che294f04hmwrk12soln - CHE/MAE 294 F04 Homework 12 Solution...

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CHE/MAE 294 F04 Homework 12 Solution 22.Nov.04 > restart: Problem 1 Formulation Fermi distribution function > eqnFermiDist := proc(DEeV,TC) 1/(1 + exp(DEeV*q/(kb*(TC+273.15)))) end proc: Givens and Solution > readmylib(pchemconstants): Table 13.1.1 Fractional occupancies of levels at different energies above Fermi energy for different temperatures. DEeV ( ) f 25 oC ( ) f 100 oC .0100 .4039 .4228 .1000 .0199 .0427 1.0000 .1249 10 -16 .3119 10 -13 Problem 2 a) Graph Calibration Linearize the scales and report the measurements 15 - 30 n mohm is equivalent to 51 mm at 0 wt% impurities, resistivity = 17.24 n mohm 0.040 wt% P/Cu has a resistivity at 26.0 mm 0.10 wt% Al/Cu has a resistivity at 12.5 mm Givens Resistivity coefficients for pure (annealed) Cu > rho_oCu := 17.24E-9: alpha_Cu := 0.00393:
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Calculations Determine resistivities at a given wt% impurity by measuring values from ordinate (y-axis) > rhom := proc(di) 15E-9 + (15E-9)*di/51 end proc: > rho04PCu := rhom(26); rho10AlCu := rhom(12.5); := rho04PCu .2264705882 10 -7 := rho10AlCu .1867647059 10 -7 Determine the slope of the lines in units of mohm/wt% impurity > slopem := proc(rhom,wtp) (rhom - rho_oCu)/(wtp) end proc: > slopePCu := slopem(rho04PCu,0.04); slopeAlCu := slopem(rho10AlCu,0.10); := slopePCu .1351764705 10 -6 := slopeAlCu .1436470590 10 -7 Solution The equation for resistivity of alloy ρ o ( ) + 1 α ( ) - T To ( ) + 1 β ϖ at 20oC becomes ρ o ( ) + 1 β ϖ , and the slope of a plot of ρ
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