PH0101 UNIT 4 LECTURE 6 RELATION BETWEEN LATTICE CONSTANT AND DENSITY DIAMOND CUBIC STRUCTURE PROBLEMS PH 0101 UNIT 4 LECTURE 6

RELATION BETWEEN LATTICE CONSTANT AND DENSITY Consider a cubic crystal of lattice constant ‘a’. Density of the crystal = ρ Volume of the unit cell = V = a3 Number of atoms per unit cell = n Atomic weight of the material = M Avagadro number = N PH 0101 UNIT 4 LECTURE 6

RELATION BETWEEN LATTICE CONSTANT AND DENSITY In M gram of material there are ‘N’ atoms i.e., mass of N atoms is ‘M’ gram. Mass of 1 atom = Mass of ‘n’ molecules i.e., mass of an unit cell = (1) Density ‘ρ’= i.e. ρ = PH 0101 UNIT 4 LECTURE 6

RELATION BETWEEN LATTICE CONSTANT AND DENSITY ρ = mass ‘m’ = ρ a3 (2) Equating (1) and (2), we get, ρ a3 = ρ = PH 0101 UNIT 4 LECTURE 6

RELATION BETWEEN LATTICE CONSTANT AND DENSITY ρ = PH 0101 UNIT 4 LECTURE 6

Edge of the unit cell, a = 4.16Å = 4.16 × 10-10m PROBLEMS Worked Example Sodium crystallises in a cubic lattice. The edge of the unit cell is 4.16 Å. The density of sodium is 975kg/m3 and its atomic weight is 23. What type of unit cell does sodium form ? (Take Avagadro number as 6.023 1026 atoms (Kg mole)-1 Edge of the unit cell, a = 4.16Å = 4.16 × 10-10m Density of the sodium, ρ = 975 kg/m3 Atomic weight of sodium, M=23 PH 0101 UNIT 4 LECTURE 6

Avogadro's number, N = 6.023 × 1026 atoms/kg mole PROBLEMS Avogadro's number, N = 6.023 × 1026 atoms/kg mole Density of the crystal material, Number of atoms in the unit cell, PH 0101 UNIT 4 LECTURE 6

Since the body centred cubic cell contains 2 PROBLEMS = 2 atoms Since the body centred cubic cell contains 2 atoms in it, sodium crystallises in a BCC cell. PH 0101 UNIT 4 LECTURE 6

Edge of the unit cell, a = 2.88Å = 2.88 × 10-10m PROBLEMS Worked Example A metallic element exists in a cubic lattice. Each side of the unit cell is 2.88 Å. The density of the metal is 7.20 gm/cm3. How many unit cells will be there in 100gm of the metal? Edge of the unit cell, a = 2.88Å = 2.88 × 10-10m Density of the metal, ρ = 7.20 gm/cm3 = 7.2 ×103 kg/m3 Volume of the metal = 100gm = 0.1kg Volume of the unit cell = a3 = (2.88 × 10-10)3 = 23.9 × 10-30m3 PH 0101 UNIT 4 LECTURE 6

Volume of 100gm of the metal = = 1.39 × 10-5m3 PROBLEMS Volume of 100gm of the metal = = 1.39 × 10-5m3 Number of unit cells in the volume = = 5.8 × 1023 PH 0101 UNIT 4 LECTURE 6

DIAMOND CUBIC STRUCTURE It is formed by carbon atoms. Every carbon atom is surrounded by four other carbon atoms situated at the corners of regular tetrahedral by the covalent linkages. The diamond cubic structure is a combination of two interpenetrating FCC sub lattices displaced along the body diagonal of the cubic cell by 1/4th length of that diagonal. Thus the origins of two FCC sub lattices lie at (0, 0, 0) and (1/4, 1/4,1/4) PH 0101 UNIT 4 LECTURE 6

DIAMOND CUBIC STRUCTURE Z Y 2r W X PH 0101 UNIT 4 LECTURE 6

DIAMOND CUBIC STRUCTURE The points at 0 and 1/2 are on the FCC lattice, those at 1/4 and 3/4 are on a similar FCC lattice displaced along the body diagonal by one-fourth of its length. In the diamond cubic unit cell, there are eight corner atoms, six face centred atoms and four more atoms. No. of atoms contributed by the corner atoms to an unit cell is 1/8×8 =1. No. of atoms contributed by the face centred atoms to the unit cell is 1/2 × 6 = 3 There are four more atoms inside the structure. PH 0101 UNIT 4 LECTURE 6

DIAMOND CUBIC STRUCTURE No.of atoms present in a diamond cubic unit cell is 1 + 3 + 4 = 8. Since each carbon atom is surrounded by four more carbon atoms, the co-ordination number is 4. ATOMIC RADIUS(R) From the figure,in the triangle WXY, XY2 = XW2 + WY2 = PH 0101 UNIT 4 LECTURE 6

DIAMOND CUBIC STRUCTURE XY2 = Also in the triangle XYZ, XZ2 = XY2 + YZ2 = XZ2 = PH 0101 UNIT 4 LECTURE 6

DIAMOND CUBIC STRUCTURE But XZ = 2r (2r)2 = 4r2 = r2= Atomic radius r = PH 0101 UNIT 4 LECTURE 6

DIAMOND CUBIC STRUCTURE Atomic packing factor (APF) APF = v = i.e. v = APF = PH 0101 UNIT 4 LECTURE 6

DIAMOND CUBIC STRUCTURE APF = i.e. APF = 34% Thus it is a loosely packed structure. PH 0101 UNIT 4 LECTURE 6

Physics is hopefully simple but Physicists are not PH 0101 UNIT 4 LECTURE 6