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The Solid State, , The number of atoms per unit cell (z) for bee structure = 2’, , Atomic mass ‘WW = 100, Cell edge ‘a’ = 400pm= 400 x 10° cm, Avagadro number (NV) = 6.02x 107, , = ee = 519g cm?, (400 x 107!°)? x 6.02 x 10, Example 1.6 The density of a face centred cubic element (atomic mass 60.2) is 6.25 g cm”. Calculate the, edge length of the unit cell., , Solution, , BIE, , Tees or a =, a, , Since the element has fcc lattice, the number of atoms per unit cell (z) = 4, M = 602, N= 6.02x 103, d = 625gcm™, Substituting we have, , pbs et oe OO ZR G4 ALO cor, 6.25 x 6.02 x 10, , .. Edge length‘a’ = (64 x tn) = 4x 10%cm = 400x 10° cm = 400 pm., Example 1.7, , An element forms bcc structure with an edge length of 360 pm. The density of the element, is 8.96 gem™. Calculate the number of unit cells and the number of atoms present in 127g of the, , element. : ‘, Solution, , 360 pm = 360x 10° cm, , (360 x 107°) = 4.666 x 10 cm?, , Edge length of the unit cell (a), , . Volume of one unit cell (a°), , Density of the substance = 896g cm?, =, Mass of a unit cell = 4666x 10x 8.96g = 41.807x 10° ¢, ff Number of unit cells in 127g of the element = wi = 3.037 x 107, , 41.807 x 10>