HW2Solutions - MSE280B: Introduction to Engineering...

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MSE280B: Introduction to Engineering Materials Homework 2: Solutions 1. (a) For indium, and from the definition of the APF Problem 3.8 from Callister (6 points) APF = V S V C = n 4 3 π R 3 a 2 c we may solve for the number of atoms per unit cell, n , as n = (APF) a 2 c 4 3 π R 3 = (0.693)(4.59 × 10 -8 cm) 2 (4.95 × 10 -8 cm) 4 3 π (1.625 × 10 -8 cm ) 3 = 4.0 atoms/unit cell (b) In order to compute the density, we just employ Equation 3.5 as ρ = nA In a 2 c N A = (4 atoms/unit cell)(114.82 g/mol) (4.59 × 10 -8 cm) 2 (4.95 × 10 -8 cm)/unit cell [ ] (6.02 × 10 23 atoms/mol) = 7.31 g/cm 3 Problem 3.28 from Callister (6 points) Direction A is a [33 1 ] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system
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x y z Projections a b c 3 Projections in terms of a , b , and c 1 1 1 3 Reduction to integers 3 3 –1 Enclosure [33 1 ] Direction B is a [ 4 0 3 ] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x y z Projections 2 a 3 0 b c 2 Projections in terms of a , b , and c 2 3 0 1 2 Reduction to integers –4 0 –3 Enclosure [ 4 0 3 ] Direction C is a [ 3 61] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x y z Projections
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HW2Solutions - MSE280B: Introduction to Engineering...

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