Question

Gold has an FCC structure, a density of 19.3 Mg/m3 and an atomic mass of 196.67 kg/kmol. Estimate the dimensions of the unit cell and the atomic diameter of gold

Answer 1

2n FCC lattice, there are y atoms per unit cell Density = Mass/volume = 4 times mass of each atom/a^3 Mass of each atom = 196.67/6.022 times 10^26 kg = 32.658585 times 10^-26 kg 19.3 times 10^3 kg/m^3 = 4 times 32.658585 times 10^-26/a^3 a^3 = 4 times 32.658585 times 10^-26/19.3 times 10^3 a^3 = 6.768618 times 10^-29 a^3 = 67.68618 times 10^-30 a^3 = 0.06768618 times 10^-27 a = 0.4075366 times 10^-9 m a = 407.5366 pm = > unit all dimension. 4R = squareroot a 2 squareroot 2 R = a R = a/2 squareroot 2 = 407.5366/2 squareroot 2 pm R = 144.08594 pm