HW Solutions Ch 1-3 2007

HW Solutions Ch 1-3 2007 - CHAPTER 1 1. Given a positive...

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CHAPTER 1 1. Given a positive and a negative ion, the attractive potential energy between these two ions is given by – A/R and the repulsive potential energy by B/R n , where R is the separation between the two ions. A , B and n are constants. The total potential energy U is therefore equal to: U = A R + B R n (a) When the system is at equilibrium, U is a minimum. Solve for the equilibrium separation between the two ions in terms of A , B and n . (b) Hence determine the binding energy for the ion pair at this equilibrium separation in terms of A , B and n . Solution When U is a minimum, dU/dR = 0 U R = A R 2 BnR n 1 = 0 A = BnR 1 n Solving, R = A Bn 1 1 n Substituting this value of R into the original expression for U, we have: U = A A Bn 1 1 n + B A Bn 1 1 n n 3. Make sketches to indicate the following planes in a cubic unit cell: a. (111) b. (200) c. (321) Solution
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a. b. c. 7. CsCl has a body-centered cubic structure. Consider the unit cell in which Cs + is at the center and Cl at the corners. Given that the radius for Cs + is 0.169 nm and that for Cl is 0.181 nm, calculate the packing factor for this unit cell. Note: The packing factor referred to in the text for the BCC structure is for the case when all atoms in the unit cell are identical. Here, the two ions have different sizes. Solution Define a = unit cell length, r Cs = radius of cesium ion, and r Cl = radius of chloride ion. We can write: a = 2 r Cs + 2 r Cl so that a = 2 r Cs + 2 r Cl Volume of unit cell = a 3 Volume of atoms = 4 3 π r Cs 3 + 4 3 r Cl 3 Therefore, packing factor (PF) 1/3 1/2 1/2 x z y 1 1 1 x z y 1 1 1 x z y 1 1 1
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= volumeof atoms volumeof unitcell = 4 3 π r Cs 3 +
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This note was uploaded on 11/21/2011 for the course MAT SCI 201 taught by Professor Matsci during the Spring '10 term at Northwestern.

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HW Solutions Ch 1-3 2007 - CHAPTER 1 1. Given a positive...

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