![]() ![]() Also, for this unit cell VC = (3.0 108 cm)2 (4.0 108 cm) = 3.60 1023 cm3/unit cell Thus, using Equation 3.5, the density is equal to = nA VC N A (2 atoms/unit cell)(141 g/mol) -23 (3.60 10 cm3/unit cell)(6.022 1023 atoms/mol) = 13.0 g/cm3 1 1 1 1 3 1 3.25 Sketch a tetragonal unit cell, and within that cell indicate locations of the and point 2 2 coordinates. (b) The crystal structure would be called body-centered tetragonal. Solution (a) The unit cell shown in the problem statement belongs to the tetragonal crystal system since a = b = 0.30 nm, c = 0.40 nm, and = = = 90. (a) To which crystal system does this unit cell belong? (b) What would this crystal structure be called? (c) Calculate the density of the material, given that its atomic weight is 141 g/mol. 3.20 Below is a unit cell for a hypothetical metal. ![]() Therefore, Rh has the FCC crystal structure. Also, from Figure 2.6, its atomic weight is 102.91 g/mol. For FCC, n = 4, and a = 2 R 2 (Equation 3.1). Solution In order to determine whether Rh has an FCC or a BCC crystal structure, we need to compute its density for each of the crystal structures. Determine whether it has an FCC or BCC crystal structure. For FCC, n = 4 atoms/unit cell, and VC = 16R3 2 (Equation 3.4). Solution We are asked to determine the radius of an iridium atom, given that Ir has an FCC crystal structure. But a depends on R according to Equation 3.3, and 3 4R 64 R3 VC = = 3 3 3 Thus, APF = VS VC = 8 R3 /3 64 R3 /3 3 = 0.68 3.8 Calculate the radius of an iridium atom, given that Ir has an FCC crystal structure, a density of 22.4 g/cm 3, and an atomic weight of 192.2 g/mol. Solution The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or APF = VS VC Since there are two spheres associated with each unit cell for BCC 4R3 8R3 VS = 2(sphere volume) = 2 3 = 3 Also, the unit cell has cubic symmetry, that is VC = a3. Furthermore, from triangle JHM, (JM ) 2 = ( JH ) 2 ( MH ) 2 or 2 c a 2 = (JH ) 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 c 2 a2 c2 a 2 = + = + 3 2 3 4 and, solving for c/a c = a 8 = 1.633 3 3.5 Show that the atomic packing factor for BCC is 0.68. And, since atoms at points J, K, and M, all touch one another, JM = JK = 2R = a where R is the atomic radius. 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. Solution A sketch of one-third of an HCP unit cell is shown below. ![]() Home-work # 2 3.4 For the HCP crystal structure, show that the ideal c/a ratio is 1.633. ![]()
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