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## PH0101 UNIT 4 LECTURE 4

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**PH0101 UNIT 4 LECTURE 4**• RECIPROCAL LATTICE • FEATURES OF RECIPROCAL LATTICE • GRAPHICAL REPRESENTATION • GENERAL PROCEDURE • SIMPLE CUBIC STRUCTURE • BODY CENTERED CUBIC STRUCTURE PH 0101 UNIT 4 LECTURE 4**RECIPROCAL LATTICE**• The concept of ‘reciprocal lattice’ provides a device for • tabulating both the slopes and the interplanar • spacings of the planes of a crystal lattice. • In a crystal, there exist many sets of planes with • different orientations and spacing. These planes can • cause diffraction. • If we draw normals to all sets of planes, from a • common origin, then the end points of normals form a • lattice which is called as ‘reciprocal lattice’. PH 0101 UNIT 4 LECTURE 4**FEATURES OF RECIPROCAL LATTICE**• Each point in the reciprocal lattice preserves the • characteristics of the set of planes which it • represents. • Its direction with respect to the origin represents • the orientation of the planes. • Its distance from the origin represents the • interplanar spacing of the planes. PH 0101 UNIT 4 LECTURE 4**GRAPHICAL REPRESENTATION**• Consider all the planes belonging to a single zone. • Since all the planes to be considered are parallel to a • common line, known as zone line, the normals to these • planes lie in the same plane (normal to the zone axis). • In this way, these planes can be represented by a two • dimensional reciprocal lattice. PH 0101 UNIT 4 LECTURE 4**FORMATION OF RECIPROCAL LATTICE**PH 0101 UNIT 4 LECTURE 4**RECIPROCAL LATTICE**• The figure shows the unit cell of a monoclinic • crystal looking along its unique axis designated by C. • The cell edges are ‘a’ and ‘b’. • The example also shows the edge view of four (h k l) • planes, namely (100), (110), (120) and (010). PH 0101 UNIT 4 LECTURE 4**GENERAL PROCEDURE FOR LOCATING THE RECIPROCAL LATTICE**• A lattice point is taken as common origin. • From the common origin, draw a normal to each • plane. • Place a point on the normal to each plane (h k l) at a • distance from the origin equal to . • Such points form a periodic array called reciprocal • lattice. PH 0101 UNIT 4 LECTURE 4**CRYSTAL STRUCTURE OF MATERIALS**• NUMBER OF ATOMS PER UNIT CELL (n) • The total number of atoms present in an unit cell is • known as number of atoms per unit cell. • COORDINATION NUMBER (CN) • It is the number of nearest neighboring atoms to a • particular atom. • ATOMIC RADIUS (r) • It is the radius of an atom. It is also defined as half the • distance between two nearest neighboring atoms in a • crystal. PH 0101 UNIT 4 LECTURE 4**CRYSTAL STRUCTURE OF MATERIALS**• ATOMIC PACKING FACTOR (APF) • It is the ratio of volume occupied by the atoms or molecules • in an unit cell (v) to the total volume of the unit cell (V). • APF = • APF = No. of atoms present in the unit cell x Volume of the atom • Volume of the unit cell PH 0101 UNIT 4 LECTURE 4**SIMPLE CUBIC STRUCTURE**PH 0101 UNIT 4 LECTURE 4**SIMPLE CUBIC STRUCTURE**• Each and every corner atom is shared by eight adjacent • unit cells. The contribution of each and every corner atom • to one unit cell is 1/8. • The total number of atoms present in a unit cell =1/8 x 8 =1. • CORDINATION NUMBER (CN) • For SCC atom, there are four nearest neighbours in its own • plane. • There is another nearest neighbour in a plane which lies • just above this atom and yet another nearest neighbour in • another plane which lies just below this atom. • Therefore the co-ordination number is 6. PH 0101 UNIT 4 LECTURE 4**SIMPLE CUBIC STRUCTURE**• ATOMIC RADIUS (R) • Since the atoms touch along cube edges, the atomic radius for a • simple cubic unit cell is, r = • (where a = 2r, is the lattice constant) • ATOMIC PACKING FACTOR (APF) • APF = • v = 1 4/3 r3 PH 0101 UNIT 4 LECTURE 4**SIMPLE CUBIC STRUCTURE**V = a3 APF = APF = Substituting r = , we get, APF = PH 0101 UNIT 4 LECTURE 4**SIMPLE CUBIC STRUCTURE**• Therefore packing density = /6 = 0.5236 APF = 0.52 • Thus 52 percent of the volume of the simple • cubic unit cell is occupied by atoms and the • remaining 48 percent volume of the unit cell is • vacant or void space. PH 0101 UNIT 4 LECTURE 4**BODY CENTERED CUBIC STRUCTURE**• A body centred cubic structure has eight comer • atoms and one body centred atom. • The atom at the centre touches all the eight • corner atoms. • The BCC structure is shown in the figure. PH 0101 UNIT 4 LECTURE 4**BODY CENTERED CUBIC STRUCTURE**PH 0101 UNIT 4 LECTURE 4**BODY CENTERED CUBIC STRUCTURE**PH 0101 UNIT 4 LECTURE 4**BODY CENTERED CUBIC STRUCTURE**• In BCC unit cell, each and every corner atom is shared by eight • adjacent unit cells. • So, the total number of atoms contributed by the corner atoms is • × 8 = 1. • A BCC unit cell has one full atom at the centre of the unit cell. • The total number of atoms present in a BCC unit cell = 1+1 = 2. PH 0101 UNIT 4 LECTURE 4**BODY CENTERED CUBIC STRUCTURE**• CO-ORDINATION NUMBER (CN) • Let us consider a body centred atom. The nearest neighbour for a • body centred atom is a corner atom. A body centred atom is • surrounded by eight corner atoms. • Therefore, the co-ordination number of a BCC unit cell = 8. • ATOMIC RADIUS (R) • For a BCC unit cell, the atomic radius can be calculated • From figure AB = BC = AD = ‘a’ and CD = 4r. PH 0101 UNIT 4 LECTURE 4**BODY CENTERED CUBIC STRUCTURE**From the triangle, ACD, CD2 = AC2 + AD2 CD2 = 2a2 + a2 (4r)2 = 3a2 16r2 = 3a2 i.e. r2 = PH 0101 UNIT 4 LECTURE 4**BODY CENTERED CUBIC STRUCTURE**• atomic radius r = a • ATOMIC PACKING FACTOR (APF) • APF = • The number of atoms present in an unit cell = 2 • v = 2 x 4/3 r3 • V = a3 PH 0101 UNIT 4 LECTURE 4**BODY CENTERED CUBIC STRUCTURE**APF = Substituting r = we get APF = APF = = 0.68 PH 0101 UNIT 4 LECTURE 4**BODY CENTERED CUBIC STRUCTURE**• Thus 68 percent of the volume of the BCC unit cell is occupied by atoms and the remaining 32 percent volume of the unit cell is vacant or void space. PH 0101 UNIT 4 LECTURE 4**Physics is hopefully simple but Physicists are not**PH 0101 UNIT 4 LECTURE 4