Chapter 4 Crystal Structures
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This document discusses crystal structures and their properties. It describes how atoms are arranged in crystalline solids through ordered unit cells that form a repeating lattice. The main crystal structures for metals are body-centered cubic, face-centered cubic, and hexagonal close-packed. It explains how to calculate properties like density from the unit cell parameters and atomic positions. Direction vectors are used to describe crystallographic directions.
![4.5 Crystallographic Directions
• Is a vector indicating the direction in a crystal, used instead by [u,v,w] Which will be the
full number (integer number) The smallest (Projection) of core x, y and z in unit cell.
• Negative direction There will be a stroke above such as [ത100] Will be the opposite of
[100]
40](https://image.slidesharecdn.com/introductiontomaterialsscienceandengineeringenglishversionlesson4-191124085500/85/Chapter-4-Crystal-Structures-40-320.jpg)
![How to find the direction of the crystals in unit cell ( when
knowing Coordinates)
1) Vector Used to indicate the direction must always pass through the origin
2) Set Coordinate Of the beginning (𝑥1, 𝑦1, 𝑧1) And the end of the direction (𝑥2, 𝑦2, 𝑧2)
3) Use Coordinate Destination minus Coordinate Origin : (𝑥2, 𝑦2, 𝑧2) − (𝑥1, 𝑦1, 𝑧1) =
(𝑥2 − 𝑥1), (𝑦2 − 𝑦1), (𝑧2 − 𝑧1),
4) If there are fractions, the integer is the smallest value (Multiply LCM )
5) To write the crystal direction index, write [uvw] No need to have any separator, If the
value is negative, write it above the number, for example [ത100]
43](https://image.slidesharecdn.com/introductiontomaterialsscienceandengineeringenglishversionlesson4-191124085500/85/Chapter-4-Crystal-Structures-43-320.jpg)
![Direction A
1) Starting point 0,0,0 Ending point 1,0,0
2) (1,0,0) - (0,0,0) = 1,0,0
3) Direction index = [100]
Direction B
1) Starting point 0,0,0 Ending point 1,1,1
2) (1,1,1) - (0,0,0) = 1,1,1
3) Direction index = [111]
Direction C
1) Starting point 1/2,1,0 Ending point 0,0,1
2) (0,0,1) - (1/2,1,0) = -1/2,-1,1
3) 2(-1/2,-1,1) = -1, -2, 2
4) Direction index = [ത1 ഥ2 2]
44](https://image.slidesharecdn.com/introductiontomaterialsscienceandengineeringenglishversionlesson4-191124085500/85/Chapter-4-Crystal-Structures-44-320.jpg)
![Families and Symmetry
In the direction of the crystal
edge (cubic edge)
[100],[010],[001],[ത100],
[0ത10],[00ത1] =< 100 >
Diagonal direction on both sides 6
(cubic face diagonals)=< 110 >
Diagonal direction on both angle 4
(cubic face diagonals)=< 111 >
45](https://image.slidesharecdn.com/introductiontomaterialsscienceandengineeringenglishversionlesson4-191124085500/85/Chapter-4-Crystal-Structures-45-320.jpg)
![• All direction vectors parallel to each other can use the same index because the starting
point can be changed.
