MSE201_Spring2009_Homework1_SOLN

MSE201_Spring2009_Homework1_SOLN - MSE 201: Homework No. 1...

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MSE 201: omework No 1 Homework No. 1 Solutions MSE 201 Spring 2009
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Problem 2.2 Next logical element in series is Sn [Group IVB] Si Ge Sn (Z = 50 electrons) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 5s 2 5p 2 {Valence = 4} MSE 201 Spring 2009
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Problem 2.14 Consider character of atomic constituents: Element Electronegativity Number Valence Electrons 44 O 3.44 6 Na 0.93 1 F3 . 9 8 7 78 In 1.78 3 P2 . 1 9 5 Ge 2.01 4 31 Mg 1.31 2 Ca 1.00 2 Si 1.90 4 MSE 201 Spring 2009 C 2.55 4 H2 . 1 0 1
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Problem 2.14 (cont.) Primary bonding for various materials Compound Character of constituents proposed bonding O 2 electronegative element covalent NaF Δ EN = 3.05; 90% ionic ionic P N 0 41 4% i i lt InP Δ EN = 0.41; 4% ionic electropositive + electronegative atoms covalent Ge not strongly electropositive or covalent gy p electronegative Mg electropositive metallic CaF 2 Δ EN = 2.98; 89% ionic ionic SiC both with N V = 4 covalent [CH 2 ] n C-H forms covalent bonds covalent MgO Δ EN = 2.13; 68% ionic ionic MSE 201 Spring 2009 CaO Δ EN = 2.44; 77% ionic ionic
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Problem 2.23 aterial isplays ronger ond rce, Material A displays a stronger bond force, and a closer equilibrium separation distance. In addition, the slope of the force-distance curve is steeper for A, ti t ti f t suggesting that the separation of atoms with applied force (or increased temperature) is more difficult to accopmlish with Material A. It is most likely Material A has a higher melting temperature than Material B. MSE 201 Spring 2009
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Problem 2.24 The thermal expansion coefficient reflects the relative ease or difficulty associated with separating constituent atoms with increased temperature. wer TE ggests reater ifficulty parating e toms nd A lower CTE suggests greater difficulty in separating the atoms, and a correspondingly higher bond energy that must be overcome to achieve melting. On this basis, one would expect a higher melting temperature or SiC. f MSE 201 Spring 2009
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Problem 2.29 x x ) ( o o o T T x = α For Titanium , x o = 2.94A (25 ° C) α = 9 x 10 -6 ° C -1 x (625 ° C) = 2.9559A; % increase is 0.54% MSE 201 Spring 2009
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Problem 2.29
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This note was uploaded on 12/01/2010 for the course MSE 201 taught by Professor Staff during the Spring '08 term at Kentucky.

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MSE201_Spring2009_Homework1_SOLN - MSE 201: Homework No. 1...

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