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Chapter 5 Structure of Solids PowerPoint Presentation
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Chapter 5 Structure of Solids

Chapter 5 Structure of Solids

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Chapter 5 Structure of Solids

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  1. Chapter 5 Structure of Solids 6 Lectures

  2. Solids Crystalline Noncrystalline Long-range periodicity No long-range periodicity Gives sharp diffraction patterns Does not give sharp diffraction patterns Does not have a sharp meliing point Has sharp melting point Has a lower density Has higher density

  3. Factors promoting the formation of noncrystalline structures • Primary bonds do not extend in all three directions and the secondary bonds are not strong enough. • The difference in the free energy of the crystalline and non crystalline phases is small. • The rate of cooling from the liquid state is higher than a critical cooling rate. Metallic Glass: 106 K s-1

  4. Inorganic Solids Covalent Solids Metals and Alloys Ionic Solids Silica: crystalline and amorphous Polymers Classification Structure Crystallinity Mechanical Behaviour

  5. Inorganic Solids Covalent Solids Metals and Alloys Ionic Solids Silica: crystalline and amorphous Polymers Classification Structure Crystallinity Mechanical Behaviour

  6. Metals and Alloys 1. Metallic bond: Nondrectional (Fact) As many bonds as geometrically possible (to lower the energy)  Close packing 2. Atoms as hard sphere (Assumption) 3. Elements (identical atoms) 1, 2 & 3  Elemental metal crystals: close packing of equal hard spheres

  7. Close packing of equal hard spheres Arrangement of equal nonoverlapping spheres to fill space as densely as possible Sphere packing problem:What is the densest packing of spheres in 3D? Kepler’s conjecture, 1611 Kissing Number Problem What is the maximum number of spheres that can touch a given sphere? Coding Theory Internet data transmission

  8. Lecture 9 13.08.2013

  9. We are currently preparing students for jobs that do not yet exist, using technologies that haven’t been invented in order to solve problems that we don’t even know are problems yet. http://www.youtube.com/watch?v=XVQ1ULfQawk

  10. Close packing of equal hard spheres 1-D packing A chain of spheres 2 P.E.= Kissing Number= =1 Close-packed direction of atoms

  11. Close packing of equal hard spheres 2-D packing A hexagonal layer of atoms Close-packed plane of atoms Close-packed directions? 3 Kissing Number=6 P.E.= 1940 L. Fejes Toth : Densest packing of circles in plane

  12. Close packing of equal hard spheres 3-D packing First layer A A A Second layer B A A C C C B B B A A Third layer A or C A A C C C B B B A A A A C C C B B B A A A A Close packed crystals: …ABABAB… Hexagonal close packed (HCP)…ABCABC… Cubic close packed (CCP)

  13. Geometrical properties of ABAB stacking B b = a A A A A =120 C C C a A B B B A A A A c C C C B B B B A A A A A C C C B B B A A A A A and B do not have identical neighbours Either A or B as lattice points, not both Unit cell: a rhombus based prism with a=bc; ==90, =120 The unit cell contains only one lattice point (simple) but two atoms (motif) ABAB stacking = HCP crystal = Hexagonal P lattice + 2 atom motif 000 2/3 1/3 1/2

  14. Hexagonal close-packed (HCP) crystal z ½ ½ y ½ ½ Corner and inside atoms do not have the same neighbourhood x Lattice: Simple hexagonal Motif: Two atoms: 000; 2/3 1/3 1/2 hcp lattice hcp crystal

  15. c/a ratio of an ideal HCP crystal B A A A A C C C A B B B A A A A c C C C B B B B A A A A A C C C B B B A A A A A single B atom sitting on a base of three A atoms forms a regular tetrahedron with edge length a = 2R The same B atom also forms an inverted tetrahedron with three A atoms sitting above it c = 2  height of a tetrahedron of edge length a

  16. Lec 10: Structure of metals and alloys (Close Packing) continued 16.08.2013 Rescheduled Lecture for the missed lecture on Wednesday: Today 7-8 pm VLT1 Office hours for discussions: 5-6 pm on tuesdays, wednesdays and thursdays Doubt clearning class on request

  17. Geometrical properties of ABCABC stacking A A A A A C C C B B B A A A A A C C C B B B A A A A A C C C B B B A A A A A A A

  18. 3 a Geometrical properties of ABCABC stacking C B All atoms are equivalent and their centres form a lattice A C B A Motif: single atom 000 What is the Bravais lattice? ABCABC stacking = CCP crystal = FCC lattice + single atom motif 000

  19. Body diagonal C Close packed planes in the FCC unit cell of cubic close packed crystal B A B A Close packed planes: {1 1 1}

  20. Stacking sequence?

  21. Stacking sequence?

  22. Find the mistake in the following figure from a website: http://www.tiem.utk.edu/~gross/bioed/webmodules/spherefig1.gif

  23. Table 5.1 Coordination Number and Packing Efficiency CW HW Crystal Coordination Packing Structure number efficiency Diamond cubic (DC) 4 Simple cubic (SC) 6 Body-centredcubic 8 Face-centred cubic 12 0.32 0.52 0.68 0.74 Empty spaces are distributed in various voids

