A metal crystallises in a face-centred cubic (fcc) lattice. The density of the metal is 10.0 g cm ⁻³ and its molar mass is 96 g mol ⁻¹ . The square of the shortest distance between two atoms in the crystal lattice is _____ ^2 . (Nearest integer) (Given: N_A = 6.0 10²³ mol ⁻¹ )

Step-by-step solution

For a face-centred cubic (fcc) lattice, Z = 4 . Density, d = Z M N_A a^3 Rearranging for the volume of the unit cell, a^3 = Z M d N_A a^3 = 4 96 10.0 6.0 10²³ a^3 = 384 60 10²³ = 6.4 10⁻²³ cm ^3 = 64 10⁻²⁴ cm ^3 Taking the cube root gives the edge length, a = 4 10⁻⁸ cm = 4 In an fcc lattice, the shortest distance between two atoms (nearest neighbour distance) is D = a 2 . The square of this distance is D^2 = a^2 2 = 4^2 2 = 16 2 = 8 ^2 . Answer: 8