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PH0101 UNIT 4 LECTURE 6

PH0101 UNIT 4 LECTURE 6. RELATION BETWEEN LATTICE CONSTANT AND DENSITY DIAMOND CUBIC STRUCTURE PROBLEMS. RELATION BETWEEN LATTICE CONSTANT AND DENSITY. Consider a cubic crystal of lattice constant ‘a’. Density of the crystal = ρ

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PH0101 UNIT 4 LECTURE 6

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

  2. 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

  3. 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

  4. 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

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

  6. 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

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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

  12. Z 2r a/4 X Y a/4 a/4 W a DIAMOND CUBIC STRUCTURE PH 0101 UNIT 4 LECTURE 6

  13. 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

  14. 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

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

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

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

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

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

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