1. Unit 1 Fundamentals 1  Atomic Structure?  Crystal Structure?  Types of Crystals?

2. Unit 1 Fundamentals 2  Atomic Structure: Bohrs Model…

3. Unit 1 Fundamentals 3  Atomic Structure: Bohrs Model…

4. Unit 1 Fundamentals 4  Crystal Structure: Atoms self-organize in crystals, most of the time. The crystalline lattice, is a periodic array of the atoms. Ex: metals, diamond and other precious stones, ice, graphite e.t.c. 2D Structures3D Structures

5. Unit 1 Fundamentals 5 Few more 2D Structures:

6. Unit 1 Fundamentals 6  Crystal Structure: (Amorphous) When the solid is not crystalline, it is called amorphous. Ex:glass, amorphous carbon (a-C), amorphous Si, most plastics

7. Unit 1 Fundamentals 7  Crystal Structure:

8. Unit 1 Unit Cell 8  Crystal Structure: Unit Cell The smallest structural unit of a crystal that has all its symmetry and by repetition in three dimensions makes up its full lattice.

9. Unit 1 Unit Cell 9  Crystal Structure: Unit Cell The smallest structural unit of a crystal that has all its symmetry and by repetition in three dimensions makes up its full lattice.

10. Unit 1 Unit Cell 10  Crystal Structure: Unit Cell The smallest structural unit of a crystal that has all its symmetry and by repetition in three dimensions makes up its full lattice.

11. Unit 1 Unit Cell 11  Crystal Structure: Unit Cell The smallest structural unit of a crystal that has all its symmetry and by repetition in three dimensions makes up its full lattice.

12. Unit 1 Unit Cell 12  Important Parameters of a Unit Cell: Unit cell is smallest repeatable entity that can be used to completely represent a crystal structure. It can be considered that a unit cell is the building block of the crystal structure and defines the crystal structure by virtue of its geometry and the atom positions within.

13. Unit 1 Unit Cell 13  Important Parameters of a Unit Cell: The type of atoms and their radii R, Cell dimensions (Lattice spacing a, b and c) in terms of, Angle between the axis α, β, γ.

14. Unit 1 Unit Cell 14  Important Parameters of a Unit Cell: Number of atoms per unit cell (n). For an atom that is shared with m adjacent unit cells, we only count a fraction of the atom, 1/m. n = (1/8 x 8) + 1 = 2

15. Unit 1 Unit Cell 15  Important Parameters of a Unit Cell: CN, the coordination number, which is the number of closest neighbours to which an atom is bonded. CN = 6

16. Unit 1 Unit Cell 16  Important Parameters of a Unit Cell: APF, the atomic packing factor, which is the fraction of the volume of the cell actually occupied by the hard spheres. APF = Sum of atomic volumes/Volume of cell.

17. Unit 1 Crystal Systems 17  Crystal Systems Only 7 crystal systems have been identified. These 7 basic crystal systems are called Primitive lattices. Unit cell of a primitive lattice contains atoms only the corners.

18. Unit 1 Crystal Systems 18  Crystal Systems Only 7 crystal systems have been identified. These 7 basic crystal systems are called Primitive lattices. Unit cell of a primitive lattice contains atoms only the corners. 

19. Unit 1 Unit Cell 19  Crystal Systems Bravais showed that there are 14 possible arrangement of points (atoms) in the space known as Bravai’s lattices.

20. Unit 1 Crystal Systems 20  Crystal Systems Bravais showed that there are 14 possible arrangement of points (atoms) in the space known as Bravai’s lattices.

21. Unit 1 Crystal Systems 21  Crystal Systems Bravais showed that there are 14 possible arrangement of points (atoms) in the space known as Bravai’s lattices.

