Phy351 ch 3

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Phy351 ch 3

  1. 1. PHY351 Chapter 3 CRYSTAL STRUCTURE
  2. 2. Long-Range-Order (LRO) Most metals and alloys, semiconductors, ceramics and some polymers have a crystalline structure (where the atoms or ions display in a long-range-order (LRO). The atoms or ions in these material form a regular repetitive, grid like pattern in three dimension. These materials are referred as a crystalline structure. 2
  3. 3. Question 1: a. Define the following terms: i. SRO atom/ion arrangement ii. no order atom/ion arrangement iii. crystalline solid iv. crystal structure v. amorphous b. Give 2 example material of i. SRO atom/ion arrangement ii. no order atom/ion arrangement 3
  4. 4. The space lattice and unit cells Properties of SOLIDS depends upon: bonding force crystal structure 4
  5. 5. Space lattice An IMAGINARY NETWORK of lines with atoms at lines intersection that representing the arrangement of atoms is called space lattice. Unit cells Unit cell is that block of atoms which REPEATS itself to form space lattice. Space Lattice Unit Cell 5
  6. 6. Question 2: a. 6 Define the following terms: i. Lattice point ii. Motif iii. Lattice constant
  7. 7. Crystal Systems and Bravais Lattice Only SEVEN different types of unit cells are necessary to create all point lattices which:        7 Cubic Tetragonal Orthorhombic Rhombohedral Hexagonal Monoclinic Triclinic
  8. 8. According to Bravais (1811-1863), fourteen standard unit cells can describe all possible lattice networks based on FOUR basic types of unit cells which are: Simple  Body Centered  Face Centered  Base Centered  8
  9. 9. i. Cubic Unit Cell  a=b=c  α = β = γ = 900 Simple 9 Body Centered Face centered
  10. 10. ii. Tetragonal  a =b ≠ c  α = β = γ = 900 Simple 10 Body Centered
  11. 11. iii. Orthorhombic  a≠ b≠ c  α = β = γ = 900 Simple 11 Body Centered Face Centered Base Centered
  12. 12. iv. Rhombohedral a= b≠ c  a =b = c  α=β=γ≠ Simple 12 v. Hexagonal 900  α = β = 900 γ = 1200 Simple
  13. 13. vi. Monoclinic vii. Triclinic  a≠ b≠ c  a≠ b≠ c  α = γ = 900 ≠ β  α ≠ β ≠ γ ≠ 900 Simple 13 Base Centered Simple
  14. 14. Principal Metallic Crystal Structures 90% of the metals have either Body Centered Cubic (BCC), Face Centered Cubic (FCC) or Hexagonal Close Packed (HCP) crystal structure. HCP is denser version of simple hexagonal crystal structure. 14 BCC Structure FCC Structure HCP Structure
  15. 15. Body Centered Cubic (BCC) Crystal Structure Represented as: one atom at each corner of cube one at the center of cube. Examples : Chromium (a=0.289 nm)  Iron (a=0.287 nm)  Sodium (a=0.429 nm) 15 (a) Atomic-site unit cell Figure 3.4: BCC crystal structure (b) Hard-sphere unit cell (c) Isolated unit cell
  16. 16. Each unit cell has:  eight 1/8 atom at corners  1 full atom at the center. Therefore each unit cell has (8 x 1/8) + 1 = 2 atoms Atoms contact each other at cube diagonal. Therefore; 4R a 3 16
  17. 17. Face Centered Cubic (FCC) Crystal Structure FCC structure is represented as one atom each at the corner of cube at the center of each cube face. Examples : Aluminum (a = 0.405)  Gold (a = 0.408) 17 (a) Atomic-site unit cell Figure 3.6: FCC crystal structure (b) Hard-sphere unit cell (c) Isolated unit cell
  18. 18. Each unit cell has:  eight 1/8 atom at corners  six ½ atoms at the center of six faces Therefore each unit cell has (8 x 1/8) + (6 x ½) = 4 atoms Atoms contact each other across cubic face diagonal. Therefore; 4R a 2 18
  19. 19. Hexagonal Close-Packed Structure (HCP) The HCP structure is represented as 1 atom at each of 12 corners of a hexagonal prism 2 atoms at top and bottom face 3 atoms in between top and bottom face. Examples: Zinc (a = 0.2665 nm, c/a = 1.85)  Cobalt (a = 0.2507 nm, c/a = 1.62) 19 Figure 3.8: HCP crystal structure: (a) Schematic of the crystal structure (larger cell) (b) Hard-sphere model
  20. 20. Each atom has:  six 1/6 atoms at each of top and bottom layer  two ½ atoms at top and bottom layer  3 full atoms at the middle layer Therefore each HCP unit cell has: (2 x 6 x 1/6) + (2 x ½) + 3 = Ideal HCP c/a ratio is 1.633. 20 6 atoms
  21. 21. Isolated HCP unit cell also called the primitive cell. The atoms at locations marked:  ‘1’ contribute 1/6 of an atom  ‘2’ contribute 1/12 of an atom  ‘3’ contribute 1 atom Figure 3.8: HCP crystal structure: (c) Isolated unit cell schematic Therefore each HCP unit cell (primitive cell) has (4 x1/6) + (4 x 1/12) + 1 = 21 2 atoms
  22. 22. Atomic Packing Factor (APF) Formula: APF = Volume of atoms in unit cell Volume of unit cell 22
  23. 23. APF for BCC structure: APF = Volume of atoms in unit cell Volume of unit cell But; Vatoms =  4R 2.  3  3     V unit cell = a3 = = 8.373R3 3 Therefore; APF = 8.723 R3 = 0.68 12.32 R 23  4R      3  3 = 12.32 R3
