Chap 03 Solns-6E

# Chap 03 Solns-6E - CHAPTER 3 THE STRUCTURE OF CRYSTALLINE...

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CHAPTER 3 THE STRUCTURE OF CRYSTALLINE SOLIDS PROBLEM SOLUTIONS 3.1 Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the constituent electrons. On the other hand, crystal structure pertains to the arrangement of atoms in the crystalline solid material. 3.2 A crystal structure is described by both the geometry of, and atomic arrangements within, the unit cell, whereas a crystal system is described only in terms of the unit cell geometry. For example, face-centered cubic and body-centered cubic are crystal structures that belong to the cubic crystal system. 3.3 For this problem, we are asked to calculate the volume of a unit cell of aluminum. Aluminum has an FCC crystal structure (Table 3.1). The FCC unit cell volume may be computed from Equation (3.4) as V C = 16R 3 2 = (16) 0.143 x 10 -9 m ( ) 3 2 = 6.62 x 10 -29 m 3 3.4 This problem calls for a demonstration of the relationship a = 4 R 3 for BCC. Consider the BCC unit cell shown below Using the triangle NOP 1

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(NP ) 2 = a 2 + a 2 =2a 2 And then for triangle NPQ , (NQ ) 2 =(QP ) 2 +(NP ) 2 But NQ = 4 R , R being the atomic radius. Also, QP = a . Therefore, (4R) 2 2 + 2a 2 , or a = 4R 3 3.5 We are asked to show that the ideal c / a ratio for HCP is 1.633. A sketch of one-third of an HCP unit cell is shown below. Consider the tetrahedron labeled as JKLM , which is reconstructed as 2
The atom at point M is midway between the top and bottom faces of the unit cell--that is MH = c /2. And, since atoms at points J , K , and M , all touch one another, JM =JK =2R=a where R is the atomic radius. Furthermore, from triangle JHM , (JM ) 2 =(JH ) 2 + (MH ) 2 ,or a 2 = (JH ) 2 + c 2 2 Now, we can determine the JH length by consideration of triangle JKL , which is an equilateral triangle, cos 30 ° = a/2 JH 3 2 , and JH a 3 Substituting this value for JH in the above expression yields a 2 a 3 2 + c 2 2 a 2 3 c 2 4 and, solving for c / a 3

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c a = 8 3 = 1.633 3.6 We are asked to show that the atomic packing factor for BCC is 0.68. The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or APF = V S V C Since there are two spheres associated with each unit cell for BCC V S = 2(sphere volume) = 2 4 π R 3 3 8 π R 3 3 Also, the unit cell has cubic symmetry, that is V C = a 3 . But a depends on R according to Equation (3.3), and V C = 4R 3 3 = 64R 3 33 Thus, 8 π R 3 /3 64R 3 /3 3 = 0.68 3.7 This problem calls for a demonstration that the APF for HCP is 0.74. Again, the APF is just the total sphere-unit cell volume ratio. For HCP, there are the equivalent of six spheres per unit cell, and thus V S = 6 4 π R 3 3 = 8 π R 3 Now, the unit cell volume is just the product of the base area times the cell height, c . This base area is just three times the area of the parallelepiped ACDE shown below.
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## This homework help was uploaded on 02/07/2008 for the course MATENG MAT201 taught by Professor Na during the Spring '08 term at Wisconsin Milwaukee.

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Chap 03 Solns-6E - CHAPTER 3 THE STRUCTURE OF CRYSTALLINE...

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