Answer
See the answer below.
Work Step by Step
a) $8\times 1/8=1$ atom in the cell.
$l=2.r$
Volume of cell: $l^3=8.r^3$
Volume of sphere: $4\pi/3.r^3$
Coverage: $(4\pi/3)/8\times100\%=52.36\%$
b) (BCC): 2 atoms per cell, $\sqrt3.l=4.r$
Volume of cell: $l^3=64/3\sqrt3.r^3$
Volume of spheres: $2\times4\pi/3.r^3$
Coverage: $(8\pi/3)/(64/3\sqrt3)\times100\%=68.02\%$
(FCC): 4 atoms per cell, $\sqrt2.l=4.r$
Volume of cell: $l^3=64/2\sqrt2.r^3$
Volume of spheres: $4\times4\pi/3.r^3$
Coverage: $(16\pi/3)/(32/\sqrt2)\times100\%=74.05\%$
All else held equal, the FCC arrangement should provide a higher density.