  24. End of lecture 10 (16.08.2013) Beginning of lecture 11

  25. Voids in Close-Packed Crystals TETRAHEDRAL VOID OCTAHEDRAL VOID B A A A B C A A B B A A A No. of atoms defining 4 6 the void No. of voids per atom 2 1 Edge length of void 2 R 2 R Size of the void 0.225 R 0.414 R A Experiment 2 HW

  26. Locate of Voids in CCP Unit cell

  27. Solid Solution A single crystalline phase consisting of two or more elements is called a solid solution. Substitutional Solid solution of Cu and Zn (FCC) Interstitial solid solution of C in Fe (BCC)

  28. Hume-Rothery Rules for Extensive Solid Solution (Unlimited solubility) Interstitial solid solution Substitutional solid solution • Structure factor Crystal structure of the two elements should be the same • Size factor: Size of the two elements should not differ by more than 15% 3. Electronegativity factor: Electronegativity difference between the elements should be small 4. Valency factor: Valency of the two elements should be the same

  29. TABLE 5.2 System Crystal Radius of Valency Electro- structure atoms, Ǻ negativity Ag-Cu Ag FCC 1.44 1 1.9 Au FCC 1.44 1 1.9 Cu-Ni Cu FCC 1.28 1 1.9 Ni FCC 1.25 2 1.8 Ge-Si Ge DC 1.22 4 1.8 Si DC 1.18 4 1.8 All three systems exhibit complete solid solubility.

  30. BRASS Cu + Zn FCC HCP Unfavourable structure factor Limited Solubility: Max solubility of Cu in Zn: 1 wt% Cu Max Solubility of Zn in Cu: 35 wt% Zn

  31. Ordered and RandomSubstitutional solid solution Random Solid Solution Ordered Solid Solution

  32. Ordered and random substitutional solid solution β-Brass: (50 at% Zn, 50 at% Cu) Above 470˚C Disordered solid solution of β-Brass: Corner and centre both have 50% proibability of being occupied by Cu or Zn34 470˚C Below 470˚C Ordered solid solution of β-Brass: Corners are always occupied by Cu, centres always by Zn

  33. Intermediate Structures FCC Crystal structure of Cu: Crystal structure of Zn: HCP Crystal structure of random β-brass: BCC Such phases that have a crystal structure different from either of the two components are called INTERMEDIATE STRUCURES If an intermediate structure occurs only at a fixed composition it is called an INTERMETALLIC COMPOUND, e.g. Fe3C in steels.

  34. End of lecture 10 (16.08.2013) Beginning of lecture 11

  35. 4th. Group: Carbon

  36. Allotropes of C Graphite Diamond Buckminster Fullerene1985 Graphene2004 Carbon Nanotubes1991

  37. Graphite Sp2 hybridization  3 covalent bonds  Hexagonal sheets a = 2 d cos 30° = √3 d y x d = 1.42 Å a = 2.46 Å =120 b=a a

  38. Graphite a = 2.46 Å c = 6.70 Å Lattice: Simple Hexagonal Motif: 4 carbon atoms 000; 2/3 1/3 0; 2/3 1/3 1/2; 1/3 2/3 1/2 A c B y x A www.scifun.ed.ac.uk/

  39. Graphite Highly Anisotropic: Properties are very different in the a and c directions Uses: Solid lubricant Pencils (clay + graphite, hardness depends on fraction of clay) carbon fibre www.sciencemuseum.org.uk/

  40. Diamond Sp3 hybridization  4 covalent bonds  Tetrahedral bonding Location of atoms: 8 Corners 6 face centres 4 one on each of the 4 body diagonals

  41. Diamond Cubic Crystal: Lattice & motif? y 0,1 0,1 R M D C y M R L S N P Q Q T D 0,1 L S K C N K T A x B B A P x 0,1 0,1 Projection of the unit cell on the bottom face of the cube Diamond Cubic Crystal = FCC lattice + motif: 000; ¼¼¼

  42. Diamond Cubic Crystal Structure FCCLattice 2 atomMotif = + Crystal Structure = Lattice + Motif There are only three Bravais Lattices: SC, BCC, FCC. Diamond Cubic Lattice

  43. There is no diamond cubic lattice.

  44. Diamond Cubic Structure Coordination number 4 Face Inside Corners Effective number of atoms in the unit cell = Relaton between lattice parameter and atomic radius Packing efficiency

  45. Diamond Cubic Crystal Structures C Si Ge Gray Sn a (Å) 3.57 5.43 5.65 6.46

  46. Equiatomic binary AB compounds having diamond cubic like structure y 0,1 0,1 IV-IV compound: SiC III-V compound: AlP, AlAs, AlSb, GaP, GaAs, GaSb, InP, InAs, InSb II-VI compound: ZnO, ZnS, CdS, CdSe, CdTe I-VII compound: CuCl, AgI 0,1 S 0,1 0,1

  47. USES: Diamond Abrasive in polishing and grinding wire drawing dies Si, Ge, compounds: semiconducting devices SiC abrasives, heating elements of furnaces

  48. End of lecture 11 (16.08.2013) (Evening class, a postponed class for the missed class on Wednesday 14.08.2013) Beginning of lecture 12