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Chapters_Chapter_3_(Solutions)_Webct

Chapters_Chapter_3_(Solutions)_Webct - 3.5 We are asked to...

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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 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
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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 c a = 8 3 = 1.633 3.8 This problem calls for a computation of the density of iron. According to Equation (3.5) ρ = nA Fe V C N A For BCC, n = 2 atoms/unit cell, and V C = 4R 3 3 Thus,
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ρ = 2 atoms/unit cell ( 29 55.9 g/mol ( 29 4 ( 29 0.124 x 10 -7 cm ( 29 3 / 3 3 / unit cell ( 29 6.023 x 10 23 atoms/mol ( 29 = 7.90 g/cm 3 The value given inside the front cover is 7.87 g/cm 3 . 3.11 For the simple cubic crystal structure, the value of n in Equation (3.5) is unity since there is only a single atom associated with each unit cell. Furthermore, for the unit cell edge length, a = 2 R . Therefore, employment of Equation (3.5) yields ρ = nA V C N A = nA (2R ) 3 N A = (1 atom/unit cell)(74.5 g/mol) (2) 1.45 x 10 -8 cm ( 29 3 /(unit cell) 6.023 x 10 23 atoms/mol ( 29 5.07 g/cm 3 3.12. (a) The volume of the Ti unit cell may be computed using Equation (3.5) as V C = nA Ti ρ N A Now, for HCP, n = 6 atoms/unit cell, and for Ti, A Ti = 47.9 g/mol. Thus, V C = (6 atoms/unit cell)(47.9 g/mol) 4.51 g/cm 3 ( 29 6.023 x 10 23 atoms/mol ( 29 = 1.058 x 10 -22 cm 3 /unit cell = 1.058 x 10 -28 m 3 /unit cell (b) From the solution to Problem 3.7, since a = 2 R , then, for HCP V C = 3 3 a 2 c 2
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but, since c = 1.58 a V C = 3 3 (1.58)a 3 2 = 1.058 x 10 -22 cm 3
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