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Unformatted text preview: u=1/3(2u’v’)=1 v=1/3(2v’u’)=0 t=(u+v)=1 w=w’=0 [1010] (0110) 2. Derive planar density expression for Face Centered Cubic (110) plane in terms of the atomic radius R. Note: Sketch FCC unit cell and (110) plane. Label all relevant distances! PD=(number of atoms per unit area)/(area of the unit area) d=4R=a √ 2 → a =2R/ √ 2 unit area A=d· a =2R·4R/ √ 2 = 8R 2 √ 2 PD= 2 atoms / 8R 2 √ 2 = 1 atom/ 4R 2 √ 2 Diagonal of the cube: 3 a D = a=length of the cube’s edge Diagonal of the square: 2 a d = a=length of the square’s side 1 nm = 109 m, 1 m = 10 2 cm N A =6.23 x 10 23 atom/mol 1/2 a a a a a a z z z z x x y y x’ y’ 1 point 1 point 1 point 1 point 1 point a d 1 point for FCC 1 point for (110) 1 point for unit area 1 point for PD 1 point for # of atoms...
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 Fall '08
 RADOVIC
 Crystallography, Max von Laue, a2 a3 a1

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