• If the distance between atoms (atom spacing) in the direction parallel to the direction
of the vector is the same We will consider that the vector direction is equal, for
example In the direction of the crystal edge (cubic edge)
[100],[010],[001],[ത100],[0ത10],[00ത1] =< 100 >
Diagonal direction on both sides 6 (cubic face diagonals)=< 110 >
Diagonal direction on both angle 4 (cubic face diagonals)=< 111 >
• The symmetrical directions are called Family of crystallographic direction Must write
in < > Only
46](https://image.slidesharecdn.com/introductiontomaterialsscienceandengineeringenglishversionlesson4-191124085500/85/Chapter-4-Crystal-Structures-46-320.jpg)
Chapter 4 Crystal Structures
- 1. Chapter 4 Crystal Structures Mr. Pem PHAKVISETH pempvs@gmail.com Department of Materials Science and Engineering Souphanouvong University Faculty of Engineering 1
- 2. Objective of Learning • Describe crystal structure and arrangement of atoms in crystal structure, formation of microstructure • Recognizing the relationship between atomic bonding, crystal structure and material properties • Describes the X-Ray Diffraction 2
- 3. 4.1 Crystal 3 • A crystal or crystalline solid is a solid material whose constituents (such as atoms, molecules, or ions) are arranged in a highly ordered microscopic structure, forming a crystal lattice that extends in all directions
- 4. Materials and Arrangement The materials are arranged into Crystalline • The atoms have an orderly arranged in a 3D . • a sample: metal, ceramic ,some polymers Long range order The materials are arranged into Amorphous • The atoms have an not order arrangement. • Occurs for : Complex structure, fast cooling Short range order “Amorphous” = Non-crystalline 4
- 5. Energy and Arrangement • Non dense, random packing • Dense, ordered packing The ordered packing is Dense, the structure will have lower energy. 5
- 6. Unit Cell • Unit Cell : Is the unit of the smallest basic structure of a crystal or a specific unit that produces crystal structure that can simulate the arrangement of atoms both crystals and has a definite shape • That will consist of atoms which is in a specific position within that cell 6
- 7. Space Lattice or Crystal Lattice • So the word Crystal Lattice or Space Lattice therefore referring to the shape formed by the arrangement of unit cells in 3 dimensions. When looking at each dimension, it will look like all similar meshes. • Lattice referring to the Network (lattice) • Any complete crystal consists of a number of unit cells arranged in 3D, which we call “Crystal Lattice or Space Lattice”. 7
- 8. Lattice Parameter • Lattice Parameter : The distance or size of the unit cell of each side • When changing to 3D on each side and the direction of the lattice parameter, it must be able to fit into the 3D structure without any space remaining. • a, b, c Is the distance or size of Lattice Parameter 8
- 9. Crystalline Solid Formation How do crystals occur? • All materials are formed by the binding of atoms by bonds such as ionic, covalent or metallic bonding. • Crystalline solid, like metals, ceramics and some polymers. • A non-crystalline solid is called a non-crystalline solid or amorphous, which is just a group of atoms. Or molecules without certain structures 9
- 10. Mechanism of Crystallization • In metals that are in a molten state the binding of atoms will have a sequence short range order or disordered. • The bonding and separation of atoms is random and can occur at any time. Due to the high activation energy (due to high temperatures) 10
- 11. • There are two types of energy of atoms: - Kinetic Energy is the energy related to the speed at which the atoms move and vary directly with the temperature - Potential energy is the energy that accumulates within when the atoms are apart will have more energy of this type • In a liquid state toms have higher energy values 11
- 12. • When the temperature drops below the hardness of that metal And begin to change to solid state, the energy will be lower • Atoms must emit latent heat called "Latent heat of fusion" in order to bond between atoms to form a nuclei formation and grow into the nucleus and become atoms. 12
- 13. • Then the nucleus formed to become an atom And each atom comes together From the incorporation of only a few atoms will begin to expand by holding on to the next atom at all times Three-dimensional lattice arrangement with a certain distance and position of atoms along the crystal axis 13
