CHAPTER 4IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS Vacancies and Self-Interstitials 4.1 Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327°C (600 K). Assume an energy for vacancy formation of 0.55 eV/atom. Solution In order to compute the fraction of atom sites that are vacant in lead at 600 K, we must employ Equation 4.1. As stated in the problem, Q v = 0.55 eV/atom. Thus, N v N = exp − Q v kT | . ` · = exp − 0.55 eV/atom (8.62 × 10 −5 eV/atom-K)(600 K) ] ] ] = 2.41 × 10 -5 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. 4.2 Calculate the number of vacancies per cubic meter in iron at 850°C. The energy for vacancy formation is 1.08 eV/atom. Furthermore, the density and atomic weight for Fe are 7.65 g/cm 3 and 55.85 g/mol, respectively. Solution Determination of the number of vacancies per cubic meter in iron at 850°C (1123 K) requires the utilization of Equations 4.1 and 4.2 as follows: N v = N exp − Q v kT | . ` · = N A ρ Fe A Fe exp − Q v kT | . ` · And incorporation of values of the parameters provided in the problem statement into the above equation leads to N v = (6.022 × 10 23 atoms /mol )(7.65 g/cm 3 ) 55.85 g/mol exp − 1.08 eV/atom (8.62 × 10 −5 eV/atom−K)(850°C+273 K) ] ] ] = 1.18 × 10 18 cm -3 = 1.18 × 10 24 m -3 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. 4.3 Calculate the activation energy for vacancy formation in aluminum, given that the equilibrium number of vacancies at 500°C (773 K) is 7.57 × 10 23 m -3 . The atomic weight and density (at 500°C) for aluminum are, respectively, 26.98 g/mol and 2.62 g/cm 3 . Solution Upon examination of Equation 4.1, all parameters besides Q v are given except N, the total number of atomic sites. However, N is related to the density, (ρ Al ), Avogadro's number (N A ), and the atomic weight (A Al ) according to Equation 4.2 as N = N A ρ Al A Al = (6.022 × 10 23 atoms /mol )(2.62 g/cm 3 ) 26.98 g/mol = 5.85 × 10 22 atoms/cm 3 = 5.85 × 10 28 atoms/m 3 Now, taking natural logarithms of both sides of Equation 4.1, lnN v = lnN − Q v kT and, after some algebraic manipulation Q v = − kT ln N v N | . ` · = − (8.62 × 10 -5 eV/atom -K)(500°C+273K) ln 7.57 × 10 23 m −3 5.85 × 10 28 m −3 ] ] ] = 0.75 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 4.4 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 Cu 0.1278 FCC 1.9 +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 Ni 0.1246 FCC 1.8 +2 Pd 0.1376 FCC 2.2 +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 copper: (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 Cu the three possible solid solution types. For complete substitutional solubility the following criteria must be met: 1) the difference in atomic radii between Cu 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 Cu FCC 2+ C –44 H –64 O –53 Ag +13 FCC 0 1+ Al +12 FCC -0.4 3+ Co -2 HCP -0.1 2+ Cr -2 BCC -0.3 3+ 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. Fe -3 BCC -0.1 2+ Ni -3 FCC -0.1 2+ Pd +8 FCC +0.3 2+ Pt +9 FCC +0.3 2+ Zn +4 HCP -0.3 2+ (a) Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions 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 Cu 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 Cu. 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. 