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12 Modern Materials Solutions to Exercises transformed into a "standard" cell by renaming the axes so that the arbitrary angle is between a and c, and a right-handed coordinate system is maintained. The cell described is still monoclinic, by virture of its geometry (actually, by virtue of its symmetry). (b) Hexagonal, a = b # c, α = = 90°, γ = 120°. A two dimensional hexagonal unit cell has a = b and = 120°. If the third lattice vector is perpendicular to the other two, the other two angles are 12.21 Plan. Refer to Figure 12.6 to find geometric characteristics of the seven three dimensional primitive lattices. If no lattice vectors are perpendicular to each other, none of the unit cell angles (α, ß, γ) are 90°. This is characteristic of two of the three dimensional primitive lattices: triclinic and rhombohedral. 12.22 If all three lattice vectors have the same length, a = b = C. This is characteristic of two of the three dimensional lattices: cubic and rhombohedral. 12.23 Analyze. Given an element with a body centered cubic lattice, find the minimum number of atoms in a unit cell. Plan. Refer to Figures 12.11 and 12.12 to visualize an element (all the same kind of atoms) with a body centered cubic lattice. Solve. A body centered cubic lattice is composed of body centered cubic unit cells. A unit cell contains the minimum number of atoms when it has atoms only at the lattice points. A body centered cubic unit cell like this is shown in Figure 12.12(b). There is one atom totally inside the cell (1 1) and one at each corner (8 1/8) for a total of 2 (metal) atoms in the unit cell. (Only metallic elements have body-centered cubic lattices and unit cells.) 12.24 A face centered cubic lattice is composed of face centered cubic unit cells. A unit cell contains the minimum number of atoms when it has atoms only on the lattice points. A face centered cubic unit cell like this is shown in Figure 12.12(c). There is one atom centered on each face (6 X 1/2) and one at each corner (8 1/8) for a total of 4 metal atoms in the unit cell. 12.25 Analyze. Given a diagram of the unit cell dimensions and contents of nickel arsenide, determine what kind of lattice this crystal possess, and the empirical formula of the compound. Plan. Refer to Figure 12.6 to find geometric characteristics of the seven three dimensional primitive lattices. Decide where atoms of the two elements are located in the unit cell and use Table 12.1 to help determine the empirical formula. (a) a = b = 3.57 Å. c = 5.10 Å # a or b. α = = 90°, = 120°. This unit cell is hexagonal. There are no atoms in the exact middle of the cell or on the face centers, so it is a primitive hexagonal unit cell. Nickel arsenide has a primitive hexagonal unit cell and crystal lattice. (b) There are Ni atoms at each corner of the cell (8 1/8) and centered on four of the unit cell edges (4 1/4) for a total of 2 Ni atoms. There are 2 As atoms totally inside the cell. The unit cell contains 2 Ni and 2 As atoms; the empirical formula is NiAs. 338