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CENG151 Introduction to Materials Science and Selection Tutorial 1 14th September, 2007 Teaching Assistants Candy Lin Rm 7250 Email: [email protected] Bryan Wei Rm 7111 Email: [email protected] Crystal In chemistry and mineralogy, a crystal is a solid in which the constituent atoms, molecules, or ions are packed in a regularly ordered, repeating pattern extending in all three spatial dimensions. Insulin crystals Synthetic bismuth crystal Gallium, a metal that easily forms large single crystals Quartz crystal 7 crystal system &14 Bravais lattice triclinic monoclinic orthorhombic cubic hexagonal rhombohedral tetragonal In all, there are 14 possible Bravais lattices that fill 3D space, and they classified by 7 kind of crystal system. Comparing Different Atomic Arrangement Different atom arrangement can alter chemical and physical properties. Carbon is a very good example of this! Carbon - Graphite Large parallel sheets of hexagonal rings. The sheets are held together by weak Van der Waals forces. Can be broken easily. Because the layers of carbon rings can rub over each other, graphite is a good lubricant. Carbon - Diamond Each carbon atom covalently bonded to four other carbons in a tetrahedral arrangement to form a tight tetrahedron lattice. Very strong bonds strongest natural material on earth. (a) Tetrahedron (b) The diamond cubic (DC) unit cell. (c) 2003 Brooks/Cole Publishing / Thomson Learning Diamond can be cleaved along its planes, but it cannot flake apart into layers because of this tetrahedral arrangement of carbons. Carbon Buckminsterfullerene The 1996 Nobel Prize in Chemistry has been awarded to three chemists (R. Smalley, R. Curl, H. Kroto) for discovery of fullerenes. Shaped like a soccer ball, with 20 hexagons and 12 pentagons. A total of 60 carbons in this molecules. Molecule is extremely stable and can withstand very high temperatures and pressures. Can react with other atoms and molecules. Packing Factor (Packing Density) Packing – geometrical arrangement of these (atomic) spheres. Packing Factor/Density – proportion of space occupied by these solid spheres in a unit cell. Determining the Relationship between Atomic Radius and Lattice Parameters SC, BCC, and FCC structures when one atom is located at each lattice point. (c) 2003 Brooks/Cole Publishing / Thomson Learning™ The relationships between the atomic radius and the Lattice parameter in cubic systems. Atoms touch along the edge of the cube in SC structures. a0 2r In BCC structures, atoms touch along the body diagonal. There are two atomic radii from the center atom and one atomic radius from each of the corner atoms on the body diagonal, so a0 4r 3 In FCC structures, atoms touch along the face diagonal of the cube. There are four atomic radii along this length—two radii from the face-centered atom and one radius from each corner, so: a0 4r 2 Packing Factor Packing Factor – the fraction of space occupied by atoms, assuming that atoms are hard spheres sized so that they touch their closest neighbor. number of atoms / cell volume of each atom Packing factor volume of unit cell number of atoms / cell atomic mass Density (of a material) – can be calculated Packing density properties of the crystal structure: volume of unit cell Avogadro' s number using number of atoms / cell atomic mass Density volume of unit cell Avogadro' s number Coordination Number = no. of touching neighbors HCP and FCC Greatest packing density obtainable (closest packed) with spheres of equal size is 74%. 1 atom surrounded by 12 others of equal size, all touching the reference sphere along the diameter. 2 related crystal structures obtain this packing density: Hexagonal close packed (HCP) Face centered cubic (FCC) Lattice Positions Coordinates of selected points in the unit cell. The number refers to the distance from the origin in terms of lattice parameters. In Class Exercise Try to find A, B, C and D. Answers A = 1, 0, 0 B = 0, 0, 1 C = ½, 1, 0 D = 1, 1, 0 Miller Indices for directions Miller Indices are shorthand notations used to describe directions. Follow these rules: Using right-handed coordinate system, determine the coordinates of the two points that lie on the direction. Substract the coordinates of the “tail” point from the coordinates of the “head” point. “head” – “tail” Clear fractions and/or reduce the results obtained from the substraction to lowest integers. Enclose the numbers in square brackets []. If negative sign is produced, put a bar over the number. Lattice Directions Example: Determine the direction of A in the figure. 