  24. 24. Question 3 3.1 Determine the atomic packing factor (APF) for: a. b. Face centered cubic structure Hexagonal Close-Packed Structure (HCP) 3.2 Calculate the volume of the zinc crystal structure unit cell by using the following data: pure zinc has HCP crystal structure with lattice constant a = 0.2665 nm and c = 0.4947nm. 3.3 By using data in question 2.2, find the volume of the larger cell. 3.4 What are the three most common metal crystal structures? List five metals that have each of these crystal structures. 24
  25. 25. Volume density Formula volume density of metal: v 25 = Mass per Unit cell Volume per Unit cell
  26. 26. Example:Copper (FCC) has atomic mass of 63.54 g/mol and atomic radius of 0.1278 nm. Calculate the volume density. 26
  27. 27. Solution: v = Mass per Unit cell Volume per Unit cell Volume of unit cell = V = a3 Known FCC unit cell has 4 atoms. Therefore; But; a= Mass of unit cell = m 4R -9 = 4 x 0.1278nm 4  0.1278 x 10 2 = 4 (63.54) 6.02 x 1023 2 Therefore; Volume of unit cell = (0.361nm)3 = 4.7 x 10-29 m3 = 4.22 x 10-22g 27 = 0.361 nm m 10  v  V  4.22xx10 4.7 -22 -29 =8.95 x 106 g/m3
  28. 28. Atom Positions in Cubic Unit Cells Cartesian coordinate system is use to locate atoms. In a cubic unit cell  y axis is the direction to the right.  x axis is the direction coming out of the paper.  z axis is the direction towards top.  Negative directions are to the opposite of positive directions. Atom positions are located using unit distances along the axes. 28
  29. 29. Figure3.9 Atomic position in a BCC unit cell. 29
  30. 30. Direction In cubic crystals, direction indices are vector components of directions resolved along each axes, resolved to smallest integers. Direction indices are position coordinates of unit cell where the direction vector emerges from cell surface, after converted to integers. 30 Fig 3.1: Some directions in cubic unit cells
  31. 31. PROCEDURE TO FIND DIRECTION INDICES Produce the direction vector till it emerges from surface of cubic cell z (1,1/2,1) Determine the coordinates of point of emergence and origin (1,1/2,1) - (0,0,0) = (1,1/2,1) y (0,0,0) Subtract coordinates of point of Emergence by that of origin Are all are integers? YES Are any of the direction vectors negative? YES 31 Represent the indices in a square bracket without comas with a over negative index (Eg: [121]) NO x 2 x (1,1/2,1) = (2,1,2) The direction indices are [212] Convert them to smallest possible integer by multiplying by an integer. NO Represent the indices in a square bracket without comas (Eg: [212] )
  32. 32. Example: Determine the direction indices of the cubic direction shown below: 32 Tips: The vector representing EF can be found by subtracting the coordinate of the tips of the vector F from the coordinate of the tail E.
  33. 33. Exercise 3.4 i. ii. 33 Draw the following direction vectors in cubic unit cells: a. [100] and [110] b. [112] c. [1 1 0] d. [3 2 1] Determine the direction indices of the cubic direction between the position coordinates (3/4 , 0 , ¼) and (1/4 , ½ , ½)
  34. 34. Lattice plane Miller Indices are are used to refer to specific lattice planes of atoms. They are reciprocals of the fractional intercepts (with fractions cleared) that the plane makes with the crystallographic x,y and z axes of three nonparallel edges of the cubic unit cell. z Miller Indices = (111) y 34 x
  35. 35. MILLER INDICES - PROCEDURE Choose a plane that does not pass through origin z y Determine the x,y and z intercepts x of the plane Find the reciprocals of the intercepts Fractions? Place a ‘bar’ over the Negative indices Yes Clear fractions by multiplying by an integer to determine smallest set of whole numbers Enclose in parenthesis (hkl)where h,k,l are miller indicesof cubic crystal plane forx,y and z axes. Eg: (111) 35 Miller Indices = (111)
  36. 36. Miller Indices - Examples z  Intercepts of the plane at x,y & z axes are 1, ∞ and ∞ (100) y  Taking reciprocals we get (1,0,0).  Miller indices are (100). x x  Intercepts are 1/3, 2/3 & 1.  taking reciprocals we get (3, 3/2, 1).  Multiplying by 2 to clear fractions, we get (6,3,2).  Miller indices are (632). 36
  37. 37. Plot the plane (101)  Taking reciprocals of the indices we get (1 ∞ 1).  The intercepts of the plane are x=1, y= ∞ (parallel to y) and z=1. Figure EP3.7 a Plot the plane (2 2 1)  Taking reciprocals of the indices we get (1/2 1/2 1).  The intercepts of the plane are x=1/2, y= 1/2 and z=1. 37 Figure EP3.7 c
  38. 38. Plot the plane (110) • The reciprocals are (1,-1, ∞) • The intercepts are x=1, y=-1 and z= ∞ (parallel to z axis) Figure EP3.7 b 38 Note: To show this plane in a single unit cell, the origin is moved along the positive direction of y axis by 1 unit.
