighboring lattice points represents a unit cell. Since each lattice point is common to four unitcells meeting at that corner, the effective number of lattice points oreffective lattice points in the unit cell is only one ( In 2-D, X 4 = 1. In 3-D,
1/8 X 8 = 1).
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UNIT I (PART-B) CRYSTAL STRUCTURES & X-RAY DIFFRACTION 2008-2009
For Downloads visit: www.charis-ancha.blogspot.com +91-9866934653
Comment at: charisisrael@gmail.com
APPLIED PHYSIC
Page 3
6. Crystallographic axes:The lines drawn parallel to the lines of
intersection of any three faces of the unit cell which donot lie in the same plane are calledcrystallographic axes .
7. Basic Lattice Parameters:(i). Primitives:
The intercepts a, b and c shown in the
adjacent figure, which define thedimensions of the Unitcell are known as itsprimitives.
(ii). Interfacial angles:
Theangles, and between the three
crystallographic axes, which define theform or shape of
the Unit cell are known as interfacial angles.
8. Bravais Lattices:There are only fourteen distinguishable ways of arranging lattice points in three dimensional
spaces. These 14 space lattices are known asBravais Lattices .
9. Nearest neighbour distance (2r):The distance between the centers of two nearest neighbouring atoms is callednearest neighbour
distance. Ifr is the radius of the atom, nearest neighbour distance is2r if the atoms are in touch with each
other.
10.Atomic radius (r):It is defined ashalfthe distance between the nearest neighbor atoms in the crystal.
11.Coordination number (Number of nearest neighbours):It is defined as the number of equidistant nearest neighbours that an atom has in a given
structure.For example, simple cubic 6, face centered cubic 12, Body centered cubic 8 etc.
12.Effective number of atoms:It is the sum of the fractional part of all the atoms contributing to the particular unit cell.
For example, simple cubic (1/8 X 8 corner atoms = 1) 1 atom, face centered cubic ((1/8 X 8 corner) +
(1/2 X 6 face centered) = 4) 4 atoms, Body centered cubic ((1/8 X 8 corner) + 1 body centered = 2)
2 atoms etc.
13.Packing fraction or Atomic packing fraction:It is the ratio of the volume occupied by the effective atoms in an unit cell (v) to thetotalvolume
of the Unit cell (V).
i.e., vPacking FractionV
=
14. Miller Indices:
Miller suggested a method of indicating the orientation of a plane by reducing the reciprocal ofthe intercepts into smallest whole numbers as h, l, k. These numbers represented in the form of(h,l,k) are
called asMiller Indices.
c
z
y
x
ab
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UNIT I (PART-B) CRYSTAL STRUCTURES & X-RAY DIFFRACTION 2008-2009
For Downloads visit: www.charis-ancha.blogspot.com +91-9866934653
Comment at: charisisrael@gmail.com
APPLIED PHYSIC
Page 4
Monoclinic, P & CTriclinic, P
Cubic, F, P & I
Tetragonal, P & I
Orthorhombic, P, C, I & F
Hexagonal, P Trigonal, P
7 CRYSTAL SYSTEMS & 14 BRAVAIS LATTICES:
The Fourteen Bravais lattices illustrated by conventional unit cells which are not always primitive
cells.
S.No. Crystal Type
Relation between
Primitives
a, b & c
Relation Between
Interfacial angles
, &
Bravais Lattices with symbols
Primitive-P, Body centered-I,
Base centered-C, Face
centered-F1. Monoclinic a b c 90o = = P, C
2. Triclinic a b c 90o P
3. Orthorhombic a b c 90o = = = P, I, F, C
4. Cubic a b c= = 90o = = = P, I, F
5. Trigonal a b c= = 90o = = P
6. Tetragonal a b c= 90o = = = P, I
7. Hexagonal a b c= 90 , 120o o
= = =
P
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UNIT I (PART-B) CRYSTAL STRUCTURES & X-RAY DIFFRACTION 2008-2009
For Downloads visit: www.charis-ancha.blogspot.com +91-9866934653
Comment at: charisisrael@gmail.com
APPLIED PHYSIC
Page 5
A A A AA A A A
B B BB B B B
B B B BA
A
AA
BB
BB
STRUCTURE AND PACKING FRACTIONS OR STACKING SEQUENCE:
(i) Simple Cubic Structure (SC):Stacking sequence: Let us consider a single layer A of atoms arranged as shown in the Fig (i),
in which all the atoms are in touch with each other. A second layer B is just above the layer A.
