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Unformatted text preview: EMSE 201 Introduction to Materials Science & Engineering Fall 2010 Case Western Reserve University 1 of 5 Case School of Engineering Solution to Homework #1 Explain how answers were obtained. Give appropriate units for all numerical answers. 1) (6 points) C&R Problem 3.3; but instead of computing the density of molybdenum from the atomic radius, compute the atomic radius of Mo from its density . Then do the same for a facecentered cubic metal of your choice, and for one of the diamond cubic elements (carbon diamond, silicon, or germanium). Solution Given information about a materials composition and crystal structure, its density equals the mass of atoms per unit cell divided by the volume of the unit cell (C&R eq. 3.5). Given that molybdenum has the bodycentered cubic structure (so the number of atoms per unit cell N C = 2 , and the cell volume V C = a 3 where a is the lattice parameter), atomic mass A of 95.94 g mol 1 , and a density of 10.22 g cm 3 (from the inside front cover of C&R), eq. 3.5 becomes (eq. 3.5) The only unknown in this equation is the lattice parameter a . But the question asks for the atomic radius of molybdenum, not the lattice parameter, so use the relationship between these two parameters that applies for the bodycentered cubic structure: (1 pt) Substituting this into eq. 3.5 and solving for r gives (1 pt) in agreement with the value given in the problem. To carry out this computation for a facecentered cubic metal, e.g. copper, N C = 4 and (1 pt) Substituting this into eq. 3.5, using values for the density and atomic mass of copper from the inside front cover of C&R, and solving for r gives (1 pt) in agreement with the value given in front inside cover of C&R. Lastly, for an element like silicon with the diamond cubic structure, N C = 8 and EMSE 201 Introduction to Materials Science & Engineering Fall 2010 Case Western Reserve University 2 of 5 Case School of Engineering (1 pt) Substituting this into eq. 3.5, using values for the density and atomic mass of silicon from the inside front cover of C&R, and solving for r gives (1 pt) again in agreement with the value given in front inside cover of C&R. again in agreement with the value given in front inside cover of C&R....
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This document was uploaded on 11/08/2011.
 Spring '09

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