  39. 39. Exercise 3.7 1. Draw the following crystallographic planes in cubic unit cells: a. (101) b. (110) c. (221) d. (110) 2. Determine the miller index for the plane given cubic unit cell. 39 (a) (b)
  40. 40. Planar Atomic Density Formula planar atomic density:  40 p = Equivalent number of atoms whose centers are intersected by selected area Selected area
  41. 41. Example:In Iron (BCC, a=0.287 nm), the (110) plane intersects center of 5 atoms (four ¼ and 1 full atom). Calculate the planar atomic density. 41
  42. 42. Solution: - Equivalent number of atoms = (4 x ¼ ) + 1 = 2 atoms - Area of 110 plane = 2a  a  2a 2 Therefore;  p =  42 2 2 0.287 2 17.2atoms nm2 1.72  101319 atoms  1.72 x 10 atoms/m2 mm2
  43. 43. Linear Atomic Density Formula linear atomic density:  l = Number of atomic diameters intersected by selected length of line in direction of interest Selected length of line 43
  44. 44. Example:For a FCC copper crystal (a=0.361 nm), the [110] direction intersects 2 half diameters and 1 full diameter. 44
  45. 45. Solution: - The number of atomic diameters intersected by this length of line are: ½ + ½ + 1 = 2 atomic diameters - The length of the line is: 2  0.361nm Therefore;  l  2atoms 2  0.361nm 45  3.92atoms nm 3.92  106 atoms  3.92 x 109 atoms/m mm
  46. 46. Interplanar Spacing The distance between two adjacent parallel planes of atoms with the same Miller indices is called the interplanar spacing, dhkl . The interplanar spacing in cubic materials is given by the general equation: 46
  47. 47. Crystal Structure Analysis – X-ray Diffraction Information about crystal structure are obtained using X-Rays diffraction technique. The X-rays source: - The wavelength (0.05-0.25 nm) which same as distance between crystal lattice planes. 47 Figure 3.22 Schematic diagram of the cross section of a sealed-off filament Xray tube.
  48. 48. Crystal planes of target metal act as mirrors reflecting X-ray beam. If rays leaving a set of planes are out of phase (as in case of arbitrary angle of incidence) no reinforced beam is produced. If rays leaving are in phase, reinforced beams are produced. Figure 3.25 The reflection of an X-ray beam by the (hkl) planes of a crystal 48 a) No reflected beam is produced at an arbitrary angle of incidence. b) At the Bragg angle , the reflected rays are in phase and reinforce one another.
  49. 49. For rays reflected from different planes to be in phase, the extra distance traveled by a ray should be a integral multiple of wave length λ . c) Similar to b) except that the wave representation has been omitted. 49
  50. 50. Let us consider incident X-rays 1 and 2 as figure 3.25c. For these rays to be in phase, the extra distance of travel of ray 2 is equal to MP + PN which must be an integral number of wavelength. Thus; nλ = MP + PN where; n = order of the diffraction = 1,2,3,….. 50
  51. 51. Since both MP and PN equal to dhklsin, the condition for constructive interference (the production of a diffraction peak of intense radiation) must be: nλ = 2 dhkl sinθ This equation known as Bragg’s law. 51 Bragg’s law gives the relationship among the angular positions of the reinforced diffracted beams in terms of the wavelength of the incoming X-ray radiation and of the interplanar spacing dhkl of the crystal planes.
  52. 52. Example 3.15: A sample of BCC iron was placed in an X-ray diffractometer using incoming X-rays with a wavelength of 0.1541 nm. Diffraction from the 110 planes was obtained at 2 = 44.70. Calculate a value for the lattice constant, a of BCC iron. (Assume first-order diffraction with n = 1). Answer: 0.287 nm 52
  53. 53. References  A.G. Guy (1972) Introduction to Material Science, McGraw Hill.  J.F. Shackelford (2000). Introduction to Material Science for Engineers, (5th Edition), Prentice Hall.  W.F. Smith (1996). Priciple to Material Science and Engineering, (3rd Edition), McGraw Hill.  W.D. Callister Jr. (1997) Material Science and Engineering: An Introduction, (4th Edition) John Wiley. 53

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