chap2_5

chap2_5 - 材材材材材材 Fundamental of Materials Prof:...

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Unformatted text preview: 材材材材材材 Fundamental of Materials Prof: Tian Min Bo Prof: Tian Tel: 62795426 材 62772851 E-mail: tmb@mail.tsinghua.edu.cn tmb@mail.tsinghua.edu.cn Department of Material Science and Engineering Tsinghua University. Beijing 100084 Lesson five §2.5 Indices of crystal planes and directions §2.5 Ⅰ.What’s crystal planes and directions? The atomic planes and The directions passing through the crystal are called (crystal) planes and directions respectively. respectively. Ⅱ. Plane indices 1. Steps to determinate the plane indices: 1. Establish a set of coordinate axes Find the intercepts of the planes to be indexed on a, b and c axes (x, y, z). c z a x y b Take the reciprocals of the intercepts 1/x, 1/y, 1/z. Clear fractions but do not reduce to lowest integers. Enclose them in parentheses, (h k l) Example: 1/2,1,2/3 1/2,1,2/3 2,1,3/2 (423) Plane indices referred to three axes a, b and c Plane and are also called Miller Indices. are Several important aspects of the Miller indices for planes should be noted: Planes and their negatives are identical. (020) = (0 Therefore .20) Planes and their multiples are not identical. In cubic systems, a direction that has the same indices as a plane is perpendicular to that plane. 2. The important planes in cubic crystals 2. (001) (110) (111) (112) 3. A family of planes consists of equivalent planes so far as the atom arrangement is concerned. concerned. {110} = (110) + ( 1 10) + (101) + (10 1 ) + (011) + (0 1 1) Total: 6 {111} = (111) + ( 1 11) + (1 1 1) + (11 1 ) Total: 4 {112} = (112) + ( 1 12) + (1 1 2) + (112 ) + (121) + ( 1 21) + (1 21) + (12 1 ) + (211) + ( 2 11) + (2 1 1) + (21 1 ) Total: 12 {123} = (123) + ( 1 23) + (12 3) + (12 3) + (132) + ( 1 32) + (1 3 2) + (132) + (231) + ( 2 31) + (2 3 1) + (23 1 ) + (213) + ( 213) + (2 1 3) + (21 3) + (312) + ( 3 12) + (3 1 2) + (312) + (321) + ( 3 21) + (321) + (32 1 ) Total: 4×3 ! =24 Ⅲ. Direction Indices 1. Derivation for the crystallographic direction a. b. c. d. As the first above, set the origin on the direction to be indexed. Find the coordinates of another point on the direction in questions. Reduce to three smallest integers: u, v, w. Enclose in square brackets [u v w]. 2. The important direction in cubic crystals: 2. <100> : crystal axes <110> : face diagonal <111> : body diagonal <112> : apices to opposite face-centers 3. Family of directions consists of Family crystallographically equivalent directions, denoted <u v w> denoted e.g. < 100 >= [100] + [010] + [001] + [ 1 00] + [0 1 0] + [00 1 ] §2.6 Hexagonal axes for hexagonal crystals crystals Ⅰ. Why choose four-axis system? Four indices has been devised for hexagonal unit cells because of the unique symmetry of the unique system. system. c (1 1 0) (100) b a [100] [110] Ⅱ. Plane indices (hkil) It can be proved: i ≡ 材 (h 材 k) (100) → (10 1 0) (1 1 0) → (1 1 00) Important planes : Important planes : c (10 1 2) (0001) a3 (11 20) a2 a1 (10 1 1) (10 1 0) Ⅲ. Direction indices [ u v t w ] Direction To make the indices unique, an additional To condition is imposed. ---- Let t 材材 (u 材 v) condition [0001] [ 1 011] Important directions [01 1 0] [ 2 1 1 0] Transformation of indices Transformation Transformation of 3 to 4 indices, or vice versa. Suppose we have a vector, whose 3 indices [u v w], and 4 indices [u v t w]. have We have Since L = ua1 + va2 + ta3 + wc = Ua1 + Va2 + Wc a3 = − (a1 + a2 ) t = − (u + v) ∴ ua1 + va 2 +(u + v)(a1 + a2 ) + wc = Ua1 + Va2 + Wc (2u + v)a1 + (u + 2v)a 2 +wc = Ua1 + Va2 + Wc ∴ U = 2u + v V = 2v + u u = 1 3 (2U − V ) v = 1 3 (2V − U ) or: or: w =W t = −(u + v) W =w For example: For 2 u= 3 [100] : 1 v=− 3 1 t=− 3 w=0 ⇒ [2 1 1 0] Examples and Discussions 1. Quick way for indexing the directions in cubic 1. crystals: crystals: The value of a direction depends on its feature while the sign on direction. while 2. The coordinate origin can be set arbitrarily 2. (for example on apices, body-center, face(for centers etc.), but never on plane in centers questions, otherwise the intercepts would be 0,0,0 . 0,0,0 3. The coordinate system can be transferred arbitrarily, but rotation is forbidden. arbitrarily, c c´ (11 1 ) b ( 1 1 1) a a´ b´ 4. The atomic arrangement and planar density of the 4. important direction in cubic crystal. important plane indices BCC FCC atomic atomic planar density arrangement arrangement planar density a 1 4= 1 a2 a2 1 4× +1 2 4 =2 a2 a 2a 1 4× +1 1.4 4 =2 a 2a 2 {100} a a {110} {111} 2a 2a 2a 4× 1 6 = 0.58 a2 32 a 2 3× a a a 2a 2a 2a 2a 1 1 4× + 2× 4 2 = 1 .4 a2 2a 2 1 1 3× + 3× 6 2 = 2.3 a2 32 a 2 5. The atomic arrangement and linear density of the 5. important direction in cubic crystal. important linear indices <100> <110> <111> BCC atomic arrangement FCC linear density 1 2=1 a a 2× a 2a 3a 1 2 = 0 .7 a 2a 2× 1 2× +1 1.16 2 = a 3a atomic arrangement linear density 1 2=1 a a 2× a 2a 3a 1 2× +1 1.4 2 = a 2a 1 2 = 0.58 a 3a 2× Exercise Exercise 1. Calculate the planar density and planar packing fraction for the (010) and (020) planes cubic polonium, which has a lattice parameter of 0.334nm. Solution atom per face 1 planar density (010) = = area of face 0.334 2 = 8.96 ×1014 atoms/cm 2 area of atoms per face packing fraction (010) = area of face 1× (πr 2 ) = = 0.79 2 ( 2r ) However, no atoms are centered on the (020) planes. There fore, the planar density and the planar packing fraction are both zero. Thank you ! Thank you ! 5 ...
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