- 14. • The crystals will keep growing. And stop growing when colliding with other crystals of different atomic directions the crystals that stop growing and are called “Grain” • The seam that crystallizes connect with other crystals is called “grain boundary” • After that, many grains combine to form a solid. And will be called as “Polycrystalline Solid” 14
- 15. 15 Grain growth See VDO-Grain Growth grain grain grain intergranular https://www.youtube.com/watch?v=FlkrBSh4514
- 16. How many types of atoms can be arranged in solid state? • There are 7 systems of crystal systems based on the symmetry of the shape. • Each crystal system can be further divided into 14 crystal structures according to the arrangement of atoms. • There 32 point group and 230 space group. • The crystal structure is closely related to the properties such as strength, heat resistance. Difficulty in forming, etc. 16 4.2 Crystal Structure Systems of Materials
- 17. Crystal System & Unit Cell Lattice Parameters 17 • Cubic system • For other crystal structures
- 18. 18
- 19. 19
- 20. Is the length of the body The angle value of the ideal angle between the axes 20
- 21. 4.3 Crystal Structure Systems of Metals There are 3 common crystal structures of metals • Body Centered Cubic Structure (BCC) • Face Centered Cubic Structure (FCC) • Hexagonal Close Packed Structure (HCP) 21
- 22. 1. Body Centered Cubic Structure (BCC) • Found in Chromium, Lithium, Iron and Tungsten etc. • have 1 atom in the middle And arranged in a straight line diagonally with the atoms in all 8 angles and connected throughout the crystal • The direction in which the atoms are packed the most is Diagonal line of the cube 22
- 23. Determined a = lattice constant and R = atomic radius Will get a relationship Number of atoms 1 unit cell • All Angles 8 = (1/8) x 8 = 1 atom • Center point = 1 atom • Total 2 atom/unit cell 23
- 24. Example 24
- 25. 2. Face Centered Cubic Structure(FCC) • Found in copper, aluminum, silver and gold etc. • Will have atoms in all 8 corners and at the center point on all 6 sides • The direction in which the atoms are compressed most is the diagonal of each of the 6 sides of the cube. 25
- 26. from unit cell atoms will touch each other diagonally of the side will Number of atom 1 unit cell • All angles 8 = (1/8) x 8 = 1 atom • All side 6 = (1/2) x 6 = 3 atom • Total 4 atom/unit cell 26 Diagonal side is equal to
- 27. Example 27
- 28. 28
- 29. 3. Hexagonal-Close Packed (HCP) • Found in Magnesium, Titanium and Zinc etc. • The top and bottom planes have atoms close together to form a hexagon. And the central point of the plane The middle plane will have 3 atoms. • For ideal hcp , c/a = 1.633 1 Unit cell of HCP have 6 atom 29
- 30. 1 Unit cell of HCP have 6 atom • Middle atom = 3 atom • Top-bottom atom = (1/2) x 2 = 1 atom • Atoms at the corner = (1/6) x 12 = 2 atom Cut times 1 remaining (1/2) Cut times 2 remaining (1/3) X (1/2) = (1/6) 30
- 31. 31
- 32. 4.3 Atomic Packing Factor (APF) • Is the value that tells the volume of the atom in unit cell • Assuming the shape of a circle of atoms so. 𝑨𝑷𝑭 = 𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒂𝒕𝒐𝒎 𝒊𝒏 𝒖𝒏𝒊𝒕 𝒄𝒆𝒍𝒍 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒖𝒏𝒊𝒕 𝒄𝒆𝒍𝒍 • In BCC, APF = 0.68 • In FCC and HCP, APF=0.74 32
- 33. Example of finding the Atomic Packing Factor for FCC Close – packed directions: Length = 4R = 2𝑎 Unit cell Contain: 6 × 1/2 + 8 × 1/8= 4 atom/ unit cell 33
- 34. Example of finding the Atomic Packing Factor for FCC • The FCC structure in 1 cell has 4 atoms. • The volume of atoms is equal to (4/3) 𝜋𝑟3 • The volume of cell is equal to 𝑎3 34
- 35. 4.4 Calculation to Density Computation of unit cell Theoretical density calculation Metal density (𝜌) can be calculated from take the mass of all the atoms, divided by the volume of 1 unit cell If the mass of atoms, A unit is atom mass unit (amu) to be divided by the Avogadro number (6.023 × 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙) In order to get the atomic mass 𝑀 𝑎𝑡𝑜𝑚 35
- 36. 36
- 37. 37
- 38. Density of different types of materials Metals • Compact structure (Metallic bonding) • Very precious atomic mass Ceramics • Less compact structure(Covalent bonding) • Is usually a lighter element Polymers • Very compact structure(Usually an allotrope) • Is a very rare element (C,H,O) composites • Have medium properties 38
- 39. Example 39
- 40. 4.5 Crystallographic Directions • Is a vector indicating the direction in a crystal, used instead by [u,v,w] Which will be the full number (integer number) The smallest (Projection) of core x, y and z in unit cell. • Negative direction There will be a stroke above such as [ത100] Will be the opposite of [100] 40
