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Aluminum (FCC) with atom radius of 0.1431nm. Find the distance between aluminum lattice plane (100) and (111),...

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Chemistry

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 & Explanation Solved by verified expert
4.5 Ratings (936 Votes)

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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