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MSE101H1
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Scott Ramsay
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Chapter 5

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Materials Science & Engineering

MSE101H1

Scott Ramsay

Spring

Description

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,
N Q 0.90 eV/atom
v = exp v= exp
N kT (8.62 10 5 eV/atom- K )(1357 K)
-4
= 4.56 10
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 5.2 Calculate the energy for vacancy formation in silver, given that the equilibrium number of vacancies
23 –3
at 800°C (1073 K) is 3.6 × 10 m . The atomic weight and density (at 800°C) for silver are, respectively, 107.9
3
g/mol and 9.5 g/cm .
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 Q ave given except N, the total number of atomic sites.
However, N is related to the density, (), Avogadro's number AN ), and the atomic weight (A) according to Equation
5.2 as
N = N A Pb
A
Pb
(6.02 10 23 atoms /mol)(9.5 g/cm 3)
= 107.9 g/mol
= 5.30 102 atoms/cm = 5.30 10 28 atoms/m 3
Now, taking natural logarithms of both sides of Equation 5.1,
lnN = lnN Qv
v kT
and, after some algebraic manipulation
N
Q v kT ln v
N
Now, inserting values for the parameters given in the problem statement leads to
23 3
Q = (8.62 10 -5 eV/atom - K)(800C + 273 K) ln .60 10 m
v 5.30 10 28 m3
= 1.10 eV/atom
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 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.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 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.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 2+ +
5.6 (a) Suppose that CaO is added as an impurity to Li O. If 2he Ca substitutes for Li , what kind of
2+
vacancies would you expect to form? How many of these vacancies are created for every Ca added?
2– –
(b) Suppose that CaO is added as an impurity to CaCl . If the 2 substitutes for Cl , what kind of
2–
vacancies would you expect to form? How many of these vacancies are created for every O added?
Solution
2+ + 2+
(a) For Ca substituting for Li in Li 2, lithium vacancies would be created. For each Ca substituting
for Li , one positive charge is added; in order to maintain charge neutrality, a single positive charge may be
2+
removed. Positive charges are eliminated by creating lithium vacancies, and for every Ca ion added, a single
lithium vacancy is formed.
(b) For O 2- substituting for Cl in CaCl , chlorine vacancies would be created. For each O 2- substituting
2
for a Cl , one negative charge is added; negative charges are eliminated by creating chlorine vacancies. In order to
maintain charge neutrality, one O 2- ion will lead to the formation of one chlorine vacancy.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 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 Cu
CAg = 100
C Ag Cu C Cu Ag
(92.5)(63.55 g/mol)
= 100
(92.5)(63.55 g/mol) (7.5)(107.87g/mol)
= 87.9 at%
' CCu Ag
CCu = C A C A 100
Ag Cu Cu Ag
(7.5)(107.87 g/mol)
= 100
(92.5)(63.55 g/mol) (7.5)(107.87g/mol)
= 12.1 at%
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 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 (C ) is just
Fe
m Fe
CFe = 100
mFe mC m Cr
105 kg
105 kg 0.2 kg 1.0 kg 100 = 98.87 wt%
Similarly, for carbon
0.2 kg
C C 100 = 0.19 wt%
105 kg 0.2 kg 1.0 kg
And for chromium
1.0 kg
CCr = 105 kg 0.2 kg 1.0 kg 100 = 0.94 wt%
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 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
3 3
and vanadium are 7.87g/cm and 6.10 g/cm , respectively, (as taken from inside the front cover), the average density
is just

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