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

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

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Planar density of FCC (100) = 1/[4r²]

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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 12.2 x 10¹⁸ m^-2

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

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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 14.1 x 10¹⁸ m^-2

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