Chapter 3 Structure of Cystalline Metallic Solids

1、Chapter 3 The Structure of Cystalline Metallic SolidsBraggs Law was derived by physicist Sir William Lawrence Bragg in 1912 and first presented on 11 November 1912 to the Cambridge Philosophical Society. Although simple, Braggs law confirmed the existence of real particles at the atomic scale, as we

2、ll as providing a powerful new tool for studying crystals in the form of X-ray and neutron diffraction. William Lawrence Bragg and his father, Sir William Henry Bragg, were awarded the Nobel Prize in physics in 1915 for their work in determining crystal structures beginning with NaCl, ZnS, and diamo

3、nd. They are the only father-son team to jointly win. W. L. Bragg was 25 years old, making him the youngest Nobel laureate.Sir William Henry Bragg (left) and Sir William Lawrence Bragg (right). Winners of 1915 Nobel Prize in Physics for their services in the analysis of crystal structure by means of

4、 X-ray.n = 2d sin Important terms and conceptsAllotropyAmorphousAnisotropyAtomic packing factor (APF)Body-cnetered cubic (BCC)Braggs lawCoordination numberCrystal structureCrystal systemCrystallineDiffractionFace-centered cubic (FCC)GrainGrain booundaryIsotropicHexagonal close packed (HCP)LatticeLat

5、tice parameterMiller indicesNoncrystallinePolycrystallinePolymorphismSingel crystal Unit cellWarming up . Elastic Scattering takes place when X-rays strik on atoms in lattice. These waves have the same frequency.Encountering of two propagating waves will result in constructive or destructive interfe

6、rence due to phase shifting.In the lattice with plane space of d, these waves with the wavelengh of add constructively in a few specific directions of incident angle , determined by Braggs law:n = 2d sinPage 33Learning ObjectivesAfter careful study of this chapter you should be able to do the follow

7、ing:1. Describe the difference in atomic/molecular structure between crystalline and noncrystalline materails .2. Draw unit cells for face-centered cubic, body-centered cubic, and hexagonal close-packed crystal structures.3. Derive the relationships between unit cell edge length and atomic radius fo

8、r the three crystal structures.4. Compute the densities for metals having faced-centered cubic and body-centered cubic crystal structures given their unit cell dimensions.5. Give three direction index integers, sketch the direction corresponding to these indices within a unit cell.6. Specify the Mil

9、ler indices for a plane that has been drawn within a unit cell.7. Describe how face-centered cubic and hexagonal close-packed stacking of close-packed planes of atoms.8. Distinguish between single crystals and polycrystalline materials.9. Define isotropy and anisotroy with respect to material proper

10、ties.Page 443.1Fundamental Concepts.Crystalline solidAtoms in crystalline solids are stacked together in a regular, repeating pattern over large atomic distances; this is, long-range order exists. Most metals and ceramics are crystalline solids as well as in some polymers.Noncrystalline (Amorphous)

11、solidAmorphous solid lacks long-range order, may has short-range order. A typical example is glass.New YorkLondonSEM image of grains in AlN ceramicsPage 553.1Fundamental Concepts.Coordination numberThe number of nearest-neighbor or touching atoms.Atomic Packing Factor (APF)The sum of the sphere volu

12、mes of all atoms within a unit cell divided by the unit cell volume.Unit CellThe small repetitive atoms group in crysatls.(a) An infinitely repeating box.(b) The unit cell must be infinitely stackable using only translation.(c)The three smallest non-coplanar vectors a, b, and c describe the unit cel

13、l with inter-edge angles of , , and Page 663.2Metallic Crystal Structures.Face-Centered Cubic (FCC)Cube edge length aRa22Coordination number 12Atomic Packing Factor (APF) 74. 0The atomic bonding is metallic and thus nondirectional in nature. Consequently, there are minimal restrictions as to the num

14、br and position of nearest-neighbour atoms; this lead to relatively large numbers of nearest neighbors and dense atomic packings.Atom number per unit cell 4Page 773.2Metallic Crystal Structures.Example3.1 (a) Derive the relationship of atomic radius R with cube edge length a and volume Vc, n FCC uni

