Bsurfacedensityns1002a26781014atomscm2ns11042a2

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Unformatted text preview: The atomic volume is Ω = a3 /nνc = 2.00 × 10−29 m3 ; density ρ = W/V = 2.33×103 kg/m3. b) Surface density NS,(100) = 2/a2 = 6.78×1014 atoms/cm2, NS,(110) = 4/√2a2 = 9.60×1014 atoms/cm2 , and NS,(111) = 4/√3a2 = 7.83×1014 atoms/cm2 . d) Height of a monolayer h(100) = a/4, h(110) = √2a/4, and h(111) = √3a/12 or √3a/4. Hint: Consider (111) planes like x + y + z = 1, x + y + z = 3/4 + 1/4 + 3/4 = 7/4, x + y + z = 2; while the total spacing is √3a/3. Homework #1 Solution EMSE 426, Spring 201...
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This note was uploaded on 03/04/2013 for the course MASE 426 taught by Professor Pirouz during the Spring '09 term at Case Western.

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