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**Unformatted text preview: **2 ) 2 = 60 o (1 0 0) (110) Area = a x a = a 2 No of atoms = ¼ x 4 = 1 Planar density = 1/a 2 Area = a x a = a 2 No of atoms = (¼ x 4) + 1 = 2 Planar density = 2/a 2 = /a 2 Atoms at (1,0,0) (0,1,0) (1,0,1) (0,1,1) ( ½, ½, ½) 6.) Calculate the radius of an iridium atom, given that Ir has an FCC crystal structure, a density of 22.4g/cm 3 , and an atomic weight of 192.2 gmol-1 . For FCC, n = 4 atoms/unit cell, and V C = 16 R 3 2 . Now, ρ = nA Ir V C N A And solving for R from the above two expressions yields R = nA Ir 16 ρ N A 2 1/3 = 4 atoms/unit cell ( 29 192.2 g/mol ( 29 2 ( 29 16 ( 29 22.4 g/cm 3 ( 29 6.023 x 10 23 atoms/mol ( 29 1/3 = 1.36 x 10-8 cm = 0.136 nm...

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- Fall '07
- keith
- Vectors, Dot Product, Crystallography, Crystal, Crystal system, Atomic packing factor