15、t cell.(b) Calculate the atomic packing factor (APF) .(c) Explain why the coordination number is 12.Relationship of R with a 222)4( RaaRa22Relationship of R with Vc 216)22(333RRaVcAtomic packing factor74. 0216)34(4)34(43333RRaRVnVAPFcatomAtom per FCC unit cell4818216nR4RaaSolution Page 883.2Metallic

16、 Crystal Structures.Body-Centered Cubic (BCC)Cube edge length a34Ra Coordination number 8Atomic Packing Factor (APF)68. 0Atom number per unit cell2Page 993.2Metallic Crystal Structures.Example3.2 (a) Derive the relationship of atomic radius R with cube edge length a and volume Vc, in BCC unit cell.(

17、b) Calculate the atomic packing factor (APF) .(c) Explain why the coordination number is 8.Relationship of R with a 222)4()2(Raa34Ra Relationship of R with Vc 3364)34(333RRaVcAtomic packing factor68. 0)34(233aRVnVAPFcatomAtom per FCC unit cell281811nSolution a24RaPage 10103.2Metallic Crystal Structu

18、res.Hexagonal Close Packing (HCP)Cube edge length a and cRa2Coordination number 12Atomic Packing Factor (APF) 74. 0Atom number per unit cell 638/ac(Ideal)Page 11113.2Metallic Crystal Structures.Example3.3 (a) Derive the relationship of atomic radius r with cube edge length a & c, and volume Vc,

19、in HCP unit cell.(b) Calculate the atomic packing factor (APF) .(c) Explain why the coordination number is 12. Solution Please finish the geometric calculations by yourselves. .2ch Regular tetrahedron ABCDABDC2cPage 1212Summary: Common Metal Structureshcpccp (fcc)bccABABABABCABCnot close-packedHexag

20、onal unit cellSpecify: a, cHexagonal implies:|a1| = |a2| = ag = 120; a = b = 90Cubic unit cellsSpecify: aCubic implies:|a1| = |a2| = |a3| = aa = b = g = 90Page 13133.3Density Computations.The knowledge of the crystal structure of a metallic solid permits computation of its theoretical dentisty throu

21、gh the relationship:AcNVnAVmwhere n = number of the atoms associated with each unit cellA = atomic weightVc = volume of the unit cellNA = Avogadros number (6.023 1023 atoms/mol) Example3.4 Copper has an atomic radius of 0.123 nm, FCC crystal structure, and atomic weight of 63.5 g/mol. Compute its th

22、eoretical density.Solution 323383/89. 8)/10023. 6(/ )1028. 1 (216)/5 .63)(/4(216cmgmolatomsunitcellcmmolgunitcellatomsNRnANVnAACuAcCuPage 14143.4Allotropy.AllotropyAllotropy or allotropism is the property of some chemical elements (both metallic and nonmetallic) to exist in two or more different for

23、ms. Allotropes are different structural modifications of an element, that is, the atoms of the element are bonded together in a different manner.Eight allotropes of carbon: a) Diamond, b) Graphite, c) Lonsdaleite, d) C60 (Buckminsterfullerene or buckyball), e) C540, f) C70, g) Amorphous carbon, and

24、h) single-walled carbon nanotube or buckytube.Page 15153.4Allotropy.Example3.5 At room temperature, pure iron (Fe, atomic weight 55.85 g/mol) has a BCC crystal structure with a unit cell volume of 0.02464 nm3. It changes to FCC structure at 912 C and the volume of unit cell increases to 0.0486 nm3.

25、Name this phenomenon and computer the change of density during the process. Solution It is Allotropy, that is, some chemical elements exist in two or more different forms.The density of BCC Fe 323321/53. 710023. 61002464. 0/85.552cmgcmmolgNVnAABCCFeBCCThe density of FCC Fe 323321/636. 710023. 610048

26、6. 0/85.554cmgcmmolgNVnAAFCCFeFCCThe density Change%4 . 1%10053. 753. 7636. 7BCCBCCFCCPage 16163.5Crystal System.Can you memorize all the names?Page 17173.5Crystal System.Whats in the box? Adding atoms to the unit cell7 Crystal System4 Types of Unit CellP = PrimitiveI = Body-CenteredF = Face-Centere

27、dC = Side-Centered14 Bravais Lattices+Page 18183.6Crystallographic Directions and Planes.Directions1. A vector of convenienet length is positioned such that it passes through the origin of the coordinate system. Any vector may be translated throughout the crystal lattice without alteration, if paral

