CHAPTER 5
IMPERFECTIONS IN SOLIDS
PROBLEM SOLUTIONS
Point Defects in Ceramics
5.4 Using the data given below that relate to the formation of Schottky defects in some oxide ceramic (having the
chemical formula MO), determine the following:
(a) the energy for defect formation (in eV),
(b) the equilibrium number of Schottky defects per cubic meter at 1000°C, and
(c) the identity of the oxide (i.e., what is the metal M?)
T (°C)
ρ (g/cm3)
Ns (m–3)
750
3.50
5.7 × 109
1000
3.45
?
1500
3.40
5.8 × 1017
Solution
The (a) portion of the problem asks that we compute the energy for defect formation. To begin, let us combine a
modified form of Equation 5.2 and Equation 5.4 as
 Q 
N s  N exp  s 
 2kT 

 N  
 Q 
A
s
 
A + A 
 exp  2kT 


 M
O 
Inasmuch as this is a hypothetical oxide material, we don't know the atomic weight of metal M, nor the value of Qs in

the above equation. Therefore, let us write equations of the above form for two temperatures, T1 and T2. These are
as follows:
 N 



A 1 exp  Qs 
N s1  
A + A 
 2kT 
 M

O 
1 
(5.S1a)
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 N 



A 2 exp  Qs 
N s2  
A + A 
 2kT 
 M

O 
2 
(5.S1b)
Dividing the first of these equations by the second leads to

 N 


A 1  exp 

A + A 

N s1
 M

O 

 N 


N s2
A 2

A + A 
 exp 

 M

O 
Qs 

2kT1 

Qs 

2kT2 

which, after some algebraic manipulation, reduces to the form

N s1
N s2

 Q  1
1 

exp  s 


2
T2 


 2k T1

1
(5.S2)
Now, taking natural logarithms of both sides of this equation gives

N 
 


s1  ln 1   Qs  1  1 
ln 
N 

  2k T
T2 
 s2 
 2 
 1

and solving for Qs leads to the expression

 N 
 
s1  ln1 
2k ln
N 

 

 2 
  s2 

Qs =
1
1

T1
T2
Let us take T1 = 750°C and T2 = 1500°C, and we may compute the value of Qs as

 5.7  109 m-3 
3.50 g/cm3 
(2)(8.62  10-5 eV/K) ln 
  ln 

17
-3

3.40 g/cm3 
 5.8  10 m 

Qs =
1
1

750 + 273 K
1500 + 273 K

= 7.70 eV
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(b) It is now possible to solve for Ns at 1000°C using Equation 5.S2 above. This time let's take T1 = 1000°C and T2
= 750°C. Thus, solving for Ns1, substituting values provided in the problem statement and Qs determined above
yields
N s1 

N s2 1
2
 Q  1
1 

exp  s 


T2 


 2k T1




(5.7  109 m-3)( 3.45 g/cm3)
1
1
7.70 eV
exp






750 + 273 K 
 (2)(8.62  10-5 eV/K) 1000 + 273 K
3.50 g/cm3


= 3.0  1013 m-3

(c) And, finally, we want to determine the identity of metal M. This is possible by computing the atomic weight of
M (AM) from Equation 5.S1a. Rearrangement of this expression leads to
 N 

A 1  = N exp

A + A 
s1
 M
O 
 Q 
s

2kT 

 1 
And, after further algebraic manipulation





N A1


= AM + AO

 Q 
N exp  s 
2kT 
 s1
 1 

And, solving this expression for AM gives





N A1


AM = 
 AO


N exp  Qs 
2kT 
 s1
 1 

Now, assuming that T1 = 750°C, the value of AM is

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





(6.02  1023 ions/mol)( 3.50 g/cm3 )(106 cm3 / m3 )
AM = 
  16.00 g/mol

7.7 eV
(5.7  109 ions/m3 ) exp 



(2)(8.62  10-5 eV/K)(750 + 273 K)


= 24.45 g/mol

Upon consultation of the periodic table in Figure 2.6, the divalent metal (i.e., that forms M 2+ ions) that has an atomic
weight closest to 24.45 g/mol is magnesium. Thus, this metal oxide is MgO.
Impurities in Solids
5.5 Below, atomic radius, crystal structure, electronegativity, and the most common valence are tabulated, for
several elements; for those that are nonmetals, only atomic radii are indicated.
Element
Atomic
Radius (nm)
Crystal
Structure
Electronegativity
Valence
Ni
0.1246
FCC
1.8
+2
C
0.071
H
0.046
O
0.060
Ag
0.1445
FCC
1.9
+1
Al
0.1431
FCC
1.5
+3
Co
0.1253
HCP
1.8
+2
Cr
0.1249
BCC
1.6
+3
Fe
0.1241
BCC
1.8
+2
Pt
0.1387
FCC
2.2
+2
Zn
0.1332
HCP
1.6
+2
Which of these elements would you expect to form the following with nickel:
(a) A substitutional solid solution having complete solubility
(b) A substitutional solid solution of incomplete solubility
(c) An interstitial solid solution
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Solution
In this problem we are asked to cite which of the elements listed form with Ni the three possible solid solution types.
For complete substitutional solubility the following criteria must be met: 1) the difference in atomic radii between
Ni 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.
Element
Ni
C
H
O
Ag
Al
Co
Cr
Fe
Pt
Zn
R%
Crystal
Structure
Electronegativity
FCC
–43
–63
–52
+16
+15
+0.6
+0.2
-0.4
+11
+7
FCC
FCC
HCP
BCC
BCC
FCC
HCP
Valence
2+
+0.1
-0.3
0
-0.2
0
+0.4
-0.2
1+
3+
2+
3+
2+
2+
2+
(a) Pt is the only element that meets all of the criteria and thus forms a substitutional solid solution having
complete solubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal
structure, and thus display complete solid solubility at these temperatures.
(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 Ni 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 Ni.
5.12 Some hypothetical alloy is composed of 25 wt% of metal A and 75 wt% of metal B. If the densities of
metals A and B are 6.17 and 8.00 g/cm 3, respectively, whereas their respective atomic weights are 171.3 and 162.0
g/mol, determine whether the crystal structure for this alloy is simple cubic, face-centered cubic, or body-centered
cubic. Assume a unit cell edge length of 0.332 nm.
Solution
In order to solve this problem it is necessary to employ Equation 3.5; in this expression density and atomic weight
will be averages for the alloy—that is
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ave =
nAave
VC N A
Inasmuch as for each of the possible crystal structures, the unit cell is cubic, then VC = a3, or

