HW4Solutions - MSE280: Introduction to Engineering...

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MSE280: Introduction to Engineering Materials Homework 4: Solutions 1. Problem 5.2 from Callister (8 points) This problem calls for the computation of the activation energy for vacancy formation in silver. Upon examination of Equation 5.1, all parameters besides Q v are given except N , the total number of atomic sites. However, N is related to the density, ( ), Avogadro's number ( N A ), and the atomic weight ( A ) according to Equation 5.2 as N = N A ρ Pb A Pb = (6.02 × 10 23 atoms /mol)(9.5 g/cm 3 ) 107.9 g/mol = 5.30 × 10 22 atoms/cm 3 = 5.30 × 10 28 atoms/m 3 Now, taking natural logarithms of both sides of Equation 5.1, ln N v = ln N Q v kT and, after some algebraic manipulation Q v = kT ln N v N Now, inserting values for the parameters given in the problem statement leads to Q v = ( 8.62 × 10 -5 eV/atom- K ) (800 ° C + 273 K) ln 3.60 × 10 23 m 3 5.30 × 10 28 m 3 = 1.10 eV/atom Problem 5.5 from Callister In this problem we are asked to cite which of the elements listed form with Ni the three possible solid solution types. For complete substitutional solubility the following criteria must be met: 1) the difference in atomic radii between Ni and the other element ( R % ) must be less (8 points)
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than ±15%, 2) the crystal structures must be the same, 3) the electronegativities must be similar, and 4) the valences should be the same, or nearly the same. Below are tabulated, for
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This note was uploaded on 03/15/2010 for the course MSE 280 taught by Professor Johnson during the Fall '08 term at University of Illinois, Urbana Champaign.

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HW4Solutions - MSE280: Introduction to Engineering...

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