CHAPTER 4
IMPERFECTIONS IN SOLIDS
PROBLEM SOLUTIONS
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.
Element
R%
Cu
Crystal
Electro-
Structure
negativity
FCC
Valence
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+
(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+.
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(c)
C, H, and O form interstitial solid solutions.
These elements have atomic radii that are
significantly smaller than the atomic radius of Cu.
4.7 In order to compute composition, in atom percent, of a 92.5 wt% Ag-7.5 wt% Cu alloy, we employ
Equation (4.6) as
C 'Ag =
=
C Ag ACu
C Ag ACu  C CuA Ag
x 100
(92.5)(63.55 g / mol)
x 100
(92.5)(63.55 g/ mol)  (7.5)(107.87g / mol)
= 87.9 at%
'
CCu
=
=
C CuA Ag
C Ag ACu  CCuAAg
x 100
(7.5)(107.87 g / mol)
x 100
(92.5)(63.55 g/ mol)  (7.5)(107.87g / mol)
= 12.1 at%
4.13 This problem calls for a conversion of composition in atom percent to composition in weight percent.
The composition in atom percent for Problem 4.11 is 44.8 at% Ag, 46.2 at% Au, and 9.0 at% Cu.
Modification of Equation (4.7) to take into account a three-component alloy leads to the following
C Ag =
=
' A
C Ag
Ag
' A
' A
C 'Ag AAg  C Au
 CCu
Au
Cu
x 100
(44.8)(107.87 g / mol)
x 100
(44.8)(107 .87 g / mol)  (46.2)(196.97 g / mol)  (9.0)(63.55 g / mol)
= 33.3 wt%
C Au =
' A
C Au
Au
C' A
Ag Ag
 C' A
Au Au
58
 C' A
Cu Cu
x 100
=
(46.2)(196.97 g / mol)
x 100
(44.8)(107 .87 g / mol)  (46.2)(196.97 g / mol)  (9.0)(63.55 g / mol)
= 62.7 wt%
CCu =
' A
C Cu
Cu
C' A
Ag Ag
=
 C' A
Au Au
 C' A
x 100
Cu Cu
(9.0)(63.55 g / mol)
x 100
(44.8)(107 .87 g / mol)  (46.2)(196.97 g / mol)  (9.0)(63.55 g / mol)
= 4.0 wt%
4.25 (a) The Burgers vector will point in that direction having the highest linear density. From Section
3.11, the linear density for the [110] direction in FCC is 1/2R, the maximum possible; therefore for
FCC
b =
a
[110]
2
From Problem 3.46 the linear density for the [111] direction in BCC is also 1/2R, and
therefore for BCC
b =
a
[111]
2
For simple cubic, a unit cell of which is shown in Figure 3.19, the atom spheres touch one
another along the cube edges (i.e., in [100] directions) and therefore, the atomic packing is greatest
in these directions. Therefore, the Burgers vector is
b =
a
[100]
2
(b) For Cu which has an FCC crystal structure, R = 0.1278 nm (Table 3.1) and a = 2R 2 =
0.3615 nm [Equation (3.1)]; therefore
59
b =
=
0.3615 nm
2
a
2
2
(1 )
h
2
 k
2
 (1 )
2
2
 l
2
 (0) = 0.2556 nm
For Fe which has a BCC crystal structure, R = 0.1241 nm (Table 3.1) and a 
4R
= 0.2866
3
nm [Equation (3.3)]; hence
b =
0.2866 nm
2
2
2
(1)  (1)  (1)
2
= 0.2482 nm
4.29 (a) The interfacial defect that exists for this stacking sequence is a twin boundary, which occurs at
the following position
The stacking sequence on one side of this position is mirrored on the other side.
(b) The interfacial defect that exists within this FCC stacking sequence is a stacking fault, which
occurs over the region indicated
For this region, the BCBC stacking sequence is HCP.
4.31 (a) This portion of the problem calls for a determination of the average grain size of the specimen
which microstructure is shown in Figure 9.22a.
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
60 mm
= 9.5 mm
6.3
Hence, the average grain diameter, d, is
d =
ave. line length int er sected
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 100x according to Equation 4.16. However, the magnification of this micrograph is
not 100x, but rather 90x. Consequently, it is necessary to use the following equation:
2
 M 
NM 
  2 n  1
100 
where NM = the number of grains per square inch at magnification M, and n is the ASTM grain size
number. (The above equation makes use of the fact that, while magnification is a length parameter,
area is expressed in terms of units of length squared. As a consequence, the number of grains per
unit area increases with the square of the increase in magnification.) Solving the above expression
for n leads to
 M 
log NM  2 log 

100
n
1
log 2
From Figure 9.22a, 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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