Question

In: Chemistry

Aluminum (FCC) with atom radius of 0.1431nm. Find the distance between aluminum lattice plane (100) and...

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.

Solutions

Expert Solution

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


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