Chap 4 Solns
CHAPTER 5
IMPERFECTIONS IN SOLIDS
PROBLEM SOLUTIONS
Vacancies and Self-Interstitials 5.1 Calculate the fraction of atom sites that are vacant for copper at its melting temperature of 1084°C (1357 K). Assume an energy for vacancy formation of 0.90 eV/atom. Solution In order to compute the fraction of atom sites that are vacant in copper at 1357 K, we must employ Equation 5.1. As stated in the problem, Qv = 0.90 eV/atom. Thus,
Q 0.90 eV /atom = exp v = exp N (8.62 105 eV /atom - K) (1357 K) kT
Nv
= 4.56 10-4
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5.2 Calculate the energy for vacancy formation in silver, given that the equilibrium number of vacancies at 800°C (1073 K) is 3.6 × 1023 m–3. The atomic weight and density (at 800°C) for silver are, respectively, 107.9 g/mol and 9.5 g/cm3. Solution This problem calls for the computation of the activation energy for vacancy formation in silver. Upon examination of Equation 5.1, all parameters besides Qv are given except N, the total number of atomic sites. However, N is related to the density, (), Avogadro's number (NA), and the atomic weight (A) according to Equation 5.2 as
N =
=
N A P b AP b
(6.02 1023 atoms / mol)(9.5 g /cm3 )
107.9 g / mol
= 5.30 1022 atoms/cm3 = 5.30 1028 atoms/m3
Now, taking natural logarithms of both sides of Equation 5.1,
Q ln N v = ln N v kT
and, after some algebraic manipulation
N Qv = kT ln v N
Now, inserting values for the parameters given in the problem statement leads to
3.60 1023 m3 Qv = (8.62 10-5 eV/atom- K) (800C + 273 K) ln 5.30 1028 m3
= 1.10 eV/atom
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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
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.
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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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5.6 (a) Suppose that CaO is added as an impurity to Li2O. If the Ca2+ substitutes for Li+, what kind of vacancies would you expect to form? How many of these vacancies are created for every Ca 2+ added? (b) Suppose that CaO is added as an impurity to CaCl2. If the O2– substitutes for Cl–, what kind of vacancies would you expect to form? How many of these vacancies are created for every O2– added? Solution (a) For Ca2+ substituting for Li+ in Li2O, lithium vacancies would be created. For each Ca2+ substituting for Li+, one positive charge is added; in order to maintain charge neutrality, a single positive charge may be removed. Positive charges are eliminated by creating lithium vacancies, and for every Ca 2+ ion added, a single lithium vacancy is formed. (b) For O2- substituting for Cl- in CaCl2, chlorine vacancies would be created. For each O2- substituting for a Cl-, one negative charge is added; negative charges are eliminated by creating chlorine vacancies. In order to maintain charge neutrality, one O2- ion will lead to the formation of one chlorine vacancy.
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Specification of Composition 5.7 What is the composition, in atom percent, of an alloy that consists of 92.5 wt% Ag and 7.5 wt% Cu? Solution In order to compute composition, in atom percent, of a 92.5 wt% Ag-7.5 wt% Cu alloy, we employ Equation 5.9 as
' CAg =
CAg ACu CAg ACu CCu AAg
100
(92.5)(63.55 g / mol) = 100 (92.5)(63.55 g / mol) (7.5)(107.87g / mol)
= 87.9 at%
' CCu =
CCu AAg CAg ACu CCu AAg
100
(7.5)(107.87 g / mol) = 100 (92.5)(63.55 g / mol) (7.5)(107.87g / mol)
= 12.1 at%
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5.8 Calculate the composition, in weight percent, of an alloy that contains 105 kg of iron, 0.2 kg of carbon, and 1.0 kg of chromium. Solution The concentration, in weight percent, of an element in an alloy may be computed using a modified form of Equation 5.6. For this alloy, the concentration of iron (CFe) is just
CFe =
mFe
mFe mC mCr
100
105 kg = 100 = 98.87 wt% 105 kg 0.2 kg 1.0 kg
Similarly, for carbon
And for chromium
CC =
0.2 kg 100 = 0.19 wt% 105 kg 0.2 kg 1.0 kg
CCr =
1.0 kg 100 = 0.94 wt% 105 kg 0.2 kg 1.0 kg
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5.16 Iron and vanadium both have the BCC crystal structure, and V forms a substitutional solid solution for concentrations up to approximately 20 wt% V at room temperature. Compute the unit cell edge length for a 90 wt% Fe–10 wt% V alloy. Solution 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 5.13a and 5.14a. 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