22. Unit 1 Metallic Crystalline Structures 22  Crystal Structure of Metals Most of the metals crystallize into three forms of crystal systems: 1) Face-Centered Cubic Structure (FCC) 2) Body-Centered Cubic Structure (BCC) 3) Hexagonal Close Paced Structure (HCP)

23. Unit 1 Metallic Crystalline Structures 23  Simple Cubic Cell: Number of atoms (n) = Effective length of unit cell (a) = Co-ordination Number (CN) = Volume of unit cell (Vc) = Volume of all atoms in the unit cell ( Vs) Atomic Packing Factor (Efficiency) of the cell (  ) = Void = 100 - 

24. Unit 1 Metallic Crystalline Structures 24  Simple Cubic Cell: Number of atoms (n) = Effective length of unit cell (a) = 2R Co-ordination Number (CN) = Volume of unit cell (Vc) = Volume of all atoms in the unit cell ( Vs) Atomic Packing Factor (Efficiency) of the cell (  ) = Void = 100 -  a R

25. Unit 1 Metallic Crystalline Structures 25  Simple Cubic Cell: Number of atoms (n) = 1/8 x 8 = 1 Effective length of unit cell (a) = 2R Co-ordination Number (CN) = 6 Volume of unit cell (Vc) = a 3 = (2R) 3 = 8R 3 Volume of all atoms in the unit cell ( Vs) = n x 4/3 πR 3 = 4/3 πR 3 Atomic Packing Factor (Efficiency) of the cell (  )= Vs/Vc = 52.4% Void = 100 -  a R

26. Unit 1 Metallic Crystalline Structures 26  Body Centered Cubic Cell: Number of atoms (n) = Effective length of unit cell (a) = Co-ordination Number (CN) = Volume of unit cell (Vc) = Volume of all atoms in the unit cell ( Vs) Atomic Packing Factor (Efficiency) of the cell (  ) = Void = 100 - 

27. Unit 1 Metallic Crystalline Structures 27  Body Centered Cubic Cell: Number of atoms (n) = Effective length of unit cell (a) = Co-ordination Number (CN) = Volume of unit cell (Vc) = Volume of all atoms in the unit cell ( Vs) Atomic Packing Factor (Efficiency) of the cell (  ) = Void = 100 -  a 4R R a

28. Unit 1 Metallic Crystalline Structures 28  Body Centered Cubic Cell: Number of atoms (n) = (1/8 x 8) + 1 = 2 Effective length of unit cell (a) = 4R/3 ½ Co-ordination Number (CN) = 8 Volume of unit cell (Vc) = a 3 = (4R/3 ½ ) 2 Volume of all atoms in the unit cell ( Vs) = n x 4/3 πR 3 = 8/3 πR 3 Atomic Packing Factor (Efficiency) of the cell (  ) = Vs/Vc = 68 % Void = 100 -  32 % a 4R R a

29. Unit 1 Metallic Crystalline Structures 29  Face Centered Cubic Cell: Number of atoms (n) = Effective length of unit cell (a) = Co-ordination Number (CN) = Volume of unit cell (Vc) = Volume of all atoms in the unit cell ( Vs) Atomic Packing Factor (Efficiency) of the cell (  ) = Void = 100 - 

30. Unit 1 Metallic Crystalline Structures 30  Face Centered Cubic Cell: Number of atoms (n) = Effective length of unit cell (a) = Co-ordination Number (CN) = Volume of unit cell (Vc) = Volume of all atoms in the unit cell ( Vs) = Atomic Packing Factor (Efficiency) of the cell (  ) = Void = 100 - 

31. Unit 1 Metallic Crystalline Structures 31  Face Centered Cubic Cell: Number of atoms (n) = (1/8 x 8) + (1/2 x 6 ) = 4 Effective length of unit cell (a) = 4R/2 ½ Co-ordination Number (CN) = 12 Volume of unit cell (Vc) = a 3 = (4R/2 ½ ) 3 Volume of all atoms in the unit cell ( Vs) = n x 4/3 πR 3 = 16/3 πR 3 Atomic Packing Factor (Efficiency) of the cell (  ) = Vs/Vc = 74 % Void = 100 -  26 %

32. Unit 1 Metallic Crystalline Structures 32  Hexagonal Closed Packed Cell:

33. Unit 1 Metallic Crystalline Structures 33  Hexagonal Closed Packed Cell: Number of atoms (n) = Effective length of unit cell (a) = Co-ordination Number (CN) = Volume of unit cell (Vc) = Volume of all atoms in the unit cell ( Vs) = Atomic Packing Factor (Efficiency) of the cell (  ) = Void = 100 -  R R

