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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,

Nv

N= exp

Qv

kT

= exp

0.90 eV /atom

(8.62 105 eV /atom - K) (1357 K)

= 4.56 10-4


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 APb

APb

= (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,

ln Nv = ln N Qv

kT

and, after some algebraic manipulation

Qv = kT ln N v

N

Now, inserting values for the parameters given in the problem statement leads to

Qv = (8.62 10-5 eV/atom- K) (800C + 273 K) ln 3.60 1023 m3

5.30 1028 m3

= 1.10 eV/atom


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.

Atomic Crystal Electro-

Element Radius (nm) Structure negativity 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.


Crystal Electro-

Element R% Structure negativity Valence

Ni FCC 2+

C –43

H –63

O –52

Ag +16 FCC +0.1 1+

Al +15 FCC -0.3 3+

Co +0.6 HCP 0 2+

Cr +0.2 BCC -0.2 3+

Fe -0.4 BCC 0 2+

Pt +11 FCC +0.4 2+

Zn +7 HCP -0.2 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.


5.6 (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 Ca2+

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 Ca2+ 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.


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' =

CAgACu

CAgACu CCu AAg

100

= (92.5)(63.55 g /mol)

(92.5)(63.55 g /mol) (7.5)(107.87g /mol) 100

= 87.9 at%

CCu' =

CCu AAg

CAgACu CCu AAg

100

= (7.5)(107.87 g /mol)

(92.5)(63.55 g /mol) (7.5)(107.87g /mol) 100

= 12.1 at%


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

105 kg 0.2 kg 1.0 kg 100 = 98.87 wt%

Similarly, for carbon

CC = 0.2 kg

105 kg 0.2 kg 1.0 kg 100 = 0.19 wt%

And for chromium

CCr = 1.0 kg

105 kg 0.2 kg 1.0 kg 100 = 0.94 wt%


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 /cm3

90 wt%

7.87 g /cm3

= 7.65 g/cm3

And for the average atomic weight

Aave = 100

CV

AV

CFe

AFe

= 100

10 wt%

50.94 g /mole

90 wt%

55.85 g /mol

= 55.32 g/mol

Now, VC is determined from Equation 3.5 as

VC = nAave

aveNA


= (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 1023cm3/unit cell)1/3

= 2.89 x 10-8 cm = 0.289 nm


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

solution to Problem W3.46, planar densities for FCC (100) and (111) planes are

1

4R2 and

1

2R2 3, respectively—

that is

0.25

R2 and

0.29

R2 (where R is the atomic radius). Thus, since the planar density for (111) is greater, it will have

the lower surface energy.


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.


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

6.3 = 9.5 mm

Hence, the average grain diameter, d, is

d = ave. line length intersected

magnification =

9.5 mm

90= 0.106 mm

(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

N MM

100

2

2n1

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:

log N M 2 log M

100

(n 1) log 2

Solving this expression for n gives

n

log N M 2 logM

100

log 2 1


From Figure 10.29(a), NM is measured to be approximately 4, which leads to

n

log 4 2 log90

100

log 2 1

= 2.7


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

aveNA

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

AInAs

(100 CInAs)

AGaAs

= 100

CInAs

189.74 g /mol

(100 CInAs)

144.64 g /mol

whereas for ave


ave = 100

CInAs

InAs

(100 CInAs)

GaAs

= 100

CInAs

5.668 g /cm3

(100 CInAs)

5.316 g /cm3

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

aveN A

=

(4 fu /unit cell)100

CInAs

189.74 g /mol

(100 CInAs)

144.64 g /mol

100

CInAs

5.668 g /cm3

(100 CInAs)

5.316 g /cm3

(6.02 1023 fu /mol)

And solving this expression for CInAs leads to CInAs = 46.1 wt%.

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