CME 260 Midterm1_070815_solution

CME 260 Midterm1_070815_solution - CME 260 Mid-Term 1 Name...

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CME 260 Mid-Term 1 07/08/15 Name: ____________________________________ 1. List four major classes of Materials (8 points) Metals, ceramics, polymers, and composites 2. Calculate the percent ionic character of the following: (12 points) a) GaAs %IC = 1- exp [-(X A -X B ) 2 /4] = 3.92 % b) CH 4 3.92 % c) NaCl %IC = 1- exp [-(X A -X B ) 2 /4] = 59.4 % d) Describe the bonding nature of GaAs The nature of the bonding is the same as that of methane (CH 4 ), therefore, it is covalent. 3. Give the electron configurations for: (10 points) a) Te 2- 52+2 = 54 electrons [Xe] or 1S 2 2S 2 2P 6 3S 2 3P 6 4S 2 3d 10 4P 6 5S 2 4d 10 5P 6 b) Pt 2+ 78-2 = 76 electrons [Xe]6S 2 4f 14 5d 6 4. Determine the index of the following: (10 points) a) We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows: x 1 = 0 a y 1 = 0 b z 1 = 2 c/ 3 Whereas head coordinates are as follows: x 2 = a y 2 = b z 2 = c/ 3 From Equations 3.10a, 3.10b, and 3.10c assuming a value of 3 for the parameter n 1
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u n x 2 x 1 a (3) a 0 a a 3 v n y 2 y 1 b (3) b 0 b b 3 w n z 2 z 1 c (3) c /3 2 c /3 c   1 Therefore, Direction A is [33 1] . b) We determine the indices of this direction vector using Equations 3.10a-3.10c—that is, by subtracting vector tail coordinates from head coordinates. Tail coordinates are as follows: x 1 = a y 1 = 0 b z 1 = c/ 3 Whereas head coordinates are as follows: x 2 = a/ 2 y 2 = b z 2 = c/ 2 From Equations 3.10a, 3.10b, and 3.10c assuming a value of 3 for the parameter n u n x 2 x 1 a (3) a /2 a a   3 2 v n y 2 y 1 b (3) b 0 b b 3 2
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w n z 2 z 1 c
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