1．The atomic weight of vanadium (V) is 50.94 g/mole. Its density is ρ=5.96 g/cm3. It
has the BCC crystal structure. Calculate the atomic radius of vanadium. Confirm that
this matches the value given in Appendix A.
3a = 4 R
a
2a
V has the BCC crystal structure,
3
1
& 4R #
then Vv= $
! , and there are 1+ ! 8 =2 atoms in one unit cell
8
% 3"
(R is the Radius of V atom，Vv= volume of the BCC vanadium unit cell)
ρ=
nA
nA
=
3
Vv N A
' 4R $
%
" ! NA
& 3#
1
' nA $ 3
3
"" !
! R = %%
4
& ( ! NA #
1
'
$3
2 ! 50.94 g / mole
3
"" !
= %%
3
23
4
& 5.96 g / cm ! 6.023 ! 10 / mole #
= 1.32 " 10!8 cm
=0.132 nm
In good agreement with the tabulated values
2．Calculate the atomic packing fraction of a BCC crystal.
3a = 4 R
a
2a
1
For BCC crystal, there are 1+ ! 8 = 2 atoms in one unit cell
8
4
R
The side length of unit cell
a=
3
4
4
4
2 " !R 3
2 " !R 3
2 " !R 3
V ATOMS
3
3
3
3
! = 0.68
APF =
=
=
=
=
3
4
64 2
a
8
VUNITCELL
3
(
R)
R
3
3 3
3．Draw a face-centered cubic unit cell describing the FCC crystal structure. Identify the
eight atomic positions in this cell [e.g. the atom at the origin is at (0,0,0)]. How many
atoms per unit cell are present in an FCC material? Justify your answer.
(1,1,0)
The 8 atomic positions on the vertices : (0,0,0), (0,1,0), (1,0,0), (1,1,0), (0,0,1), (0,1,0),
(1,1,1), (0,1,1), (1,0,1)
The 6 atomic positions on faces: (
1 1
1
1
1 1
1
1
1 1
, , 0), ( , 0, ), (1, , ), ( ,1, ), ( , ,1), (0,
2 2
2
2
2 2
2
2
2 2
1 1
, )
2 2
Atoms per unit cell:
1
1
!8+ ! 6 = 4
8
2
(Atoms on the vertices are shared by 8 cell units, while atoms on the faces are shared by
2 cell units)
4．(From test 1, summer, 2005.) Gold (Au) and silver (Ag) are both found with the the
FCC crystal structure. Their atomic weights are 196.97 g/mole and 107.87 g/mole,
respectively. The lattice parameter of gold is 0.4079 nm, and the lattice parameter of
silver is 0.4086 nm.
a) determine the radii of a single gold atom and a single silver atom.
b) determine the density of each metal. Briefly justify any density difference you
find.
a
2a = 4 R
a)
For FCC crystal structure,
R=
2
a
4
(a is the lattice parameter )
2
! 0.4079nm = 0.1442 nm
4
2
RAg =
! 0.4086nm = 0.1445 nm
4
nA
ρ=
, a = 2 2R
VC N A
RAu =
b)
# Au =
4 ! 196.97( g / mol )
16 2 ! (1.442 ! 10"8 )3 (cm)3 ! 6.023 ! 1023 (atoms / mol )
= 19.28
# Ag =
g/cm3
4 ! 107.87( g / mol )
16 2 ! (1.445 ! 10"8 )3 (cm)3 ! 6.023 ! 1023 (atoms / mol )
= 10.49
g/cm3
Though Au and Ag atoms are essentially the same size, the density of Au sample is much
higher than that of a Ag sample. The difference is due to the fact that a Au atom has
many more protons and neutrons, so its atoms weight is much heavier.
5．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 3.23.
Alloy
Atomic Weight
(g/mol)
43.1
184.4
91.6
A
B
C
Density(g/cm3)
Atomic Radius
(nm)
0.122
0.146
0.137
6.40
12.30
9.60
Common solution one:
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,
4R
2, and 4, whereas the expressions for a (since Vc=a3) are 2R, 2R 2 ,and
.
3
For alloy A, let us calculate ρ assuming a BCC crystal structure.
ρ=
=
nAA
nAA
=
VC N A & 4 R #3
$
! NA
% 3"
(2atoms / unitcell )(43.1g / mol )
(". (4 )1.22 ! 10 /8 cm + 3
%"
23
(
)
/
unitcell
',
$ 6.023 ! 10 atoms / mol
)
3
"&"#
*
(
)
(
)
=6.40 g/cm 3
Therefore, its crystal structure is BCC.
