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Unformatted text preview: 1 eV = 0.160 x 10-18 J n/N = exp (- 0.68x0.160x10-18 /13.8x10-24 x 933) = 2.1x10-4 . 1. A crystal has an enthalpy of formation for vacancies of 1.5 eV, establishing a cer-tain equilibrium concentration of vacancies at the temperature 1200 K. By how much does the temperature have to be raised to increase the vacancy concentra-tion by a factor of 10? n/N = exp (-1.5 x0.160x10-18 /13.8x10-24 x1200) = exp(-14.49) = 5.094x10-7 10n/N = 5.094x10-6 = exp [-1.5 x0.160x10-18 /13.8x10-24 x (1200+t)] = exp [- 17391/(1200+t)]-12.187 = - 17391/(1200+t); 1200+t = 1427 t = 227 C 2. The figure on the next page represents a (100) section through the structure of CaO, which has the NaCl (rock salt) structure type. Five defects are shown in the structure. Write the defect symbols using the common convention, e.g. V Na V O .. A B . V ca , V O .. , F O . , Fe Ca . , Li Ca...
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- Spring '08