1. 2. 3. 4. (c) 2003 Brooks/Cole Publishing / Thomson Learning™ Direction A Two points are 1, 0, 0, and 0, 0, 0 1, 0, 0, - 0, 0, 0 = 1, 0, 0 No fractions to clear or integers to reduce [100] Lattice Directions Determine the direction of B in the figure. 1. 2. 3. 4. (c) 2003 Brooks/Cole Publishing / Thomson Learning™ Direction B Two points are 1, 1, 1, and 0, 0, 0 1, 1, 1, - 0, 0, 0 = 1, 1, 1 No fractions to clear or integers to reduce [111] Lattice Directions Determine the direction of C in the figure. Direction C 1. Two points are 0, 0, 1, and ½, 1, 0 2. 0, 0, 1, - ½, 1, 0 = -½, -1, 1 3. Fractions for clearing: 2(-½, -1, 1) = -1, -2, 2 (c) 2003 Brooks/Cole Publishing / Thomson Learning™ 4. [ 1 22] Exercise: Try to determine the directions for A, B, C, and D. (c) 2003 Brooks/Cole Publishing / Thomson Learning A: 0,0,1 – 1,0,0 = -1,0,1 = [101] B: 1,0,1 – ½,1,0 = ½,-1,1 = [122] C: 1,0,0 – 0,¾,1 = 1,-¾,-1 = [434] D: 0,1,½ – 1,0,0 = -1,1,½ = [221] Miller Indices for planes Miller Indices are used to describe planes as well. Follow these rules: Identify the points at which the plane intercepts the x, y, z coordinates. If the plane passes through the origin, the origin must be moved! Take reciprocals of these intercepts. Clear fractions and do NOT reduce to lowest integers. Enclose the numbers in parentheses (). If negative sign is produced, put a bar over the number, like before. Lattice Planes Intercepting points at: x= ∞, y=1, z= ∞ Take reciprocals: x=0, y=1, z=0 No fractions to clear, Enclose numbers in parentheses: (010) Intercepting points at: x=-1, y= ∞, z= ∞ Take reciprocals: x=1, y=0, z=0 No fractions to clear, Enclose numbers in parentheses: (100) Lattice positions no brackets Lattice directions square brackets Lattice planes parentheses Plane 1. x = 1, y = 1, z = ∞ 2.1/x = 1, 1/y = 1, 1 /z = 0 3. No fractions to clear 4. (110) Plane 1. X = ⅔, y = 1, z = ∞ 2. 1/x = 3/2, 1/y=1, 1/z=0 3. Clear fractions: 1/x = 3, 1/y = 2, 1/z = 0 4. (320) Plane (origin in blue) 1. x = 1, y = 1, z = ∞ 2.1/x = 1, 1/y = 1, 1/z = 0 3. No fractions to clear 4. (110) Plane (origin in red) 1. x = -1, y = -1, z = ∞ 2.1/x = -1, 1/y = -1, 1/z = 0 3. No fractions to clear 4. (110) y Plane (origin in green) 1. x = -1, y = 1, z = ∞ 2.1/x = -1, 1/y = 1, 1/z = 0 3. No fractions to clear 4. (110) Plane (origin in orange) 1. x = 1, y = -1, z = ∞ 2.1/x = 1, 1/y = -1, 1/z = 0 3. No fractions to clear 4. (110) Exercise: Determining Miller Indices of Planes Determine the Miller indices of planes A, B, and C in Figure. Plane A 1. x = 1, y = 1, z = 1 2.1/x = 1, 1/y = 1, 1/z = 1 3. No fractions to clear 4. (111) Plane B 1. The plane never intercepts the z axis, so x = 1, y = 2, and z =∞ 2.1/x = 1, 1/y =1/2, 1/z = 0 3. Clear fractions: 1/x = 2, 1/y = 1, 1/z = 0 4. (210) Exercise: Determining Miller Indices of Planes Determine the Miller indices of planes A, B, and C in Figure. Plane C (origin in red) 1. We must move the origin, since the plane passes through 0, 0, 0. Let’s move the origin one lattice parameter in the y-direction. Then, x= ∞, y=-1, and z= ∞ 2. 1/x = 0, 1/y = -1, 1/z = 0 3. No fractions to clear. 4. (010) (c) 2003 Brooks/Cole Publishing / Thomson Learning™ More Exercise! Try these! (c) 2003 Brooks/Cole Publishing / Thomson Learning Plane A (origin in red) 1. x = 1, y = -1, z = 1 2.1/x = 1, 1/y = -1, 1/z = 1 3. No fractions to clear 4. (111) Plane B (origin in orange) 1. x = ∞, y = ⅓, and z =∞ 2.1/x = 0, 1/y =3, 1/z = 0 3. No fractions to clear 4. (030) Plane C (origin in green) 1. 2. 3. Note: When origin in orange 4. x=1, y=∞, and z=-½ 1/x = 1, 1/y = 0, 1/z = -2 No fractions to clear. (102), (102) End of Tutorial 1 Thank you!