(a) Calculate the packing efficiency in a Face Centered Cubic lattice. (b) If a metal with atomic mass 209 crystallizes in a simple cubic lattice what is the edge length of its unit cell. (Given d = 91.5 kg m-3).

(a) Calculate the packing efficiency in a Face Centered Cubic lattice.

1.	(a) Calculate the packing efficiency in a Face Centered Cubic lattice. (b) If a metal with atomic mass 209 crystallizes in a simple cubic lattice what is the edge length of its unit cell. (Given d = 91.5 kg m-3).
Answer» Solution : (a) The number of atoms per unit cell in fcc structures is four. Each ATOM is considered as one sphere. So, the volume of four atoms (four spheres) - `4xx 4/3 pi r^(3) =16/3 pi r^(3)` From this FIGURE it is clear that there are three spheres touching each other ALONG the face diagonal. Let .a. be the edge length of the cube and .r. be the radius of the sphere. The face diagonal AC = b, then b = 4r. In ABC, `AC^(2) = BC^(2) + AB^(2)` `b^(2)= a^(2)+ a^(2)` ` b^(2)= 2a^(2)` `b= sqrt(2a)= 4r`` b= sqrt(2a) =4r` ` a= 4r/sqrt(2) =2sqrt(2r)` the volume of the cubic unit cell = `a_(3) = (2sqrt(2r)^(3)` therefore, packing efficiency=volume of four atoms(four spheres) in unit cell/ volume of unit cell `xx100 % ` packing efficiency =` (16/3pir^(3))/((2sqrt(2r)^(3)))xx100%` packing efficiency`=(16/3 xx22/7 r^(3))/(8(sqrt(2))^(3) r^(3)) xx100% ` packing efficiency = `(16/3xx22/7)/(8(sqrt2)^(3))xx100% =74%` therefore 74% of unit cell is occupied by atoms and the rest 26% is empty space in fcc structure . the packing fractionin fcc structure (ccp)= 0.74 the fraction the fraction of fraction of empty space in fcc structure (ccp)0.26 (b) `a^(3)= (ZM)/(N_(A)d)= (1xx209)/(6.023xx10^(23)xx91.5xx10^(-3))`, `a^(3)= 3.359xx10^(-6) cm`.