4.5 For both FCC and BCC crystal structures, there are two different types of interstitial sites. In each case, one site is larger than the other, and is normally occupied by impurity atoms. For FCC, this larger one is located at the center of each edge of the unit cell; it is termed an octahedral interstitial site. On the other hand, with BCC the larger site type is found at 0 1 2 1 4 positions—that is, lying on {100} faces, and situated midway between two unit cell edges on this face and one-quarter of the distance between the other two unit cell edges; it is termed a tetrahedral interstitial site. For both FCC and BCC crystal structures, compute the radius r of an impurity atom that will just fit into one of these sites in terms of the atomic radius R of the host atom. Solution In the drawing below is shown the atoms on the (100) face of an FCC unit cell; the interstitial site is at the center of the edge. The diameter of an atom that will just fit into this site (2r) is just the difference between that unit cell edge length (a) and the radii of the two host atoms that are located on either side of the site (R); that is 2r = a – 2R However, for FCC a is related to R according to Equation 3.1 as a ·2R 2 ; therefore, solving for r from the above equation gives r = a−2R 2 = 2R 2 −2R 2 = 0.41R A (100) face of a BCC unit cell is shown below. 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. The interstitial atom that just fits into this interstitial site is shown by the small circle. It is situated in the plane of this (100) face, midway between the two vertical unit cell edges, and one quarter of the distance between the bottom and top cell edges. From the right triangle that is defined by the three arrows we may write a 2 | . ` · 2 + a 4 | . ` · 2 = (R +r) 2 However, from Equation 3.3, a = 4R 3 , and, therefore, making this substitution, the above equation takes the form 4R 2 3 | . ` · 2 + 4R 4 3 | . ` · 2 = R 2 + 2Rr + r 2 After rearrangement the following quadratic equation results: r 2 + 2Rr −0.667 R 2 = 0 And upon solving for r: 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. r · −(2R) t (2R) 2 − (4)(1)(−0.667R 2 ) 2 · −2Rt 2.582R 2 And, finally r(+) · −2R +2.582R 2 · 0.291R r(−) · −2R −2.582R 2 · −2.291R Of course, only the r(+) root is possible, and, therefore, r = 0.291R. Thus, for a host atom of radius R, the size of an interstitial site for FCC is approximately 1.4 times that for BCC. 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 4.6 Derive the following equations: (a) Equation 4.7a (b) Equation 4.9a (c) Equation 4.10a (d) Equation 4.11b Solution (a) This problem asks that we derive Equation 4.7a. To begin, C 1 is defined according to Equation 4.3 as C 1 = m 1 m 1 + m 2 × 100 or, equivalently C 1 = m 1 ' m 1 ' + m 2 ' × 100 where the primed m's indicate masses in grams. From Equation 4.4 we may write m 1 ' = n m1 A 1 m 2 ' = n m2 A 2 And, substitution into the C 1 expression above C 1 = n m1 A 1 n m1 A 1 + n m2 A 2 × 100 From Equation 4.5 it is the case that 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. n m1 = C 1 ' (n m1 + n m2 ) 100 n m2 = C 2 ' (n m1 + n m2 ) 100 And substitution of these expressions into the above equation leads to C 1 = C 1 ' A 1 C 1 ' A 1 + C 2 ' A 2 × 100 which is just Equation 4.7a. (b) This problem asks that we derive Equation 4.9a. To begin, C 1 " is defined as the mass of component 1 per unit volume of alloy, or C 1 " = m 1 V If we assume that the total alloy volume V is equal to the sum of the volumes of the two constituents--i.e., V = V 1 + V 2 --then C 1 " = m 1 V 1 +V 2 Furthermore, the volume of each constituent is related to its density and mass as 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. V 1 = m 1 ρ 1 V 2 = m 2 ρ 2 This leads to C 1 " = m 1 m 1 ρ 1 + m 2 ρ 2 From Equation 4.3, m 1 and m 2 may be expressed as follows: m 1 = C 1 (m 1 + m 2 ) 100 m 2 = C 2 (m 1 + m 2 ) 100 Substitution of these equations into the preceding expression yields C 1 " = C 1 (m 1 +m 2 ) 100 C 1 (m 1 +m 2 ) 100 ρ 1 + C 2 (m 1 +m 2 ) 100 ρ 2 · C 1 C 1 ρ 1 + C 2 ρ 2 If the densities ρ 1 and ρ 2 are given in units of g/cm 3 , then conversion to units of kg/m 3 requires that we multiply this equation by 10 3 , inasmuch as 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. 