ave =
100 CV V
CFe
Fe
100
=
10 wt% 6.10 g /cm 3
90 wt% 7.87 g /cm 3 3
= 7.65 g/cm
And for the average atomic weight
Aave =
100 CV AV
=
CFe 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
ave N A
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=
(2 atoms / unit cell)(55.32 g / mol) (7.65 g /cm3 )(6.02 1023 atoms / mol)
= 2.40 10-23 cm3/unit cell
And, finally
a = (VC )1/3 = (2.40 10 23cm 3/unit cell )1/3
= 2.89 x 10-8 cm = 0.289 nm
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Interfacial Defects 5.17 For an FCC single crystal, would you expect the surface energy for a (100) plane to be greater or less than that for a (111) plane? Why? (Note: You may want to consult the solution to Problem W3.46 at the end of Chapter 3.) Solution 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 the 1 1 solution to Problem W3.46, planar densities for FCC (100) and (111) planes are and , respectively— 2 2 2R 3 4R 0.25 0.29 that is and (where R is the atomic radius). Thus, since the planar density for (111) is greater, it will have 2 R R2 the lower surface energy.
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5.18 (a) For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary energy? Why? (b) The grain boundary energy of a small-angle grain boundary is less than for a high-angle one. Why is this so? Solution (a) The surface energy will be greater than the grain boundary energy. For grain boundaries, some atoms on one side of a boundary will bond to atoms on the other side; such is not the case for surface atoms. Therefore, there will be fewer unsatisfied bonds along a grain boundary. (b) The small-angle grain boundary energy is lower than for a high-angle one because more atoms bond across the boundary for the small-angle, and, thus, there are fewer unsatisfied bonds.
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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 (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 n1 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
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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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DESIGN PROBLEMS 5.D2 Gallium arsenide (GaAs) and indium arsenide (InAs) both have the zinc blende crystal structure and are soluble in each other at all concentrations. Determine the concentration in weight percent of InAs that must be added to GaAs to yield a unit cell edge length of 0.5820 nm. The densities of GaAs and InAs are 5.316 and 5.668 g/cm3, respectively. Solution This problem asks that we determine the concentration (in weight percent) of InAs that must be added to GaAs to yield a unit cell edge length of 0.5820 nm. The densities of GaAs and InAs were given in the problem statement as 5.316 and 5.668 g/cm3, respectively. To begin, it is necessary to employ Equation 3.6, and solve for the unit cell volume, VC, for the InAs-GaAs alloy as
VC =
n' Aave
ave N A
where Aave and ave are the atomic weight and density, respectively, of the InAs-GaAs alloy. Inasmuch as both of these materials have the zinc blende crystal structure, which has cubic symmetry, VC is just the cube of the unit cell length, a. That is VC = a3 = (0.5820 nm)3
= (5.820 108 cm)3 1.971 1022 cm3 It is now necessary to construct expressions for Aave and ave in terms of the concentration of indium arsenide, CInAs using Equations 5.14a and 5.13a. For Aave we have
Aave =
=
100 CInAs (100 CInAs) AInAs AGaAs
100 CInAs 189.74 g / mol
whereas for ave
(100
CInAs) 144.64 g / mol
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ave =
100
CInAs
InAs
=
(100
100 CInAs 5.668 g /cm 3
CInAs) GaAs
(100
CInAs) 5.316 g /cm 3
Within the zinc blende unit cell there are four formula units, and thus, the value of n' in Equation 3.6 is 4; hence, this expression may be written in terms of the concentration of InAs in weight percent as follows: VC = 1.971 x 10-22 cm3
=
n' Aave
ave N A
100 (4 fu / unit cell) C InAs (100 CInAs) 189.74 g / mol 144.64 g / mol = 100 (6.02 1023 fu / mol) C InAs (100 CInAs) 5.316 g /cm3 5.668 g /cm3
And solving this expression for CInAs leads to CInAs = 46.1 wt%.
