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Unformatted text preview: 11) planes. (a) The spacing between (100) planes is the lattice constant, or 0.528 nm. (b) The spacing between (110) planes is the lattice constant divided by 21/2, or 0.373 nm. (c) The spacing between (111) planes is the lattice constant divided by 31/2, or 0.305 nm. 1.19 The lattice constant of a single crystal is 0.473 nm. Calculate the surface density of of atoms on the (i) (100), (ii) (110), and (iii) (111) planes for (a) simple cubic, (b) body centered cubic, and (c) face centered cubic lattice. (i) For the cubic, and body centered cubic, the (100) plane has only 4 atoms in the surface unit cell. Each of these atoms is shared with 4 neighboring cells, so tha there is 1 atom per surface cell. Hence 1
ρC (100 ) = ρ BC (100 ) =
= 4.47...
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 Fall '14

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