Example 1: The lattice constant “a” of BCC iron is 2.86 Å. Determine the specific gravity. Atomic mass of Fe = g/mole. Å 2.86 Å # of atoms/cell = 1/8*8+1=2 Mass of single Fe atom = / 6.02x10 23 = 9.28x gr Specific gravity = 2 * 9.28x / (2.86x10-8cm)3 = 7.93 g/cm 3 Due to presence of imperfections in the crystal structures the actual density is always slightly less than the theoretical density. The density of Fe ≈ 7.86 g/cm 3

Example 2: Specific gravity of a metal is If the metal has an FCC structure and its atomic mass is g/mole, determine: a) The lattice constant, a=? b) The radius of a single atom? c) Determine δ [110] (linear density) # of atoms/cell = 1/2*6+1/8*8=4 Mass of single atom = / 6.02x10 23 = 6.65x gr

4 * 6.65x = a3a3 a 3 = 1.72x cm 3 = 5.56 Å a a = 4r = 1.97 Å 4 r =a2 a2 z [110] y x δ [110] = 2.54x10 7 atoms/cm δ [110] = 2 atoms a 2 = 2 2 * 5.56x10 -8

Example 3: In a cubic unit cell, determine; a) ‹110› → family of directions b) {110} → family of planes a. ‹110› Ξ [110] [101] [011] z [110] y x [011] [101]

b. {110} Ξ (110) (101) (011) z y x (110) z y x (101) z y x (011)

Example 4: Determine the planar density of Pb along the following planes a) (100) b) (110) Note that Pb has an FCC structure and its atomic radius is r Pb = 1.75 Å a) z y x (100) r 2r r a a a 2 = 4r → a = 4.95 Å # of atoms = 1+1/4*4 =2 δ (100) = 2 (4.95x10 -8 ) 2 = 8.16x10 14 atoms/cm 2

b) FCC structure, (110) plane 2a a # of atoms = 1/4*4+1/2*2 =2 atoms δ (110) = 2 (4.95x10 -8 ) 2 = 5.77x10 14 atoms/cm 2 2 z y x (110)

Example 5: Determine the linear density of [110] for the previous problem. z y x 4r δ [110] = 2 4*1.75x10 -8 = 2.86x10 7 atoms/cm