ave =
nAave
a3N A
And, in order to determine the crystal structure it is necessary to solve for n, the number of atoms per unit

cell. For n =1, the crystal structure is simple cubic, whereas for n values of 2 and 4, the crystal structure will be
either BCC or FCC, respectively. When we solve the above expression for n the result is as follows:
n =
avea 3 N A
Aave
Expressions for Aave and aveare found in Equations 5.14a and 5.13a, respectively, which, when incorporated into

the above expression yields




 100
a 3 N
A
C A
C B 



 B 
 A
n =




 100

C A
C B 



AB 
 AA
Substitution of the concentration values (i.e., CA = 25 wt% and CB = 75 wt%) as well as values for the

other parameters given in the problem statement, into the above equation gives




100

(3.32  10-8 nm)3 (6.02  1023 atoms/mol)
 25 wt%  75 wt% 


8.00 g/cm3 
6.17 g/cm3
n =




100


75 wt% 
 25 wt% 


162.0 g/mol 
171.3 g/mol
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
by Sections
= 1.00 atom/unit cell
Therefore, on the basis of this value, the crystal structure is simple cubic.
5.15 Germanium forms a substitutional solid solution with silicon. Compute the weight percent of
germanium that must be added to silicon to yield an alloy that contains 2.43 × 10 21 Ge atoms per cubic centimeter.
The densities of pure Ge and Si are 5.32 and 2.33 g/cm 3, respectively.
Solution
To solve this problem, employment of Equation 5.22 is necessary, using the following values:
N1 = NGe = 2.43 x 1021 atoms/cm3
1 = Ge = 5.32 g/cm3
2 = Si = 2.33 g/cm3
A1 = AGe = 72.64 g/mol
A2 = ASi = 28.09 g/mol
Thus
CGe =

=
1

1
100
N ASi
N Ge AGe

Si
Ge
100
2.33 g /cm3 

atoms / mol)(2.33 g /cm3 )



(2.43  1021 atoms /cm3 ) (72.64 g / mol) 5.32 g /cm3 
(6.02
1023
= 11.7 wt%
Grain Size Determination
5.20 (a) Employing the intercept technique, determine the average grain size for the steel specimen whose
microstructure is shown in Figure 10.29(a); use at least seven straight-line segments.
(b) Estimate the ASTM grain size number for this material.
Solution
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(a) This portion of the problem calls for a determination of the average grain size of the specimen which
microstructure is shown in Figure 10.29(a). Seven line segments were drawn across the micrograph, each of which
was 60 mm long. The average number of grain boundary intersections for these lines was 6.3. Therefore, the
average line length intersected is just
60 mm
= 9.5 mm
6.3
Hence, the average grain diameter, d, is

d =
ave. line length intersected
9.5 mm
=
= 0.106 mm
magnification
90
(b) This portion of the problem calls for us to estimate the ASTM grain size number for this same material.

The average grain size number, n, is related to the number of grains per square inch, N, at a magnification of 100
according to Equation 5.19.
However, the magnification of this micrograph is not 100x, but rather 90.
Consequently, it is necessary to use Equation 5.20
 M 2
N M    2 n1
100 
where NM = the number of grains per square inch at magnification M, and n is the ASTM grain size number. Taking

logarithms of both sides of this equation leads to the following:
 M 
log N M  2 log   (n  1) log 2
100 
Solving this expression for n gives

 M 
log N M  2 log  
100 
n
1
log 2
From Figure 10.29(a), NM is measured to be approximately 4, which leads to

 90 
log 4  2 log  
100 
n
1
log 2

= 2.7
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5.21 For an ASTM grain size of 6, approximately how many grains would there be per square inch at
(a) a magnification of 100, and
(b) without any magnification?
Solution
(a) This part of problem asks that we compute the number of grains per square inch for an ASTM grain size of 6 at a
magnification of 100. All we need do is solve for the parameter N in Equation 5.19, inasmuch as n = 6. Thus
N  2 n1

= 2 61 = 32 grains/in.2
(b) Now it is necessary to compute 
the value of N for no magnification. In order to solve this problem it is
necessary to use Equation 5.20:
 M 2
N M    2 n1
100 
where NM = the number of grains per square inch at magnification M, and n is the ASTM grain size number.

Without any magnification, M in the above equation is 1, and therefore,
 1 2
N 1    2 61  32
100 
And, solving for N1, N1 = 320,000 grains/in.2.

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