34. Unit 1 Metallic Crystalline Structures 34  Hexagonal Closed Packed Cell: Number of atoms (n) = (1/6 x 12) + (1/2 x 2) + 3 = 6 Effective length of unit cell (a) = 2R Co-ordination Number (CN) = Volume of unit cell (Vc) = Volume of all atoms in the unit cell ( Vs) = Atomic Packing Factor (Efficiency) of the cell (  ) = Void = 100 -  R R

35. Unit 1 Metallic Crystalline Structures 35  Hexagonal Closed Packed Cell: Number of atoms (n) = (1/6 x 12) + (1/2 x 2) + 3 = 6 Effective length of unit cell (a) = 2R Co-ordination Number (CN) = Volume of unit cell (Vc) = Volume of all atoms in the unit cell ( Vs) = Atomic Packing Factor (Efficiency) of the cell (  ) = Void = 100 - 

36. Unit 1 Metallic Crystalline Structures 36  Hexagonal Closed Packed Cell: Number of atoms (n) = (1/6 x 12) + (1/2 x 2) + 3 = 6 Effective length of unit cell (a) = 2R Co-ordination Number (CN) = 12 Volume of unit cell (Vc) = Volume of all atoms in the unit cell ( Vs) = Atomic Packing Factor (Efficiency) of the cell (  ) = Void = 100 - 

37. Unit 1 Metallic Crystalline Structures 37  Hexagonal Closed Packed Cell: Number of atoms (n) = (1/6 x 12) + (1/2 x 2) + 3 = 6 Effective length of unit cell (a) = 2R Co-ordination Number (CN) = 12 Volume of unit cell (Vc) = Volume of all atoms in the unit cell ( Vs) = Atomic Packing Factor (Efficiency) of the cell (  ) = Void = 100 - 

38. Unit 1 Metallic Crystalline Structures 38  Hexagonal Closed Packed Cell: Volume of unit cell (Vc) = Base consists of 6 triangles. Area of Base = 6 x ½ x a x L = 3a 2 sin 60 o Volume of HCP cell = Area of base x height = 3a 2 sin 60 o x c ⇒ c/a = 1.633. Volume of HCP cell (Vc) = = 3a 2 sin 60 o x c = 4.2426 a 3 = 4.2426 (2R) 3 a a L

39. Unit 1 Metallic Crystalline Structures 39  Hexagonal Closed Packed Cell: a a L Number of atoms (n) = (1/6 x 12) + (1/2 x 2) + 3 = 6 Effective length of unit cell (a) = 2R Co-ordination Number (CN) = 12 Volume of unit cell (Vc) = 4.2426 (2R) 3 Volume of all atoms in the unit cell ( Vs) = = n x 4/3 πR 3 = 6 x 4/3 πR 3 Atomic Packing Factor (Efficiency) of the cell (  ) = Vs/Vc = 74 % Void = 100 - 

40. Unit 1 40 Metallic Crystalline Structures  Crystal Structure of Metals

41. Unit 1 41 Density Computations  Density of Materials Density = Mass/Volume Mass of Unit Cell = n. A gm NANA Where n = number of atoms effective atoms in unit cell, A = Atomic mass (in gm), N A = Avagadro’s number = 6.023 x 10 23 /mole Volume of Unit Cell = a 3 Where a = Lattice Constant, cm ⇒ Density = ρ = gm a 3. N A n. A ⇒ Lattice constant = a = cm ρ N A n. A 3√3√

42. Unit 1 42  Problems : 1)Iron has BCC structure and the atomic radius is 1.24 Ǻ. Calculate the lattice constant ‘a’ for the cubic structure of the iron unit cell. Atomic radius: R = 1.24 Ǻ. Lattice constant: a = 4R/(3) ½ = 4 x 1.24/ (3) ½ = 2.8637 Ǻ a 4R R a Density Computations

43. Unit 1 43  Problems : 2) Cu has FCC structure and the atomic diameter is 2.556 Ǻ. Calculate the lattice parameter. Atomic radius: R = D/2 = 2.556/2 = 1.278 Ǻ. Lattice constant: a = 4R/(2) ½ = 4 x 1.278/ (2) ½ = 3.6147 Ǻ Density Computations

