•• In the CsCl crystal the Cs+ ions and Cl− ions are arranged on two identical simple cubic lattices. The Cl− lattice is offset from the Cs+ lattice, so that each Cl− is at the exact center of a cube of Cs+ ions, and vice versa. Thus a unit cell of CsCl looks just like the BCC cell in Fig. 13.10 (except that the center ion is Cl−, whereas those at the corners are Cs+, or vice versa). (a) The density of CsCl is 3.97 g/cm3. Use the result of Problem 13.14 to find the edge length of the unit cube of CsCl. (b) What is the nearest-neighbor distance?
Problem 1
•• Extend the reasoning of Problem 2 to prove that the number of atoms per unit cell of the BCC lattice is 2.
Problem 2
• One can think of a simple cubic lattice [Fig. 13.8(b)] as a stack of many unit cubes [Fig. 13.8(c)]. However, in counting atoms there is a danger of miscounting, since adjacent cubes share several atoms. (a) Consider any one atom in the lattice. How many cubes share this atom? (b) Consider any one cube. How many atoms does the cube have a share of? (c) Prove that in the lattice as a whole, there is one atom per unit cube.
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