EGR 221 – Materials Science, HW 2, CH 3: 3, 6, 7, 11, 20, 22, 30a-c, 32
3.3 Show for the body-centered cubic crystal structure that the unit cell edge length a and the atomic radius R are
related through a =4R/
3.
Solution
Consider the BCC unit cell shown below
Using the triangle NOP
(NP) 2 = a 2 + a 2 = 2a 2
And then for triangle NPQ,
(NQ) 2 = (QP) 2 + ( NP) 2
But NQ = 4R, R being the atomic radius. Also, QP = a. Therefore,
(4R) 2 = a 2 + 2a 2
or
a =
4R
3
3.6 Show that the atomic packing factor for HCP is 0.74.
Solution
The APF is just the total sphere volume-unit cell volume ratio. For HCP, there are the equivalent of six
spheres per unit cell, and thus
 4π R 3 
 = 8π R 3
VS = 6 
3


Now, the unit cell volume is just the product of the base area times the cell height, c. This base area is just three
times the area of the parallelepiped ACDE shown below.
The area of ACDE is just the length of CD times the height BC . But CD is just a or 2R, and
BC = 2R cos (30°) =
2R 3
2
Thus, the base area is just
2 R 3 
2
AREA = (3)(CD)(BC) = (3)(2 R) 
 = 6R 3
 2 
and since c = 1.633a = 2R(1.633)
VC = (AREA)(c) = 6 R 2 c 3
(3.S1)
= (6 R 2 3 ) (2)(1.633)R = 12 3 (1.633) R 3
Thus,
APF =
VS
VC
8π R 3
=
12 3 (1.633) R 3
= 0.74
Density Computations
3.7 Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol.
Compute and compare its theoretical density with the experimental value found inside the front cover.
Solution
This problem calls for a computation of the density of iron. According to Equation 3.5
ρ =
nAFe
VC N A
For BCC, n = 2 atoms/unit cell, and
 4 R 3
VC =  
 3
Thus,
ρ =
=
nAFe
 4 R 3

 NA
 3
(2 atoms/unit cell)(55.85 g/mol)
[(4) (0.124 × 10-7 cm) / 3] /(unit cell) (6.022 × 1023 atoms/mol)
3
= 7.90 g/cm3
The value given inside the front cover is 7.87 g/cm3.
3.11 Zirconium has an HCP crystal structure and a density of 6.51 g/cm3.
(a) What is the volume of its unit cell in cubic meters?
(b) If the c/a ratio is 1.593, compute the values of c and a.
Solution
(a) The volume of the Zr unit cell may be computed using Equation 3.5 as
VC =
nAZr
ρN A
Now, for HCP, n = 6 atoms/unit cell, and for Zr, AZr = 91.22 g/mol. Thus,
VC =
(6 atoms/unit cell)(91.22 g/mol)
(6.51 g/cm3)(6.022 × 10 23 atoms/mol)
= 1.396 × 10-22 cm3/unit cell = 1.396 × 10-28 m3/unit cell
(b) From Equation 3.S1 of the solution to Problem 3.6, for HCP
VC = 6 R 2 c 3
But, since a = 2R, (i.e., R = a/2) then
 a 2
3 3 a2 c
VC = 6   c 3 =
2
2
but, since c = 1.593a
VC =
3 3 (1.593) a 3
= 1.396 × 10 -22 cm3/unit cell
2
Now, solving for a
 (2)(1.396 × 10 -22 cm3 ) 1/3
a = 

(3) ( 3) (1.593)


= 3.23 × 10-8 cm = 0.323 nm
And finally
c = 1.593a = (1.593)(0.323 nm) = 0.515 nm
Crystal Systems
3.20 Below is a unit cell for a hypothetical metal.
(a) To which crystal system does this unit cell belong?
(b) What would this crystal structure be called?
(c) Calculate the density of the material, given that its atomic weight is 141 g/mol.
Solution
(a) The unit cell shown in the problem statement belongs to the tetragonal crystal system since a = b = 0.30
nm, c = 0.40 nm, and α = β = γ = 90°.
(b) The crystal structure would be called body-centered tetragonal.
(c) As with BCC, n = 2 atoms/unit cell. Also, for this unit cell
VC = (3.0 × 10−8 cm) 2 ( 4.0 × 10−8 cm)
= 3.60 × 10−23 cm3/unit cell
Thus, using Equation 3.5, the density is equal to
ρ =
nA
VC N A
=
(3.60
×
(2 atoms/unit cell) (141 g/mol)
cm3/unit cell)(6.022 × 10 23 atoms/mol)
10 -23
= 13.0 g/cm3
Point Coordinates
3.22 List the point coordinates for all atoms that are associated with the FCC unit cell (Figure 3.1).
Solution
From Figure 3.1b, the atom located of the origin of the unit cell has the coordinates 000. Coordinates for
other atoms in the bottom face are 100, 110, 010, and
1 1
2 2
0. (The z coordinate for all these points is zero.)
For the top unit cell face, the coordinates are 001, 101, 111, 011, and
11
22
1.
Coordinates for those atoms that are positioned at the centers of both side faces, and centers of both front
and back faces need to be specified. For the front and back-center face atoms, the coordinates are 1
respectively. While for the left and right side center-face atoms, the respective coordinates are
3.30 Within a cubic unit cell, sketch the following directions:
(a)
[1 10],
(b)
[1 2 1],
(c)
[01 2],
Solution
The directions asked for are indicated in the cubic unit cells shown below.
1
2
0
1
2
11
22
and
and 0
1 1
1 .
2 2
1 1
2 2
,
3.32 Determine the indices for the directions shown in the following cubic unit cell:
Solution
Direction A is a [ 430] direction, which determination is summarized as follows. We first of all position
the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system
x
y
2a
b
3
2
2
1
3
2
Projections
–
Projections in terms of a, b, and c
–
z
0c
0
Reduction to integers
–4
Enclosure
3
0
[ 430]
Direction B is a [23 2] direction, which determination is summarized as follows. We first of all position
the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system
x
Projections
Projections in terms of a, b, and c
Reduction to integers
2a
3
2
3
2
Enclosure
y
–b
–1
–3
z
2c
3
2
3
2
[23 2]
Direction C is a [13 3 ] direction, which determination is summarized as follows. We first of all position
the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system
x
Projections
Projections in terms of a, b, and c
Reduction to integers
a
3
1
3
1
Enclosure
y
z
–b
–c
–1
–1
–3
–3
[13 3 ]
Direction D is a [136 ] direction, which determination is summarized as follows. We first of all position the
origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system
Projections
Projections in terms of a, b, and c
Reduction to integers
Enclosure
x
y
a
b
6
1
2
1
6
2
1
3
[136 ]
z
–c
–1
–6