44. Unit 1 44  Problems : 3)Determine the percentage volume change that occurs when iron changes from BCC structure to FCC cubic structure. For BCC structure, a = 4R/(3) ½ and n = 2 and volume/ atom V = a 3 /n = 4R/(3) ½ x ½ = 6.16 R 3 For FCC structure, a = 4R/(2) ½ and n = 4 and volume/ atom V = a 3 /n = 4R/(2) ½ x ¼ = 5.66 R 3. Density Computations Assuming there is no change in the size of iron atom, % of Δ V = Final Volume – Initial Volume Initial Volume ) ( 100 % of Δ V = 5.66 R 3 – 6.16 R 3 5.66 R 3 )( 100 ⇒ % of Δ V = - 8.1 % ⇒

45. Unit 1 45  Problems : Density Computations 4) Cu has a FCC structure and an atomic radius of 0.1278 nm. The atomic mass of Cu is 63.5 g/mol & Avogadro's No. is 6.023 x 10 23. Calculate the density of Cu. Atomic Radius R = 0.1278 nm = 0.1278 x 10 -7 cm Atomic Mass = 63.5 g/mol. For FCC, n = 4. a = 4R/2 ½ = 4 x 0.1278 x 10 -7 2 ½

46. Unit 1 46  Problems : Density Computations ⇒ 4) Cu has a FCC structure and an atomic radius of 0.1278 nm. The atomic mass of Cu is 63.5 g/mol & Avogadro's No. is 6.023 x 10 23. Calculate the density of Cu. Atomic Radius R = 0.1278 nm = 0.1278 x 10 -7 cm Atomic Mass = 63.5 g/mol. For FCC, n = 4. a = 4R/2 ½ = 4 x 0.1278 x 10 -7 2 ½ Density = ρ = gm a 3. N A n. A Density of Cu = 2 ½ 4 x 0.1278 x 10 -7 6.023 x 10 23 4 x 63.5 ) ( 3 = 8.933 g /cm 3

47. Unit 1 47  Problems : Density Computations 5) The lattice constant of iron (in BCC) is 2.86 Ǻ. What is the density of iron taking the atomic weight of iron as 55.85 amu Lattice Constant: a = 2.86 Ǻ = 2.86 x 10 -8 cm. Density of Iron : For Iron (BCC): n = 2, A = 55.86 amu (g/mol) and N = 6.023 x 10 23 ρ = gm a 3. N A n. A ( 2.86 x 10 -7 ) 3. 6.023 x 10 23 2 x 55.85 ρ = = 7.93 g/cc

48. Unit 1 48 Density Computations  Problems : 6) NaCl crystals have FCC structure. The density of NaCl is 2.18 g/cc. Calculate the distance between two adjacent atoms. Density (ρ) = 2.18 g/cc; and for FCC n = 4. Molecular Weight of NaCl = atomic weight of Na + atomic Weight of Cl = 23 + 35.5 = 58.5 (i.e. atomic mass. A = 58.5 g/mol) Density : ρ = gm a 3. N A n. A 2.18 x. 6.023 x 10 23 4 x 58.5 a = = 5.63 x 10 -8 cm = 5.63 Ǻ ⇒ Lattice constant = a = cm ρ N A 3√3√ n. A 3√3√

49. Unit 1 Polymorphism & Allotropy 49 Some Metals and non-metals may have more than one crystal structure. This phenomenon is called as Polymorphism. If similar behavior prevails in elemental solids, it is called Allotropy. The prevailing crystal structure depends on both temperature & Pressure  Definition : Carbon forms Graphite under ambient conditions Where as, the same carbon forms Diamond under extreme temperature & pressure. (~ 5 GPa @ 1500 oC )

50. Unit 1 Polymorphism & Allotropy 50 Most of the time there will be changes in the density and other physical properties of the materials, with polymorphism. Polymorphism : Pure Iron (Fe) has BCC crystal structure @ room temperature. Whereas, pure Iron (Fe) changes to FCC when heated @ 912 o C