BacktestKey - Materials Science for Engineers Answer Key...

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Materials Science for Engineers ENGR 1600 Answer Key Prepared by Carl C. Hansen Final Exam, Spring 2004 Question 1: True or False 1. F 4. F 7. T 10. F 12. T 14. T 2. T 5. F 8. F 11. F 13. F 15. F 3. F 6. F 9. F (a) Potential Energy originates from positive nucleus repulsion when close, and nucleus/electron field attraction at farther separations. (b) (c) Young’s modulus is proportional to the second derivative (“curvature”) of the potential energy curve at its lowest point. The tighter the curvature, the higher Young’s modulus. (d) The thermal expansion coefficient is determined by how asymmetric the dip in the potential energy curve is. At one value for energy (a particular temperature), there are two radius values it oscillates between. The average of these two is the inter-atomic separation at that temperature. Question 3: Unit cells (a) - 1 of 7 - (0,0,0) (0,½,½) (½,½,0) (½,0,½) (½,½,1) (½,1,½) (0,1,0) (1,1,0) (1,½,½) (1,0,0) (1,0,1) (0,0,1) (1,1,1) (0,1,1)
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Materials Science for Engineers ENGR 1600 Answer Key Prepared by Carl C. Hansen Final Exam, Spring 2004 (b) (c) (111) (d) 12 (e) a=2r√2 r=a/(2√2) 0.128 nm (f) n = 4 atoms / unit cell, V = a 3 , M=63.5 g/mol, N A = 6.022*10 23 atoms/mol m = n*M/N A ρ = n* (M/N A ) / V V = (0.361 nm * (100 cm / 10 9 nm)) 3 = 4.705*10 –23 m = 4*63.5 g/mol / 6.022*10 23 atoms/mol = 4.218*10 –22 ρ = 8.965 g/cm 3 (g) yes ***Cu is FCC n*λ = 2d*sin(θ) assume n = 1 λ = 0.1542nm sin(θ) = ½ λ/d θ = sin –1 (½ λ/d) if it is FCC, we examine the plane (1 1 1) d = 3 a 2 2 2 2 2 2 1 1 1 a L k h a = = + + + + a = 0.361 nm d = 0.2084 nm θ = 21.71°, 2*θ = 43.42° 43.42° is given in the graph, therefore it is Copper Question 4: Polymers (a) - 2 of 7 -
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BacktestKey - Materials Science for Engineers Answer Key...

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