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Test1_Solutions

Test1_Solutions - MMAT 201-02 Materials Science and...

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MMAT 201-02 Materials Science and Engineering I Spring 2006 Test 1 Feb. 16, 2006 Name: KEY 1. For a K + -Cl - ion pair, the net energy is given by: 9 r B r A E N + = [in eV (per K + -Cl - pair)] where A =1.436 eV·nm and B =5.86x10 -6 eV·nm 9 . a) Determine the equilibrium interionic seperation and the bonding energy [ 15% ]. b) Calculate the percent ionic character of this bond given the electronegativities X K =0.8 and X Cl =3.0 [ 5% ]. ANSWER 1 (a) Differentiation of E N yields: dE N dr = A r (1 + 1) - nB r (n + 1) = 0 Now, solving for r (= r o ) A r o 2 = nB r o (n + 1) or r o = A nB 1/(1 - n) Substitution for r o in E N and solving for E (= E o ) E o = - A r o + B r o n = - A A nB 1/(1 - n) + B A nB n/(1 - n) Substituting values gives: r o = A nB 1/(1 - n) = 1.436 (9)(5.86 x 10 -6 ) 1/(1 - 9) = 0.279 nm E o = - 1.436 1.436 (9)(5.86 x 10 -6 ) 1/(1 - 9) + 5.86 x 10 -6 1.436 (9)(5.86 x 10 -6 ) 9/(1 - 9) = - 4.57 eV b) %IC= 18 . 70 100 ]} 4 / ) 3 8 . 0 ( exp[ 1 { 100 ]} 4 / ) ( exp[ 1 { 2 2 = = x x X X Cl K Page 1 of 5

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