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Unformatted text preview: respectively. 4. What is the activation ener~ for diffusion if the diffusion rate doubles between 100°C and 120°C? The gas constant is 8.31 J mole.- K-1 5, D = Doe-Q/RT Dl = Doe-Q/S.3l (373) Solving Q = 42.2kJ/mole There is one activation energy quoted for self diffusion in FCC aluminum, so how many are quoted for self diffusion in a) diamond, b) graphite? 6. I expect you to recognize that diamond has a ~ crystal structure and graphite a hexagonal structure, because I have used them often for illustrations of properties. Here diamond, a = b = c, need 1 Here graphite, a = a * c, need 2...
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- Fall '11
- Atom, Crystal, 3L, 7.5 W, 0.484 nm