Chapter 5 homework solutions.pdf

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Department
Materials Science & Engineering
Course
MSE101H1
Professor
Scott Ramsay
Semester
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)(800C + 273 K) ln .60  10 m  v 5.30  10 28 m3  = 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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