1 g/cm 3 = 10 3 kg/m 3 Therefore, the previous equation takes the form C 1 " = C 1 C 1 ρ 1 + C 2 ρ 2 × 10 3 which is the desired expression. (c) Now we are asked to derive Equation 4.10a. The density of an alloy ρ ave is just the total alloy mass M divided by its volume V ρ ave = M V Or, in terms of the component elements 1 and 2 ρ ave = m 1 + m 2 V 1 + V 2 [Note: here it is assumed that the total alloy volume is equal to the separate volumes of the individual components, which is only an approximation; normally V will not be exactly equal to (V 1 + V 2 )]. Each of V 1 and V 2 may be expressed in terms of its mass and density as, V 1 · m 1 ρ 1 V 2 · m 2 ρ 2 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. When these expressions are substituted into the above equation, we get ρ ave = m 1 + m 2 m 1 ρ 1 + m 2 ρ 2 Furthermore, from Equation 4.3 m 1 = C 1 (m 1 + m 2 ) 100 m 2 = C 2 (m 1 + m 2 ) 100 Which, when substituted into the above ρ ave expression yields ρ ave = m 1 + m 2 C 1 (m 1 + m 2 ) 100 ρ 1 + C 2 (m 1 + m 2 ) 100 ρ 2 And, finally, this equation reduces to = 100 C 1 ρ 1 + C 2 ρ 2 (d) And, finally, the derivation of Equation 4.11b for A ave is requested. The alloy average molecular weight is just the ratio of total alloy mass in grams M’ and the total number of moles in the alloy N m . That is A ave = MÕ N m = m 1 ' + m 2 ' n m1 + n m2 But using Equation 4.4 we may write m 1 ' = n m1 A 1 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. m 2 ' = n m2 A 2 Which, when substituted into the above A ave expression yields A ave = M' N m = n m1 A 1 + n m2 A 2 n m1 + n m2 Furthermore, from Equation 4.5 n m1 = C 1 ' (n m1 + n m2 ) 100 n m2 = C 2 ' (n m1 + n m2 ) 100 Thus, substitution of these expressions into the above equation for A ave yields A ave = C 1 ' A 1 (n m1 + n m2 ) 100 + C 2 ' A 2 (n m1 + n m2 ) 100 n m1 + n m2 = C 1 ' A 1 + C 2 ' A 2 100 which is the desired result. 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. 4.7 What is the composition, in atom percent, of an alloy that consists of 30 wt% Zn and 70 wt% Cu? Solution In order to compute composition, in atom percent, of a 30 wt% Zn-70 wt% Cu alloy, we employ Equation 4.6 as C Zn ' = C Zn A Cu C Zn A Cu +C Cu A Zn × 100 = (30)(63.55 g/mol ) (30)(63.55 g/mol ) +(70)(65.41 g/mol ) × 100 = 29.4 at% C Cu ' = C Cu A Zn C Zn A Cu +C Cu A Zn × 100 = (70)(65.41 g/mol ) (30)(63.55 g/mol ) +(70)(65.41 g/mol ) × 100 = 70.6 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. 4.8 What is the composition, in weight percent, of an alloy that consists of 6 at% Pb and 94 at% Sn? Solution In order to compute composition, in weight percent, of a 6 at% Pb-94 at% Sn alloy, we employ Equation 4.7 as C Pb = C Pb ' A Pb C Pb ' A Pb +C Sn ' A Sn × 100 = (6)(207.2g/mol ) (6)(207.2g/mol ) +(94)(118.71 g/mol ) × 100 = 10.0 wt% C Sn = C Sn ' A Sn C Pb ' A Pb +C Sn ' A Sn × 100 = (94)(118.71 g/mol ) (6)(207.2g/mol )+(94)(118.71 g/mol ) × 100 = 90.0 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. 4.9 Calculate the composition, in weight percent, of an alloy that contains 218.0 kg titanium, 14.6 kg of aluminum, and 9.7 kg of vanadium. Solution The concentration, in weight percent, of an element in an alloy may be computed using a modified form of Equation 4.3. For this alloy, the concentration of titanium (C Ti ) is just C Ti = m Ti m Ti +m Al +m V × 100 = 218 kg 218 kg +14.6kg +9.7kg × 100 = 89.97wt% Similarly, for aluminum C Al = 14.6kg 218 kg +14.6kg +9.7kg × 100 = 6.03wt% And for vanadium C V = 9.7kg 218 kg +14.6kg +9.7kg × 100 = 4.00wt% 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. 