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IMPERFECTIONS IN SOLIDS
PROBLEM SOLUTIONS
Vacancies and Self-Interstitials 5.1 Calculate the fraction of atom sites that are vacant for copper at its melting temperature of 1084°C (1357 K). Assume an energy for vacancy formation of 0.90 eV/atom. Solution In order to compute the fraction of atom sites that are vacant in copper at 1357 K, we must employ Equation 5.1. As stated in the problem, Qv = 0.90 eV/atom. Thus,
Q 0.90 eV /atom = exp v = exp N (8.62 105 eV /atom - K) (1357 K) kT
Nv
= 4.56 10-4
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5.2 Calculate the energy for vacancy formation in silver, given that the equilibrium number of vacancies at 800°C (1073 K) is 3.6 × 1023 m–3. The atomic weight and density (at 800°C) for silver are, respectively, 107.9 g/mol and 9.5 g/cm3. Solution This problem calls for the computation of the activation energy for vacancy formation in silver. Upon examination of Equation 5.1, all parameters besides Qv are given except N, the total number of atomic sites. However, N is related to the density, (), Avogadro's number (NA), and the atomic weight (A) according to Equation 5.2 as
N =
=
N A P b AP b
(6.02 1023 atoms / mol)(9.5 g /cm3 )
107.9 g / mol
= 5.30 1022 atoms/cm3 = 5.30 1028 atoms/m3
Now, taking natural logarithms of both sides of Equation 5.1,
Q ln N v = ln N v kT
and, after some algebraic manipulation
N Qv = kT ln v N
Now, inserting values for the parameters given in the problem statement leads to
3.60 1023 m3 Qv = (8.62 10-5 eV/atom- K) (800C + 273 K) ln 5.30 1028 m3
= 1.10 eV/atom
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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
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.
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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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5.6 (a) Suppose that CaO is added as an impurity to Li2O. If the Ca2+ substitutes for Li+, what kind of vacancies would you expect to form? How many of these vacancies are created for every Ca 2+ added? (b) Suppose that CaO is added as an impurity to CaCl2. If the O2– substitutes for Cl–, what kind of vacancies would you expect to form? How many of these vacancies are created for every O2– added? Solution (a) For Ca2+ substituting for Li+ in Li2O, lithium vacancies would be created. For each Ca2+ substituting for Li+, one positive charge is added; in order to maintain charge neutrality, a single positive charge may be removed. Positive charges are eliminated by creating lithium vacancies, and for every Ca 2+ ion added, a single lithium vacancy is formed. (b) For O2- substituting for Cl- in CaCl2, chlorine vacancies would be created. For each O2- substituting for a Cl-, one negative charge is added; negative charges are eliminated by creating chlorine vacancies. In order to maintain charge neutrality, one O2- ion will lead to the formation of one chlorine vacancy.
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Specification of Composition 5.7 What is the composition, in atom percent, of an alloy that consists of 92.5 wt% Ag and 7.5 wt% Cu? Solution In order to compute composition, in atom percent, of a 92.5 wt% Ag-7.5 wt% Cu alloy, we employ Equation 5.9 as
' CAg =
CAg ACu CAg ACu CCu AAg
100
(92.5)(63.55 g / mol) = 100 (92.5)(63.55 g / mol) (7.5)(107.87g / mol)
= 87.9 at%
' CCu =
CCu AAg CAg ACu CCu AAg
100
(7.5)(107.87 g / mol) = 100 (92.5)(63.55 g / mol) (7.5)(107.87g / mol)
= 12.1 at%
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5.8 Calculate the composition, in weight percent, of an alloy that contains 105 kg of iron, 0.2 kg of carbon, and 1.0 kg of chromium. Solution The concentration, in weight percent, of an element in an alloy may be computed using a modified form of Equation 5.6. For this alloy, the concentration of iron (CFe) is just
CFe =
mFe
mFe mC mCr
100
105 kg = 100 = 98.87 wt% 105 kg 0.2 kg 1.0 kg
Similarly, for carbon
And for chromium
CC =
0.2 kg 100 = 0.19 wt% 105 kg 0.2 kg 1.0 kg
CCr =
1.0 kg 100 = 0.94 wt% 105 kg 0.2 kg 1.0 kg
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5.16 Iron and vanadium both have the BCC crystal structure, and V forms a substitutional solid solution for concentrations up to approximately 20 wt% V at room temperature. Compute the unit cell edge length for a 90 wt% Fe–10 wt% V alloy. Solution 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 5.13a and 5.14a. 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
