Question

Aluminum (FCC) with atom radius of 0.1431nm.

Find the distance between aluminum lattice plane (100) and (111), and calculate the density of that planes.

Answer 1

Given that atomic radius , r = 0.1431 nm = 0.1431 x 10^-9 m

For FCC,

edge length a = r (8)^1/2 = 0.1431 x 10^-9 m x (8)^1/2 = 0.405 m

--------------------------------------------------------------------------------------

Planar density of FCC (100) = 1/[4r²]

                                             = 1/ [4 x (0.1431 x 10^-9)²]

                                              = 12.2 x 10¹⁸ m^-2

Planar density of FCC (111) = 1/ [2r²(3)^1/2]

                                            = 1/ [2.(0.1431 x 10^-9)².(3)^1/2]

                                           = 14.1 x 10¹⁸ m^-2