solutiontosamplequestion7

solutiontosamplequestion7 - t 1 2245 0 t 2 2245 1 t 3<<...

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Solution to Problem #7 of Sample Exam Questions 7. In CaF 2 , the predominant ionic defects are the anti-Frenkel defects whose concentrations are controlled from the following defect reaction: F F × V F + F i ' and the equilibrium constant is given by K 1 = V F [] F i ' [] The fluoridation reaction is 1 2 F 2 = F i ' + h with the equilibrium constant given by K 2 = F i ' [] h [] p F 2 12 The electrons and holes are also at equilibrium with a equilibrium constant given by K 3 = e ' [] h [] From class, we know, k t 1 + t 2 (29 t 3 σ tot where t 1 and t 2 are the transference numbers for cation and anion, respectively, t 3 is the transference number for electronic defects, and σ tot is the total conductivity. From the problem, we get
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Unformatted text preview: t 1 2245 0, t 2 2245 1, t 3 << 1. Hence, k ∝ t 3 tot = 3 Since electrons are the major electronic defect, the main contribution to σ 3 is by electrons. Therefore, in order to reduce the rate of fluoridation, we have to reduce the electron concentration in CaF 2 . For this purpose, we may dope lower valence cations, for example, NaF. The doping reaction equation is given by NaF → Na Ca ' + V F • + F F × From the equilibrium constants, K 1 , K 2 and K 3 , we can see that NaF doping results in V F • [ ] ↑ ⇒ F i ' [ ] ↓ ⇒ h • [ ] ↑ ⇒ e ' [ ] ↓...
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