HW 2 Solutions

# HW 2 Solutions - Defects 1 Chung 2.5 As the temperature is...

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Defects: 1. Chung 2.5 As the temperature is raised from 800 to 1000K, the unit cell size of a given metal crystal (as measured by x-ray diffraction) increases 0.05% due to thermal expansion (i.e., increased equilibrium distance between atoms). In the same temperature range, the density decreases by 0.16%, due to the combination of thermal expansion and vacancy production. (a) What is the expected percentage decrease in density due to normal thermal expansion alone? Hint: When the unit cell size increases by x % (<<100%), the volume per unit cell increases by 3 x %. (b) Hence evaluate the net density decrease due to vacancy production alone. (c) If the vacancy concentration at 800K is 10 -6 , estimate the activation energy for vacancy production. Note: Depending on how a vacancy is produced, equation 2.18 should be written as exp vv NQ A N kT    where A is a constant. Therefore, to determine the value of Q , one needs to use data from two different temperatures (see Appendix of Chapter 2). Solution (a) An increase of 0.05% in the lattice constant decreases the density by 0.15% (as given by the hint above). (b) The net density decrease due to vacancy production = 0.16% - 0.15% = 0.01% (c) The vacancy concentration at 800 K = 10 -6 , while the vacancy concentration at 1000 K is 10 -4 (= 0.01%). We can write: 10 4 10 6 exp Q k B (1000) Q k B (800) , where Q is the activation energy for vacancy production, and k B Boltzmann constant. Taking natural log on both sides, we have: 4.605 Q k B 1 800 1 1000 Solving, Q = 1.59 eV Alternate solution: if Nv/N(800)/Nv/N(1000) = 0.0001, then Nv/N(1000) = 0.01, solving for Q yields 3.17eV

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2. Chung 2.6 A TiO 2 crystal is grown with some Cr 2 O 3 impurities. The valence of Ti is +4, and the valence of Cr is +3. Chromium ions occupy titanium sites substitutionally. Explain what type of vacany will be introduced. When a Cr 3+ sits on a Ti +4 that the site becomes relatively negative to what it would have been if a Ti +4 were occupying it. Therefore, you need a vacancy which is relatively positively charged to maintain neutrality. In this case you would create oxygen vacancies because if you remove something with a valence of -2 then the vacancy left behind has a relative charge of +2. For every one Cr 2 O 3 added to the system, two Cr go onto two Ti sites, three oxygen go onto oxygen sites, and one oxygen site remains empty. (since in TiO 2 you have two oxygen sites for every Ti site if you fill 2 Ti sites you have 4 associated oxygen sites.). Thus you have two Cr on Ti sites with a relative charge of -1 each and one oxygen vacancy with a relative charge of +2, so you maintain charge neutrality. If you were to create Ti vacancies, they would have a relative charge of -4 and would not help in your quest for neutrality.
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## This note was uploaded on 04/30/2008 for the course MATSCI 201 taught by Professor Chung during the Spring '08 term at Northwestern.

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HW 2 Solutions - Defects 1 Chung 2.5 As the temperature is...

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