ave =
100 CV V
CFe
Fe
100
=
10 wt% 6.10 g /cm 3
90 wt% 7.87 g /cm 3 3
= 7.65 g/cm
And for the average atomic weight
Aave =
100 CV AV
=
CFe 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
ave N A
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=
(2 atoms / unit cell)(55.32 g / mol) (7.65 g /cm3 )(6.02 1023 atoms / mol)
= 2.40 10-23 cm3/unit cell
And, finally
a = (VC )1/3 = (2.40 10 23cm 3/unit cell )1/3
= 2.89 x 10-8 cm = 0.289 nm
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Interfacial Defects 5.17 For an FCC single crystal, would you expect the surface energy for a (100) plane to be greater or less than that for a (111) plane? Why? (Note: You may want to consult the solution to Problem W3.46 at the end of Chapter 3.) Solution 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 the 1 1 solution to Problem W3.46, planar densities for FCC (100) and (111) planes are and , respectively— 2 2 2R 3 4R 0.25 0.29 that is and (where R is the atomic radius). Thus, since the planar density for (111) is greater, it will have 2 R R2 the lower surface energy.
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5.18 (a) For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary energy? Why? (b) The grain boundary energy of a small-angle grain boundary is less than for a high-angle one. Why is this so? Solution (a) The surface energy will be greater than the grain boundary energy. For grain boundaries, some atoms on one side of a boundary will bond to atoms on the other side; such is not the case for surface atoms. Therefore, there will be fewer unsatisfied bonds along a grain boundary. (b) The small-angle grain boundary energy is lower than for a high-angle one because more atoms bond across the boundary for the small-angle, and, thus, there are fewer unsatisfied bonds.
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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 (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 n1 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
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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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DESIGN PROBLEMS 5.D2 Gallium arsenide (GaAs) and indium arsenide (InAs) both have the zinc blende crystal structure and are soluble in each other at all concentrations. Determine the concentration in weight percent of InAs that must be added to GaAs to yield a unit cell edge length of 0.5820 nm. The densities of GaAs and InAs are 5.316 and 5.668 g/cm3, respectively. Solution This problem asks that we determine the concentration (in weight percent) of InAs that must be added to GaAs to yield a unit cell edge length of 0.5820 nm. The densities of GaAs and InAs were given in the problem statement as 5.316 and 5.668 g/cm3, respectively. To begin, it is necessary to employ Equation 3.6, and solve for the unit cell volume, VC, for the InAs-GaAs alloy as
VC =
n' Aave
ave N A
where Aave and ave are the atomic weight and density, respectively, of the InAs-GaAs alloy. Inasmuch as both of these materials have the zinc blende crystal structure, which has cubic symmetry, VC is just the cube of the unit cell length, a. That is VC = a3 = (0.5820 nm)3
= (5.820 108 cm)3 1.971 1022 cm3 It is now necessary to construct expressions for Aave and ave in terms of the concentration of indium arsenide, CInAs using Equations 5.14a and 5.13a. For Aave we have
Aave =
=
100 CInAs (100 CInAs) AInAs AGaAs
100 CInAs 189.74 g / mol
whereas for ave
(100
CInAs) 144.64 g / mol
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ave =
100
CInAs
InAs
=
(100
100 CInAs 5.668 g /cm 3
CInAs) GaAs
(100
CInAs) 5.316 g /cm 3
Within the zinc blende unit cell there are four formula units, and thus, the value of n' in Equation 3.6 is 4; hence, this expression may be written in terms of the concentration of InAs in weight percent as follows: VC = 1.971 x 10-22 cm3
=
n' Aave
ave N A
100 (4 fu / unit cell) C InAs (100 CInAs) 189.74 g / mol 144.64 g / mol = 100 (6.02 1023 fu / mol) C InAs (100 CInAs) 5.316 g /cm3 5.668 g /cm3
And solving this expression for CInAs leads to CInAs = 46.1 wt%.
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