4.10 What is the composition, in atom percent, of an alloy that contains 98 g tin and 65 g of lead? Solution The concentration of an element in an alloy, in atom percent, may be computed using Equation 4.5. However, it first becomes necessary to compute the number of moles of both Sn and Pb, using Equation 4.4. Thus, the number of moles of Sn is just n m Sn = m Sn ' A Sn = 98 g 118.71 g/mol = 0.826mol Likewise, for Pb n m Pb = 65 g 207.2g/mol = 0.314mol Now, use of Equation 4.5 yields C Sn ' = n m Sn n m Sn +n m Pb × 100 = 0.826 mol 0.826 mol + 0.314 mol × 100 = 72.5at% Also, C Pb ' = = 0.314 mol 0.826 mol + 0.314 mol × 100 = 27.5at% 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. 4.11 What is the composition, in atom percent, of an alloy that contains 99.7 lbm copper, 102 lbm zinc, and 2.1 lbm lead? Solution In this problem we are asked to determine the concentrations, in atom percent, of the Cu-Zn-Pb alloy. It is first necessary to convert the amounts of Cu, Zn, and Pb into grams. m Cu ' =(99.7lb m )(453.6 g/lb m )=45, 224 g m Zn ' =(102lb m )(453.6g/lb m )=46, 267g m Pb ' =(2.1lb m )(453.6g/lb m )=953g These masses must next be converted into moles (Equation 4.4), as n m Cu = m Cu ' A Cu = 45,224 g 63.55 g/mol = 711.6mol n m Zn = 46,267 g 65.41 g/mol = 707.3mol n m Pb = 953 g 207.2g/mol = 4.6mol Now, employment of a modified form of Equation 4.5, gives C Cu ' = n m Cu n m Cu +n m Zn + n m Pb × 100 = 711.6mol 711.6mol + 707.3mol + 4.6mol × 100 = 50.0at% C Zn ' = 707.3mol 711.6mol + 707.3mol + 4.6mol × 100 = 49.7at% 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. C Pb ' = 4.6mol 711.6mol + 707.3mol + 4.6mol × 100 = 0.3at% 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. 4.12 What is the composition, in atom percent, of an alloy that consists of 97 wt% Fe and 3 wt% Si? Solution We are asked to compute the composition of an Fe-Si alloy in atom percent. Employment of Equation 4.6 leads to C Fe ' = C Fe A Si C Fe A Si + C Si A Fe × 100 = 97(28.09 g/mol ) 97(28.09 g/mol ) + 3(55.85 g/mol ) × 100 = 94.2 at% C Si ' = C Si A Fe C Si A Fe + C Fe A Si × 100 = 3(55.85 g/mol ) 3(55.85 g/mol ) + 97(28.09 g/mol ) × 100 = 5.8 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. 4.13 Convert the atom percent composition in Problem 4.11 to weight percent. Solution The composition in atom percent for Problem 4.11 is 50.0 at% Cu, 49.7 at% Zn, and 0.3 at% Pb. Modification of Equation 4.7 to take into account a three-component alloy leads to the following C Cu = C Cu ' A Cu C Cu ' A Cu + C Zn ' A Zn + C Pb ' A Pb × 100 = (50.0)(63.55 g/mol ) (50.0)(63.55 g/mol ) +(49.7)(65.41 g/mol ) +(0.3)(207.2g/mol ) ×100 = 49.0 wt% C Zn = C Zn ' A Zn C Cu ' A Cu + C Zn ' A Zn + C Pb ' A Pb × 100 = (49.7)(65.41 g/mol ) (50.0)(63.55 g/mol ) +(49.7)(65.41 g/mol ) +(0.3)(207.2g/mol ) ×100 = 50.1 wt% C Pb = C Pb ' A Pb C Cu ' A Cu + C Zn ' A Zn + C Pb ' A Pb × 100 = (0.3)(207.2g/mol ) (50.0)(63.55 g/mol ) +(49.7)(65.41 g/mol ) +(0.3)(207.2g/mol ) ×100 = 1.0 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. 4.14 Calculate the number of atoms per cubic meter in aluminum. Solution In order to solve this problem, one must employ Equation 4.2, N = N A ρ Al A Al The density of Al (from the table inside of the front cover) is 2.71 g/cm 3 , while its atomic weight is 26.98 g/mol. Thus, N = (6.022 × 10 23 atoms /mol )(2.71 g/cm 3 ) 26.98 g/mol = 6.05 × 10 22 atoms/cm 3 = 6.05 × 10 28 atoms/m 3 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. 4.15 The concentration of carbon in an iron-carbon alloy is 0.15 wt%. What is the concentration in kilograms of carbon per cubic meter of alloy? Solution In order to compute the concentration in kg/m 3 of C in a 0.15 wt% C-99.85 wt% Fe alloy we must employ Equation 4.9 as C C " = C C C C ρ C + C Fe ρ Fe ×10 3 From inside the front cover, densities for carbon and iron are 2.25 and 7.87 g/cm 3 , respectively; and, therefore C C " = 0.15 0.15 2.25 g/cm 3 + 99.85 7.87 g/cm 3 ×10 3 = 11.8 kg/m 3 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. 