For alloy B, let us calculate assuming a simple cubic crystal structure.
ρ=
nAB
(2a )3 N A
=
(1atom / unitcell )(184.4 g / mol )
{[(2)(1.46 !10"8 cm)]3 /(unitcell )}(6.023 !1023 atoms / mol )
=12.3g/cm 3
Therefore, its crystal structure is simple cubic.
For alloy C, let us calculate assuming a BCC crystal structure.
ρ=
nAC
4R 3
( ) NA
3
=
(2atom / unitcell )(91.6 g / mol )
(". 4 ! (1.37 ! 10/8 cm) + 3
%"
23
/
unitcell
',
$ ! 6.023 ! 10 atoms / mol
)
3
"&"#
*
= 9.60 g/cm3
Therefore, its crystal structure is BCC.
Common solution two:
For FCC crystal,
4
4
4 " !R 3
4 " !R 3
V
3
3
APF = s =
=
=
a3
Vc
(2 2R )3
2
! = 0.74
6
For BCC crystal,
4
4
4
2 " !R 3
2 " !R 3
2 " !R 3
V
3
3
3
APF = s =
=
=
=
3
4
64
a
Vc
3
2
(
R)
R
3
3 3
For SC crystal,
4
4
1 " !R 3
1 " !R 3
1
V
3
3
APF = s =
=
= ! = 0.52
3
3
(2R )
a
6
Vc
From
ρ=
nAA
VC N A
V
then APF = s =
Vc
! Vc =
nAA
!N A
4
n ! #R 3
3
=
nAA
"N A
4 3
"R !N A
3
AA
3
! = 0.68
8
For alloy A，
3
4
4 3
# ! 1.22 ! 10"8 cm ! 6.40 g / cm3 ! 6.023 ! 1023 atoms / mol
"R !N A
APF = 3
= 3
43.1g / mol
AA
(
)
= 0.68
Therefore, its crystal structure is BCC.
For alloy B，
3
4
4 3
# ! 1.46 ! 10"8 cm ! 12.3 g / cm3 ! 6.023 ! 1023 atoms / mol
"R !N A
APF = 3
= 3
184.4 g / mol
AA
(
)
= 0.52
Therefore, its crystal structure is SC.
For alloy C，
3
4
4 3
# ! 1.37 ! 10"8 cm ! 9.60 g / cm3 ! 6.023 ! 1023 atoms / mol
"R !N A
APF = 3
= 3
91.6 g / mol
AA
(
= 0.68
Therefore, its crystal structure is BCC.
)
6．Explain why the properties of polycrystalline materials are most often isotropic.
Each individual grain in a polycrystalline material may be anisotropic. In a
polycrystalline assembly, however, the grains can adopt random orientations. If there is
a large number of grains randomly oriented in a solid, the anisotropy of the individual
grains becomes averaged, so the properties of the assembly become isotropic. Note that
anisotropy can persist in a polycrystalline material, if: (1) the number of grains is small;
or (2) the orientation of the grains is not random but instead has some preferential
orientation.
7．Suppose you are trying to select a material for use as the hood of a car. List the
important properties required by this application and briefly indicate why a plain-carbon
steel sheet is or is not a good material to choose for this application.
Open your mind to imagine what the important properties are. Your answer should
include the following properties among possibly others: relatively high strength; good
formability (related to ductility); high toughness (resistance to brittle fracture during
impact); and paintability. Among the reasons why a plain-carbon steel is not a good
material are: poor corrosion resistance if unpainted and relatively high density.
8. (From test 1, Fall, 1994) Make an argument why hydrochloric acid (HCl) boils at T =
-84.9 ℃ whereas hydroflouric acid (HF) boils at T = 19.54 ℃.
F has higher electronegativity value than Cl, and F thus attracts the H 1s electron more
strongly than Cl does. Consequently, a stronger permanent dipole is formed in the F-H
bond than in the Cl-H bond. Stronger permanent dipoles will give rise to stronger
dipole-dipole bonding. A higher temperature is thus needed to break the HF-HF
dipole-dipole bonds than to break the HCl-HCl dipole-dipole bonds. The boiling point
of HF is consequently higher than that of HCl.