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CHAPTER 4
IMPERFECTIONS IN SOLIDS
PROBLEM SOLUTIONS
4.1
In order to compute the fraction of atom sites that are vacant in lead at 600 K, we must employ
Equation (4.1).
As stated in the problem,
Q
v
= 0.55 eV/atom.
Thus,
N
V
N
= exp
−
Q
V
kT
⎛
⎝
⎜
⎞
⎠
⎟
−
0.55 eV /atom
8.62 x 10
−
5
eV /atom K
()
(600 K)
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
= 2.41 x 10
5
4.2
Determination of the number of vacancies per cubic meter in gold at 900
°
C (1173 K) requires the
utilization of Equations (4.1) and (4.2) as follows:
N
V
= N exp
−
Q
V
kT
⎛
⎝
⎜
⎞
⎠
⎟
=
N
A
ρ
Au
A
Au
exp
−
Q
V
kT
⎛
⎝
⎜
⎞
⎠
⎟
6.023 x 10
23
atoms/mol
19.32 g/ cm
3
196.9 g/mol
exp
−
0.98 eV / atom
−
5
eV /atom
−
K
(1173 K)
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
= 3.65 x 10
18
cm
3
= 3.65 x 10
24
m
3
4.3
This problem calls for the computation of the energy for vacancy formation in silver.
Upon
examination of Equation (4.1), all parameters besides
Q
v
are given except
N
, the total number of
atomic sites. However,
N
is related to the density, (
ρ
), Avogadro's number (
N
A
), and the atomic
weight (
A
) according to Equation (4.2) as
N =
N
A
ρ
Ag
A
Ag
23
atoms /mol
10.49 g/cm
3
(
)
107.87 g/mol
1
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View Full Document= 5.86 x 10
22
atoms/cm
3
= 5.86 x 10
28
atoms/m
3
Now, taking natural logarithms of both sides of Equation (4.1), and, after some algebraic
manipulation
Q
V
=
−
RT ln
N
V
N
⎛
⎝
⎜
⎞
⎠
⎟
−
8.62 x 10
5
eV/atom  K
()
(1073 K) ln
3.60 x 10
23
m
−
3
5.86 x 10
28
m
−
3
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
= 1.11 eV/atom
4.4
In this problem we are asked to cite which of the elements listed form with Cu the three possible solid
solution types.
For complete substitutional solubility the following criteria must be met:
1) the
difference in atomic radii between Cu and the other element (
∆
R%
) must be less than ±15%, 2) the
crystal structures must be the same, 3) the electronegativities must be similar, and 4) the valences
should be the same, or nearly the same.
Below are tabulated, for the various elements, these
criteria.
Crystal
∆
Electro
Element
∆
R%
Structure
negativity
Valence
Cu
FCC
2+
C
44
H
64
O
53
Ag
+13
FCC
0
1+
Al
+12
FCC
0.4
3+
Co
2
HCP
0.1
2+
Cr
2
BCC
0.3
3+
Fe
3
BCC
0.1
2+
Ni
3
FCC
0.1
2+
Pd
+8
FCC
+0.3
2+
Pt
+9
FCC
+0.3
2+
Zn
+4
HCP
0.3
2+
2
(a)
Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions having complete
solubility.
(b) Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility.
All these
metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii
and that for Cu are greater than ±15%, and/or have a valence different than 2+.
(c)
C, H, and O form interstitial solid solutions.
These elements have atomic radii that are
significantly smaller than the atomic radius of Cu.
4.5
In the drawing below is shown the atoms on the (100) face of an FCC unit cell;
the interstitial site is
at the center of the edge.
The diameter of an atom that will just fit into this site (2
r
) is just the difference between the unit cell
edge length (
a
) and the radii of the two host atoms that are located on either side of the site (
R
);
that
is
2r = a  2R
However, for FCC
a
is related to
R
according to Equation (3.1) as
a
= 2
R
2
;
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 Spring '08
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