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HW 1 posted solutions - Ch 3 problems 1 3 7 12 23 27 29 36...

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Ch 3, problems 1, 3, 7, 12, 23, 27, 29, 36, 43, 44 3.1 If the atomic radius of lead is 0.175 nm, calculate the volume of its unit cell in cubic meters. Solution Lead has an FCC crystal structure (Table 3.1). The FCC unit cell volume may be computed from Equation 3.4 to be 3 -28 m 10 1.213 = × C V -------------------------------------------- 3.3 Molybdenum has a BCC crystal structure, an atomic radius of 0.1363 nm, and an atomic weight of 95.94 g/mol. Compute its theoretical density and compare it with the experimental value found inside the front cover. Solution This problem calls for a computation of the density of molybdenum. According to Equation 3.5 ρ = nA Mo V C N A For BCC, n = 2 atoms/unit cell, this results in [] ) ( ) ( atoms/mol 10 6.02 ) /( 3 / cm 10 0.1363 ) 4 ( g/mol) 4 cell)(95.9 atoms/unit 2 ( = 23 3 7 - × × cell unit = 10.22 g/cm 3 The value given inside the front cover is 10.22 g/cm 3 . -------------------------------------------
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3.7 Below are listed the atomic weight, density, and atomic radius for three hypothetical alloys. For each determine whether its crystal structure is FCC, BCC, or simple cubic and then justify your determination. A simple cubic unit cell is shown in Figure 3.42. Alloy Atomic Weight Density Atomic Radius ( g/mol ) ( g/cm 3 ) ( nm ) A 43.1 6.40 0.122 B 184.4 12.30 0.146 C 91.6 9.60 0.137 Solution For each of these three alloys we need, by trial and error, to calculate the density using Equation 3.5, and compare it to the value cited in the problem. For SC, BCC, and FCC crystal structures, the respective values of n are 1, 2, and 4, whereas the respective expressions for a (since V C = a 3 ) are 2 R , 2 R 2 , and 4 R 3 .
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This note was uploaded on 02/08/2012 for the course MAE 3324 taught by Professor Michaels during the Summer '10 term at UT Arlington.

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HW 1 posted solutions - Ch 3 problems 1 3 7 12 23 27 29 36...

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