28、lelism is maintained.2. The length of the vector projection on each of the three axes is determined; there are measured in terms of the unit cell dimensions a, b, and c.3. There three numbers are multiplie or divided by a common factor to reduce them to the smallest integer values.4. The three indic

29、es, not separated by commas, are enclosed in squared bracket, thus: uvw. the u, v, and w integers correspond to the reduced projections along the x, y, and z axes, repectively.5. Possible negative coordinateds are represented by a bar over the index. Example3.6Index the four vectors direction in the

30、 unit cell.100, 110, 111, and 011Solution Page 19193.6Crystallographic Directions and Planes.Planes1. If the plane passes through the selected origin, either another parallel plane must be constructed by appropriated translation, or a new origin point must be established.2. At the point the plane ei

31、ther intersects or parallels each of the three axes; the length of the plannar intercept for each axis is determined in terms of the lattice parameters a, b, and c.3. The reciprocals of these numbers are taken. A plane that parallels an axis my be considered to have an infinite intercept, and theref

32、or, a zero index.4. The three numbers are changed to the set of smallest integers.5. The three numbers, not separated by commas, are enclosed within parentheses, thus: (hkl), known as Miller Indices. 6. Possible negative coordinateds are represented by a bar over the index. Example3.7Draw plane(s) w

33、ith Miller indices of (111) in the unit cell.& Solution Page 20203.6Crystallographic Directions and Planes.Can you explain the solution?3.8 Index all the edges and planes of an octahedron in FCC cube. Solution ABCDEFADEBCF)111(ABFCDE) 111 (ABECDF)111(ADFBCE)111(ECAF /011EDBF /011EACF /110EBDF /1

34、01DCAB/110ADBC /101AC100BD010EF001ExamplePage 21213.7Imperfections in SolidsVacancy Abstence of an atom from normally occupied position.All crystalline solid contain vacanceis.Impurity - Two kindsSubstitutional and Interstitial.VacancySubstitutionalImpurity atomInterstitialImpurity atomLinear Defect

35、s - DislocationPoint DefectsInterfacial DefectsEdge DislocationScrew DislocationPage 22223.8Alloys and Alloying.1. Atomic size factor. The difference in atomic radii between the two atom type is less than about 15%. Otherwise the solute atoms will creat substantial lattic distroions and a new phase

36、will form.2. Crystal structure. For appreciable solid solubility the crystal structures for metals of both atom types must be the same.3. Electronegativity. The more electropositive one element and the more electronegative the other, the greater is the likelihood that will form an intermetallic comp

37、ound instead of a substitutional solid solution.4. Valences. Other factors being equal, a metal will have more of a tendency to dissolve another metal of higher valency than one of a lower valency.AlloyAn alloy is a mixture or metallic solid solution composed of two or more elements.Solid Solution -

38、 Two KindsA solid solutin forms when, as the solute atoms are added to the host material (solvent), the crystal structure is maintained, and no new structure are formed. The composition is homogeneous and the impurity (solute) atoms are randomly and uniformly dispersed within the solid.Rules for the

39、 formationExtensive ReadingProf. Dan Shechtman, 2011 Nobel LaureateAtomic model of an Al-Pd-Mn quasicrystal surface.A Ho-Mg-Zn icosahedral quasicrystal formed as a dodecahedron, the dual of the icosahedronA quasiperiodic crystal, or, in short, quasicrystal, is a structure that is ordered but not per

40、iodic. A quasicrystalline pattern can continuously fill all available space, but it lacks translational symmetry. While crystals, according to the classical crystallographic restriction theorem, can possess only two, three, four, and six-fold rotational symmetries, the Bragg diffraction pattern of q

41、uasicrystals shows sharp peaks with other symmetry orders, for instance five-fold.A penteract (5-cube) pattern using 5D orthographic projection to 2D using Petrie polygon basis vectors overlaid on the diffractogram from an Icosahedral Ho-Mg-Zn quasicrystalFor three millennia we have known that five-fold symmetry is incompatible with periodicity, and for almost three centuries we believed that periodicity was a prerequisite for crystallinity. The electron diffraction pattern obtained by Dan Shechtman on April 8, 1982 shows that at least one