This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1 eV = 0.160 x 1018 J n/N = exp ( 0.68x0.160x1018 /13.8x1024 x 933) = 2.1x104 . 1. A crystal has an enthalpy of formation for vacancies of 1.5 eV, establishing a certain equilibrium concentration of vacancies at the temperature 1200 K. By how much does the temperature have to be raised to increase the vacancy concentration by a factor of 10? n/N = exp (1.5 x0.160x1018 /13.8x1024 x1200) = exp(14.49) = 5.094x107 10n/N = 5.094x106 = exp [1.5 x0.160x1018 /13.8x1024 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...
View
Full
Document
 Spring '08
 Tomozawa

Click to edit the document details