- 41. Point Coordinates • Determining the point in unit cell mainly like Cartesian coordinates • Except the value of x, y and z will show as a proportion of the length of unit vectors 41
- 42. 42
- 43. How to find the direction of the crystals in unit cell ( when knowing Coordinates) 1) Vector Used to indicate the direction must always pass through the origin 2) Set Coordinate Of the beginning (𝑥1, 𝑦1, 𝑧1) And the end of the direction (𝑥2, 𝑦2, 𝑧2) 3) Use Coordinate Destination minus Coordinate Origin : (𝑥2, 𝑦2, 𝑧2) − (𝑥1, 𝑦1, 𝑧1) = (𝑥2 − 𝑥1), (𝑦2 − 𝑦1), (𝑧2 − 𝑧1), 4) If there are fractions, the integer is the smallest value (Multiply LCM ) 5) To write the crystal direction index, write [uvw] No need to have any separator, If the value is negative, write it above the number, for example [ത100] 43
- 44. Direction A 1) Starting point 0,0,0 Ending point 1,0,0 2) (1,0,0) - (0,0,0) = 1,0,0 3) Direction index = [100] Direction B 1) Starting point 0,0,0 Ending point 1,1,1 2) (1,1,1) - (0,0,0) = 1,1,1 3) Direction index = [111] Direction C 1) Starting point 1/2,1,0 Ending point 0,0,1 2) (0,0,1) - (1/2,1,0) = -1/2,-1,1 3) 2(-1/2,-1,1) = -1, -2, 2 4) Direction index = [ത1 ഥ2 2] 44
- 45. Families and Symmetry In the direction of the crystal edge (cubic edge) [100],[010],[001],[ത100], [0ത10],[00ത1] =< 100 > Diagonal direction on both sides 6 (cubic face diagonals)=< 110 > Diagonal direction on both angle 4 (cubic face diagonals)=< 111 > 45
- 46. • All direction vectors parallel to each other can use the same index because the starting point can be changed. • If the distance between atoms (atom spacing) in the direction parallel to the direction of the vector is the same We will consider that the vector direction is equal, for example In the direction of the crystal edge (cubic edge) [100],[010],[001],[ത100],[0ത10],[00ത1] =< 100 > Diagonal direction on both sides 6 (cubic face diagonals)=< 110 > Diagonal direction on both angle 4 (cubic face diagonals)=< 111 > • The symmetrical directions are called Family of crystallographic direction Must write in < > Only 46
- 47. 4.6 Crystallographic planes • Means specifying the crystal plane that occurs in a unit cell, also known as Miller indices • Miller indices is' the reciprocals of the fractional intercepts which the plane makes with the crystallographic x, y and z axes of the three nonparallel edges of the cubic unit. cell ‘ • By writing the plane in the unit cell, the symbol (hkl) is used for the x, y and z axes respectively (when writing without commas, separated). 47
- 48. Determining the crystal plane Or Miller index (Miller indices) (when knowing the intersection)) 1) Choose a plane that does not pass a point (000) 2) Find the point of intersection on all 3 axes of the plane and reverse the fractions of the point of intersection. 3) without any marks Separator and no space From picture beside The intersection of X, Y, Z axis is 1, ∞, ∞ Therefore, returning fractions to (reciprocals) 1, 0, 0 The miller indices is (100) 48
- 49. 49 3 3 (200)
- 50. From picture beside intercepts x, y, z is Τ1 2 , Τ1 2 , ∞ Reciprocals 2, 2, 0 Millers indices (220) From picture beside intercepts x, y, z is ∞, 1, 1 Reciprocal 0, 1, 1 Millers indices (011) 50
- 53. 4.8 X-Ray Diffraction x-ray crystallography, also called x ray diffraction, is used to determine crystal structure by interpreting the diffraction patterns formed when x ray are scattered by the electron of atom in crystalline solids. x-rays are sent through a crystal to reveal the patterns in which the molecules and atom contained within the atom arranged. This x-ray crystallography was develop by physicist William Lawrence Bragg and his father William Henry Bragg in 1912 -1913, the younger Bragg develop Bragg's law, which connect the observed scattering with reflections from evenly spaced planes within the crystal. 53
- 54. 𝜆: wavelength 𝑑: distance in unit cell 𝜃: angle between beam and surface in unit cell The equation is known as Bragg's law, also n is the order of reflection which may be integer (1,2,3....) consistent with not exceeding unity. Bragg's law 54
- 55. The magnitude of the distance between two adjacent and parallel planes of atom (i.e.., interplanar spacing 𝑑ℎ𝑘𝑙 ). Is a function of the Miller indices (h, k and l )as well as the lattice parameter for example , for crystal structure that have cubic symmetry 𝑑ℎ𝑘𝑙 = 𝑎 ℎ2 + 𝑘2 + 𝑙2 In which a is the lattice parameter (unit cell edge length ). Relationships similar to equation above 55
- 56. 56 Thank you for your attention !!! 감사합니다 !!! Department of Materials Science and Engineering Souphanouvong University Faculty of Engineering Next Chapter is Crystal Metals and Crystal Imperfections