2r is closest distance between two atom.
(b) The problem is to find how many face centers are equidistant from a corner atom. Point A in figure may be taken as the reference corner atom. In that same figure, B is one of the face-center points at the nearest distance to A. In plane ABD in the figure, there are three other points equally close to A: the centers of the squares in the upper right, lower left, and lower right quadrants of the plane, measured around A. Plane ACE,
Parallel to the plane of the paper, also has points in the centers of each of the squares in the four quadrants around A. Also, plane ACF, perpendicular to the plane of the paper, has points in the centers of each of the squares in the four quadrants around A. Thus there are 12 nearest neighbors in all, the number expected for a close-packed structure.
The same result would have been obtained by counting the nearest neighbours around B, a face-centered point.
(c) In fcc, n = 4, M = 197 g mol-1
a = 4.070 Å = 4.070 x 10-10 cm
(d) Since atoms at closest distance are in contact in a close-packed structure, the closest distance between centres calculated in (a), a/√2, must equal the sum of the radii of the two spherical atoms, 2r. Thus, r = a/23/2. From (c), there are 4 gold atoms per unit cell.
Then, Volume of 4 gold atoms
∴ Packing fraction = volume of 4 gold atoms/volume of unit cell
