Question

Material science

2. Barium (Ba) has a BCC crystal structure, a measured density of 3.51 g/cm3, and a lattice parameter of 0.502 nm. If the atomic mass of Ba is 137.33 g/mol calculate (a) the fraction of attice points that contain vacancies, and (b) the total number of vacancies in a cubic meter of Ba.

Answer 1

z for BCC Crystal Structure is 2. Z = 2 Density = x(atomic Mass)/(lattice parameter)^3 (6.02 times 10^23 atom/mol) a) put, lattice parameter rightarrow 0.502 nm = > 5.02 times 10^-8 cm Density (delta) = 3.51 g/cm^3 atomic Mass [= 137.33 g/mol 3.51 = x times (137.33 g/mol)/(5.02 times 10^-8 cm)^3 (6.02 times 10^23 atoms mol) x = 3.51 f/cm^3 times 5.02 times 10^-8)^3 cm^3 times (6.02 times 10^23 atom/mol/137.33 g/mol x = 1.946477 fraction = z-x/z = 1-1.946477/2 = 0.0267615 \r\n b) number = 0.007/(5.02 times 10^-8 cm)^3 vacancies/vc = 5.53 times 10^19 Vacancies/cm^3 = 5.53 times 10^13 Vacancies/m^3