mat33hw11 - 2 The conductivity of a metal decreases when...

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Mat. 33 Homework # 11 4/20/09 1) i) All solute atoms are dissolved to form a single phase solid solution of type α through a process known as solution heat treatment. In the next step, which is known as precipitation heat treatment, the supersaturated α solid solution is heated to an intermediate temperature within the α+β two- phase region where temperature diffusion rates are more appreciable. The β precipitate phase begins to form as finely dispersed particles of composition C β . The alloy is then cooled to room temperature. ii) For the first step, solution heat treatment, the solution should be brought to around 550 o C (give or take 50 o C). The temperature range for the second step, precipitation heat treatment, is between 400 and 450 o C. iii) It is not possible to age harden an alloy of composition Al-0.5 wt.% Cu because all there is is α , so you will not have a precipitation affect.
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Unformatted text preview: 2) The conductivity of a metal decreases when temperature is increased while the conductivity of an intrinsic semiconductor increases because of the thermal affect; metals are great conductors of heat with a large number of free electrons, making it harder and harder to conduct electricity as the temperature is increased. 3) A) n = 1.3N Ag = 1.3(10.5)(6.023*10 23 )/107.87 = 7.62E22 cm-3 = 7.62E28 m-3 B) μ e = σ/ (n|e|) = 6.8E7/(7.62E28*1.602E-19) = 5.57E-3 m 2 /V-s 4) A) n = ( σ-p|e| μ h )/(|e| μ e ) = [5.93E-3 – (7E17)(1.602E-19)(0.05)]/(1.602E-19 *0.14) = 2.5E17m-3 B) p-type extrinsic because p is greater than n 5) σ = n|e| μ e = 5E17(1.602E-19)(350/1000) = 0.028035(omega-m)-1 6) A) metal electrical insulator undoped semiconductor donor doped semiconductor B)...
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