4-4
Impurities in Solids
4.4 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
∆R%
Ni
C
H
O
Ag
Al
Co
Cr
Fe
Pt
Zn
–43
–63
–52
+16
+15
+0.6
+0.2
-0.4
+11
+7
Crystal
Structure
∆Electronegativity
FCC
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.
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4-25
4.18 This problem asks that we determine, for a hypothetical alloy that is composed of 25 wt% of metal A
and 75 wt% of metal B, whether the crystal structure is simple cubic, face-centered cubic, or body-centered cubic.
We are given the densities of the these metals (ρ = 6.17 g/cm3 and ρ = 8.00 g/cm3 for B), their atomic weights
A
B
(AA = 171.3 g/mol and AB = 162.0 g/mol), and that the unit cell edge length is 0.332 nm (i.e., 3.32 x 10-8 cm). 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
ρave =
nAave
VC N A
Inasmuch as for each of the possible crystal structures, the unit cell is cubic, then VC = a3, or
ρave =
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 =
ρave a 3 N A
Aave
Expressions for Aave and ρave are found in Equations 4.11a and 4.10a, respectively, which, when incorporated into
the above expression yields
⎛
⎞
⎜
⎟
100
⎜
⎟a 3 N
A
⎜ CA
CB ⎟
+
⎜
⎟
ρB ⎠
⎝ ρA
n =
⎞
⎛
⎟
⎜
100
⎟
⎜
⎜ CA
CB ⎟
+
⎟
⎜
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
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4-26
⎛
⎞
⎜
⎟
100
⎜
⎟(3.32 × 10-8 nm) 3 (6.023 × 10 23 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
= 1.00 atom/unit cell
Therefore, on the basis of this value, the crystal structure is simple cubic.
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4-34
4.25 This problems asks that we compute the unit cell edge length for a 90 wt% Fe-10 wt% V alloy. First
of all, the atomic radii for Fe and V (using the table inside the front cover) are 0.124 and 0.132 nm, respectively.
Also, using Equation 3.5 it is possible to compute the unit cell volume, and inasmuch as the unit cell is cubic, the
unit cell edge length is just the cube root of the volume. However, it is first necessary to calculate the density and
average atomic weight of this alloy using Equations 4.10a and 4.11a. Inasmuch as the densities of iron and
vanadium are 7.87g/cm3 and 6.10 g/cm3, respectively, (as taken from inside the front cover), the average density is
just
ρave =
100
CV
C Fe
+
ρV
ρ Fe
100
=
10 wt%
6.10 g /cm3
+
90 wt%
7.87 g /cm 3
= 7.65 g/cm3
And for the average atomic weight
Aave =
100
CV
AV
=
+
C Fe
AFe
100
10 wt%
90 wt%
+
50.94 g / mole
55.85 g / mol
= 55.32 g/mol
Now, VC is determined from Equation 3.5 as
VC =
=
nAave
ρave N A
(2 atoms / unit cell)(55.32 g / mol)
(7.65
g /cm3 )(6.023 × 10 23 atoms / mol)
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4-35
= 2.40 x 10-23 cm3/unit cell
And, finally
a = (VC )1/ 3
= (2.40 × 10 −23cm3/unit cell)1/3
= 2.89 x 10-8 cm = 0.289 nm
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4-38
4.28 The surface energy for a crystallographic plane will depend on its packing density [i.e., the planar
density (Section 3.11)]—that is, the higher the packing density, the greater the number of nearest-neighbor atoms,
and the more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From
3
3
the solution to Problem 3.54, the planar densities for BCC (100) and (110) are
and
, respectively—
16R 2
8R 2 2
0.19
0.27
and
. Thus, since the planar density for (110) is greater, it will have the lower surface energy.
