HW2-Solution - 11$N# 11Wp, ii #'4/($1*'...

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11$N# 11 W p, ii , #'4/($1 *' '@MWliffy,,*,ii1l$1I$W""Yii i ~ IViViM ..2 wt§m ,Wf!WI*' i! /=/JAI$i i ;f:t, ,2 ~ S ~ ti~ ~ $I , .( ~ i 6 i! i@1),'W_A -. ~, ~ 1, Calculate (a) the equilibrium lumber of vacancies per cubic meter in pure copper at:;; 500°C and Cb) the vacancy fraction at 500°C in pure copper. Assume the energy of;i~ 1'1 formationofa vacancy in pure copper is 0.90 eV.UseEq. (4.10) with C = 1. (Boltzmanns .~ constant k :::= 8.62 X 10-5 eV/K) Solution: (a) The equilibrium number of vacancies per cubic meters in pure copper at SOoec is nu = Ne-E;,jki (assume C = 1) where nu = no. of vacandes/m3 N = no. of atom sites 1m3 Eo = energy of formation of a vacancy in pure copper at sooec I eV k = Bo!tzmann'sconstant T = temperature, K First, we determine a value for N by using the equation N = NoPcu at. mass Cu (4.11) where No = Avogadro's constant and PClt = density of Cu = 8.96 Mg/m3.Thus 6.02 x 1023atoms 1 8.96 x 106g N = x x at. mass 63.54 g/at. mass m3 = 8A9 x }O2$atoms/m3 Substitutingthe values of N, Et»k, and T into Eq. (4.10a) gives nv = Ne-t::JkT 28 { [ 0.90 eV ] } =: (8.49 x 10 ) exp .- (8.62 x }O-5 eV/K)(773 K) . ;;;:: (8.49 x 1028)(e"'13.5) = (8.49 x 1028)(1.37x 10. .6) = 1.2 x 1023vacancies/m3 ~ (4.100) Cb) The vacancy fraction in pure copper at 500"Cis found from the ratio
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This note was uploaded on 09/28/2010 for the course MECHENG 382 taught by Professor Thouless during the Winter '08 term at University of Michigan.

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HW2-Solution - 11$N# 11Wp, ii #'4/($1*'...

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