Engineering Materials
Engineering M12
Moorpark College
Homework Chapter #3 Solutions
3.1
What is the difference between atomic structure and crystal structure?
A crystal structure is described by both the geometry of, and atomic arrangements within, the unit cell,
whereas a crystal system is described only in terms of the unit cell geometry.
For example, face-centered cubic and body-centered cubic are crystal structures that belong to the cubic
crystal system.
3.2
If the atomic radius of lead is 0.175nm, calculate the volume of its unit cell in cubic meters.
For this problem, we are asked to calculate the volume of a unit cell of lead. Lead 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 0.175 × 10 −9 m
) ( 2 ) = 1.213 × 10
3
−28
m3
3.3
Show for the body-center cubic crystal structure that the unit cell edge length a and the
4R
atomic radius R are related through a =
3
This problem calls for a demonstration of the relationship a = 4R 3 for BCC. Consider the
BCC unit cell shown below
Using the triangle NOP
2
2
2
(NP) = a + a = 2a
2
And then for triangle NPQ,
2
2
2
(NQ) = (QP) + (NP)
But NQ = 4R, R being the atomic radius. Also, QP = a. Therefore,
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Engineering Materials
Engineering M12
Moorpark College
2
(4R)
= a
2
a =
2
+ 2a , or
4R
3
3.7
Molybdenum has a BCC crystal structure, an atomic radius of .1363 nm, and a atomic weight
of 95.94 g/mol. Compute and compare it theoretical density with the experimental value found
inside the front cover
This problem calls for a computation of the density of molybdenum. According to Equation (3.5)
ρ=
nAMo
VC A
For BCC, n = 2 atoms/unit cell, and
 4R  3
VC = 

 3
Thus,
ρ=
(2 atoms/unit cell)(95.94 g/mol)
3


3



-7
23
(4 ) 0.1363 x 10 cm / 3  / (unit cell) 6.023 x 10 atoms/mol



(
(
)
)
= 10.21 g/cm3
3
The value given inside the front cover is 10.22 g/cm .
2
Engineering Materials
Moorpark College
Engineering M12
3.13
Niobium has a atomic radius of 0.1430 nm and a density of 8.57 g/cm3. Determine whether it
has an FCC or BCC crystal structure.
In order to determine whether Nb has an FCC or BCC crystal structure, we need to compute its density
for each of the crystal structures.
For FCC, n = 4, and a = 2 R 2 . Also, from Figure 2.6, its atomic weight is 92.91 g/mol. Thus, for FCC
ρ =
nANb
3
2R 2 N A
(
)
(4 atoms/unit cell)(92.91 g/mol)
3


-8

2  /(unitcell) 6.023 x 10 23 atoms / mol
(2) 1.43 x 10 cm



=
(
)( )
(
)
= 9.33 g/cm3
For BCC, n = 2, and a =
=
4R
, thus
3
(2 atoms/unit cell)(92.91 g/mol)
3


-8


 (4) 1.43 x 10 cm 

23

/(unitcell)
 6.023 x 10 atoms/ mol

3





(
)
(
)
= 8.57 g/cm3
which is the value provided in the problem statement. Therefore, Nb has a BCC crystal structure.
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Engineering Materials
Moorpark College
Engineering M12
3.17
Beryllium has a HCP unit cell for which the ratio of the lattice parameters c/a is 1.568. If the
radius of the Be atom is 0.1143 nm.
Determine the unit cell volume.
For HCP, there are the equivalent of six spheres per unit cell, and thus
 4π R3 
3
VS = 6 
 = 8πR
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
 2R 3 
2
AREA = (3)(CD)(BC) = (3)(2 R) 
 = 6R 3
2


VC = 6R2c 3
But, c = 1.568a, and a = 2R, or c = 3.14R, and
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Engineering M12
Moorpark College
VC = (6)(3.14) R3 3
( )[0.1143 x 10-7 cm]
3
= (6) (3.14 ) 3
= 4.87 x 10
−23
3
cm /unit cell
Calculate the theoretical density of Be and compare it with the literature value.
ρ =
nA Be
VC NA
For HCP, n = 6 atoms/unit cell, and for Be, A = 9.01 g/mol. Thus,
Be
ρ =
(6 atoms/unit cell)(9.01 g/mol)
-23
3
23
4.87 x 10
cm /unit cell 6.023 x 10
atoms/mol
(
)(
)
= 1.84 g/cm3
3
The value given in the literature is 1.85 g/cm .
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