# HW1 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

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14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK 01 Question 2.14: The net potential energy between two adjacent ions, E N , may be represented by the sum of Equations 2.8 and 2.9; that is, E N = A r B r n Calculate the bonding energy E 0 in terms of the parameters A, B, and n using the following procedure: 1. Differentiate E N with respect to r, and then set the resulting expression equal to zero, since the curve of E N versus r is a minimum at E 0 . 2. Solve for r in terms of A, B, and n, which yields r 0 , the equilibrium interionic spacing. 3. Determine the expression for E 0 by substitution of r 0 into Equation 2.11. Solution of 2.14: a) Differentiation of Equation 2.11 yields dE N dr = d A r dr d B r n dr = A r (1 + 1) nB r ( n + 1) = 0 b) Now, solving for r (= r 0 ) A r 0 2 = nB r 0 ( n + 1) or r 0 = A nB 1/(1 - n ) c) Substitution for r 0 into Equation 2.11 and solving for E (= E 0 ) E 0 = A r 0 + B r 0 n

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= A A nB 1/(1 - n ) + B A nB n /(1 - n )
Question 2.15: For a K + –Cl ion pair, attractive and repulsive energies E A and E R , respectively, depend on the distance between the ions r, according to E A   1.436 r E R 5.8 10 6 r 9 For these expressions, energies are expressed in electron volts per K + –Cl pair, and r is the distance in nanometers. The net energy E N is just the sum of the two expressions above. (a) Superimpose on a single plot E N , E R , and E A versus r up to 1.0 nm. (b) On the basis of this plot, determine (i) the equilibrium spacing r

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## This note was uploaded on 09/11/2011 for the course ENG 407 taught by Professor Tsakal during the Spring '11 term at Rutgers.

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HW1 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

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