Example 1. Determine the number of vacancies needed for a BCC iron crystal to have a density of 7.87g/cm3. The lattice parameter of the iron is 2. X 10-8cm Solution Xatoms/cell)(55.847g/mol) 7.8814g/cm (2.866×10cm)3(602×1023 atoms/mo) X atoms/cell(7.87)2866×108)(602×1 1.9971 5.847 o. there shou ald be 2.00-1.9971=0.0029 vacancies per unit cell 0.0029 vacancies/cell Vacancies/cm 1.23×10 (2.866×108cm)31. Determine the number of vacancies needed for a BCC iron crystal to have a density of 7.87g/cm3. The lattice parameter of the iron is 2.866×10-8cm. 3 8 3 23 7.8814g / cm (2.866 10 cm) (6.02 10 atoms/ mol) ( atoms/ cell)(55.847g / mol)      X  1.9971 55.847 (7.87)(2.866 10 ) (6.02 10 ) atoms/ cell 8 3 23       X So, there should be 2.00－1.9971=0.0029 vacancies per unit cell. 20 8 3 3 1.23 10 (2.866 10 cm) 0.0029 vacancies/ cell Vacancies/ cm      