Likewise all the layers can be stacked one above the other resulting in simple cubic structure such asshown in Fig (ii). The unit cell of the simple cubic structure is shown in Fig (iii).
In simple cubic lattice, there is only one lattice point at each of the eight corners of the unit cell
and the atoms touch along the unit cell edges.
Fig. (i) Fig. (ii) Fig. (iii)
To find the packing factor in simple cubic structure:
' '.
' '
volume of the vPacking Factor
total volume of the unit cell V
effective atoms=
Step 1: In simple cubic structure, 8 atoms occupy 8 corners and the contribution of each atom to
the unit cell is 1/8. Hence the effective number of atoms in simple cubic structure is 1/8 X 8 = 1atom.
Step 2: Therefore volume of the effective atoms, 34 ( )3
v r number of effective atoms=
341
3v r=
34
3v r= where r is the radius of the atom.
Step3: In simple cubic structure, the atoms touch each other along the cube edge; hence lattice
constanta isequal to 2r.Thenearest neighbor distance is also 2r, hence thenumber of nearestneighbors is 6 (i.e.,coordination number in simple cubic structure is 6) .
Step 4: Therefore the total volume of the unit cell, V= a3= (2r)
3= 8r
3.
Step 5:
3
3
2
4
' ' 3 0.52 52%.' ' 8 6
rvolume of the effective atoms v
Packing Factortotal volume of the unit cell V r
/
/= = = = =
Hence52% of the total volume of the unit cell is occupied by the atoms and the rest 48% of the
unit cell is vacant.
8/9/2019 Crystal Structures by Mr.charis israel ancha
6/7
8/9/2019 Crystal Structures by Mr.charis israel ancha
7/7
UNIT I (PART-B) CRYSTAL STRUCTURES & X-RAY DIFFRACTION 2008-2009
For Downloads visit: www.charis-ancha.blogspot.com +91-9866934653
Comment at: charisisrael@gmail.com
APPLIED PHYSIC
Page 7
' '.
' '
volume of the vPacking Factor
total volume of the unit cell V
effective atoms=
A A A A
A
AB B B
B B B
AB
C
C
CC
C
A
3
3 3
16
' ' 3 0.74 74%.' ' (2 2) 3 2
rvolume of the effective atoms v
Packing Factortotal volume of the unit cell V r
/
/= = = = =
(iii) Face Centered Cubic structure (FCC):Stacking sequence: Let us consider the arrangement of atoms in a single closest-packed layer A
by placing each atom in contact with six others as shown in the Fig (i). A second similar layer B may be
added by placing each atom ofB in contact with three atoms of the layer A. A third layer C is added
over the holes in the layer A that are not occupied by B. Likewise the layers ABCcan be stacked one
above the other, resulting the sequence ABCABCABC and the structure is called Face-centered cubicstructure.
In Face-centered cubic lattice, there is a lattice point at each of the eight corners of the unit cell
and six lattice points on the six face centers. The atoms touch each other along the face diagonals of the
unit cell.
Fig.1 Fig.2 Fig.3
To find the packing factor in Face- Centered Cubic structure:
Step 1: In face-centered cubic structure, 8 atoms occupy 8 corners and 6 atoms at the 6 face
centers and hence the contribution of each corner atom is 1/8 and due to each face centered atomis 1/2. Hence thenumber of effective atoms are (1/8 X 8) + (1/2 X 6) = 4 atoms.
Step 2: Therefore volume of the effective atoms,34 ( )
3
v r number of effective atoms=
34 43
v r= fi 316
3
v r= where r is the radius of the atom.
Step3: In face-centered cubic structure, the atoms touch each other along the face diagonal of the
unit cell; hence length of the face diagonal in terms of the radius of the atom is 4r and in terms
of the lattice constant is 2 a.
42 4 ( ) 2 2 ( )
2 2 2
aa r a r or a r or r = = = =
Each corner atom is in touch with 12 face-centered atoms, hence the number of nearest
neighbours is 12 (i.e.,coordination number in face-centered cubic structure is 12).
Step 4: Therefore the total volume of the unit cell,
( )
33 2 2V a r= = .
Step 5:
Hence 74% of the total volume of the FCC unit cell is occupied by the atoms and the rest26% of
the unit cell is vacant.
TO BE CONTINUED..
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