that is
2
R
R2
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5-6
5.6 This problem calls for the mass of hydrogen, per hour, that diffuses through a Pd sheet. It first
becomes necessary to employ both Equations 5.1a and 5.3. Combining these expressions and solving for the mass
yields
M = JAt = − DAt
∆C
∆x
⎡ 0.4 − 2.0 kg / m3 ⎤
⎥
= − (1.7 × 10-8 m2 /s)(0.25 m2 ) (3600 s/h) ⎢
⎢⎣ 6 × 10−3 m ⎥⎦
= 4.1 x 10-3 kg/h
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5-24
5.19 For this problem we are given D0 (1.1 x 10-4) and Qd (272,000 J/mol) for the diffusion of Cr in Ni,
and asked to compute the temperature at which D = 1.2 x 10-14 m2/s. Solving for T from Equation 5.9a yields
T =
=
[
Qd
R (ln D0 − ln D)
272,000 J/mol
]
(8.31 J/mol - K) ln (1.1 × 10-4 m2 /s) - ln (1.2 × 10-14 m2 /s)
= 1427 K = 1154°C
Note: this problem may also be solved using the “Diffusion” module in the VMSE software. Open the “Diffusion”
module, click on the “D vs 1/T Plot” submodule, and then do the following:
1. In the left-hand window that appears, click on the “Custom1” box.
2. In the column on the right-hand side of this window enter the data for this problem. In the window
under “D0” enter preexponential value—viz. “1.1e-4”. Next just below the “Qd” window enter the activation
energy value—viz. “272”. It is next necessary to specify a temperature range over which the data is to be plotted.
The temperature at which D has the stipulated value is probably between 1000ºC and 1500ºC, so enter “1000” in the
“T Min” box that is beside “C”; and similarly for the maximum temperature—enter “1500” in the box below “T
Max”.
3. Next, at the bottom of this window, click the “Add Curve” button.
4. A log D versus 1/T plot then appears, with a line for the temperature dependence of the diffusion
coefficient for Cr in Ni. At the top of this curve is a diamond-shaped cursor. Click-and-drag this cursor down the
line to the point at which the entry under the “Diff Coeff (D):” label reads 1.2 x 10-14 m2/s. The temperature at
which the diffusion coefficient has this value is given under the label “Temperature (T):”. For this problem, the
value is 1430 K.
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7-3
7.3 It is possible for two screw dislocations of opposite sign to annihilate one another if their dislocation
lines are parallel. This is demonstrated in the figure below.
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7-11
7.9 This problem asks that we compute the magnitudes of the Burgers vectors for copper and iron. For
Cu, which has an FCC crystal structure, R = 0.1278 nm (Table 3.1) and a = 2 R 2 = 0.3615 nm (Equation 3.1);
also, from Equation 7.1a, the Burgers vector for FCC metals is
b =
a
〈110〉
2
Therefore, the values for u, v, and w in Equation 7.10 are 1, 1, and 0, respectively. Hence, the magnitude of the
Burgers vector for Cu is
b =
=
0.3615 nm
2
a
2
u2 + v2 + w 2
(1 ) 2 + (1 ) 2 + (0) 2 = 0.2556 nm
For Fe which has a BCC crystal structure, R = 0.1241 nm (Table 3.1) and a =
4R
= 0.2866 nm (Equation
3
3.3); also, from Equation 7.1b, the Burgers vector for BCC metals is
b =
a
〈111〉
2
Therefore, the values for u, v, and w in Equation 7.10 are 1, 1, and 1, respectively. Hence, the magnitude of the
Burgers vector for Fe is
b =
0.2866 nm
2
(1) 2 + (1) 2 + (1) 2 = 0.2482 nm
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7-16
7.13 We are asked to compute the critical resolved shear stress for Zn. As stipulated in the problem, φ =
65°, while possible values for λ are 30°, 48°, and 78°.
(a) Slip will occur along that direction for which (cos φ cos λ) is a maximum, or, in this case, for the
largest cos λ. Cosines for the possible λ values are given below.
cos(30°) = 0.87
cos(48°) = 0.67
cos(78°) = 0.21
Thus, the slip direction is at an angle of 30° with the tensile axis.
(b) From Equation 7.4, the critical resolved shear stress is just
τ crss = σ y (cos φ cos λ) max
= (2.5 MPa) [cos(65°) cos(30°)] = 0.90 MPa (130 psi)
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