Page 86 - Materials Science and Engineering An Introduction

58 • Chapter 3 / The Structure of Crystalline Solids

If a and c represent, respectively, the short and long unit cell dimensions of Figure 3.4a,
the c/a ratio should be 1.633; however, for some HCP metals, this ratio deviates from the
ideal value.
The coordination number and the atomic packing factor for the HCP crystal struc-
ture are the same as for FCC: 12 and 0.74, respectively. The HCP metals include cad-
mium, magnesium, titanium, and zinc; some of these are listed in Table 3.1.

EXAMPLE PROBLEM 3.1

Determination of FCC Unit Cell Volume
Tutorial Video Calculate the volume of an FCC unit cell in terms of the atomic radius R.

Solution
In the FCC unit cell illustrated, the atoms touch one another
across a face-diagonal, the length of which is 4R. Because the
3
unit cell is a cube, its volume is a , where a is the cell edge
length. From the right triangle on the face, R
2
2
a + a = (4R) 2
a 4R
or, solving for a,
a = 2R12 (3.1)
The FCC unit cell volume V C may be computed from
a
3
3
V C = a = (2R12) = 16R 12 (3.6)
3
EXAMPLE PROBLEM 3.2

Computation of the Atomic Packing Factor for FCC
Tutorial Video Show that the atomic packing factor for the FCC crystal structure is 0.74.

Solution
The APF is defined as the fraction of solid sphere volume in a unit cell, or
volume of atoms in a unit cell V S
APF = =
total unit cell volume V C
Both the total atom and unit cell volumes may be calculated in terms of the atomic radius R.
4
3
The volume for a sphere is pR , and because there are four atoms per FCC unit cell, the total
3
FCC atom (or sphere) volume is
4 3 16 3
V S = (4) pR = 3 pR
3
From Example Problem 3.1, the total unit cell volume is
3
V C = 16R 12
Therefore, the atomic packing factor is
16
( )pR 3
V S 3
APF = = 3 = 0.74
V C 16R 12