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HW
3.2 If the atomic radius of aluminum is 0.143 nm, calculate the volume of its unit cell in cubic meters.
Solution
For this problem, we are asked to calculate the volume of a unit cell of aluminum. Aluminum has an FCC
crystal structure (Table 3.1). The FCC unit cell volume may be computed from Equation 3.4 as
VC = 16 R^3 2 = (16)(0.143 10 -9^ m)^3 ( 2 ) = 6.62 10 -29^ m^3
3.4 For the HCP crystal structure, show that the ideal c/a ratio is 1.633.
Solution
A sketch of one-third of an HCP unit cell is shown below.
Consider the tetrahedron labeled as JKLM , which is reconstructed as
The atom at point M is midway between the top and bottom faces of the unit cell--that is
MH = c /2. And, since
atoms at points J , K , and M , all touch one another,
JM = JK = 2 R = a
where R is the atomic radius. Furthermore, from triangle JHM ,
( JM )
2 = ( JH )
2 ( MH )
2
or
a
2 = ( JH )
2
c
2
2
Now, we can determine the
JH length by consideration of triangle JKL , which is an equilateral triangle,
cos 30 =
a / 2
JH
and
JH =
a
3
Substituting this value for
JH in the above expression yields
a
2
a
3
2
c
2
2
=
a
2
c
2
and, solving for c/a
c
a
3.13 Rhodium has an atomic radius of 0.1345 nm and a density of 12.41 g/cm
3
. Determine whether it has an FCC
or BCC crystal structure.
Solution
In order to determine whether Rh has an FCC or a BCC crystal structure, we need to compute its density
for each of the crystal structures. For FCC, n = 4, and a =
2 R 2 (Equation 3.1). Also, from Figure 2.6, its atomic
weight is 102.91 g/mol. Thus, for FCC (employing Equation 3.5)
nA Rh
a^3 N A
nA Rh
( 2 R 2 )^3 N
A
(4 atoms/unit cell)(102.91 g/mol)
(2) (1.345^ ^10 -8^ cm)(^2 )
3 /(unit cell)
(6.022 1023 atoms / mol)
= 12.41 g/cm
3
which is the value provided in the problem statement. Therefore, Rh has the FCC crystal structure.
3.14 Below are listed the atomic weight, density, and atomic radius for three hypothetical alloys. For each
determine whether its crystal structure is FCC, BCC, or simple cubic and then justify your determination. A simple
cubic unit cell is shown in Figure 3.24.
Alloy Atomic Weight Density Atomic Radius
( g/mol ) ( g/cm
3 ) ( nm )
A 77.4 8.22 0.
B 107.6 13.42 0.
C 127.3 9.23 0.
Solution
For each of these three alloys we need, by trial and error, to calculate the density using Equation 3.5, and
compare it to the value cited in the problem. For SC, BCC, and FCC crystal structures, the respective values of n
are 1, 2, and 4, whereas the expressions for a (since V C
= a
3 ) are 2 R ,
2 R 2 , and
4 R
For alloy A, let us calculate assuming a simple cubic crystal structure.
nA A
V C
N
A
nA A
^2 R
3 N (^) A
(1 atom/unit cell)(77.4 g/mol)
8
3 /(unit cell)
23
atoms/mol)
= 8.22 g/cm
Therefore, its crystal structure is simple cubic.
For alloy B, let us calculate assuming an FCC crystal structure.
nA B
(2 R 2)
3 N (^) A
(4 atoms/unit cell)(107.6 g/mol)
^2 2 (1.33^ ^10
cm)
3 /(unit cell)
23
atoms/mol)
= 13.42 g/cm
3
Therefore, its crystal structure is FCC.
For alloy C, let us calculate assuming a simple cubic crystal structure.
nA C
^2 R
3 N (^) A
(1 atom/unit cell)(127.3 g/mol)
cm)
3 /(unit cell)
23
atoms/mol)
= 9.23 g/cm
3
Therefore, its crystal structure is simple cubic.