4.16 Determine the approximate density of a high-leaded brass that has a composition of 64.5 wt% Cu, 33.5 wt% Zn, and 2.0 wt% Pb. Solution In order to solve this problem, Equation 4.10a is modified to take the following form: ρ ave = 100 C Cu ρ Cu + C Zn ρ Zn + C Pb ρ Pb And, using the density values for Cu, Zn, and Pb—i.e., 8.94 g/cm 3 , 7.13 g/cm 3 , and 11.35 g/cm 3 —(as taken from inside the front cover of the text), the density is computed as follows: ρ ave = 100 64.5wt% 8.94 g/cm 3 + 33.5wt% 7.13 g/cm 3 + 2.0 wt% 11.35 g/cm 3 = 8.27 g/cm 3 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. 4.17 Calculate the unit cell edge length for an 85 wt% Fe-15 wt% V alloy. All of the vanadium is in solid solution, and, at room temperature the crystal structure for this alloy is BCC. Solution 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 ρ ave = nA ave V C N A Inasmuch as the unit cell is cubic, then V C = a 3 , then ρ ave = nA ave a 3 N A And solving this equation for the unit cell edge length, leads to a = nA ave ρ ave N A | . ` · 1/3 Expressions for A ave and ρ ave are found in Equations 4.11a and 4.10a, respectively, which, when incorporated into the above expression yields a = n 100 C Fe A Fe + C V A V | . ` · 100 C Fe ρ Fe + C V ρ V | . ` · N A ] ] ] ] ] ] ] ] ] ] ] 1/ 3 Since the crystal structure is BCC, the value of n in the above expression is 2 atoms per unit cell. The atomic weights for Fe and V are 55.85 and 50.94 g/mol, respectively (Figure 2.6), whereas the densities for the Fe 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. and V are 7.87 g/cm 3 and 6.10 g/cm 3 (from inside the front cover). Substitution of these, as well as the concentration values stipulated in the problem statement, into the above equation gives a = (2atoms/unit cell) 100 85wt% 55.85g/mol + 15wt% 50.94g/mol | . ` · 100 85wt% 7.87g/cm 3 + 15wt% 6.10g/cm 3 | . ` · 6.022 × 10 23 atoms/mol ( ) ] ] ] ] ] ] ] ] ] ] ] 1/3 ·2.89×10 -8 cm=0.289nm 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. 4.18 Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27 and 6.35 g/cm 3 , respectively, whereas their respective atomic weights are 61.4 and 125.7 g/mol, determine whether the crystal structure for this alloy is simple cubic, face-centered cubic, or body-centered cubic. Assume a unit cell edge length of 0.395 nm. Solution 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 ρ ave = nA ave V C N A Inasmuch as for each of the possible crystal structures, the unit cell is cubic, then V C = a 3 , or ρ ave = nA ave a 3 N 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 = ρ ave a 3 N A A ave Expressions for A ave and ρ ave are found in Equations 4.11a and 4.10a, respectively, which, when incorporated into the above expression yields n = 100 C A ρ A + C B ρ B | . ` · a 3 N A 100 C A A A + C B A B | . ` · 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. Substitution of the concentration values (i.e., C A = 12.5 wt% and C B = 87.5 wt%) as well as values for the other parameters given in the problem statement, into the above equation gives n = 100 12.5 wt% 4.27 g/cm 3 + 87.5 wt% 6.35 g/cm 3 | . ` · (3.95 × 10 -8 nm) 3 (6.022 × 10 23 atoms/mol ) 100 12.5 wt% 61.4 g/mol + 87.5 wt% 125.7 g/mol | . ` · = 2.00 atoms/unit cell Therefore, on the basis of this value, the crystal structure is body-centered cubic. 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. 4.19 For a solid solution consisting of two elements (designated as 1 and 2), sometimes it is desirable to determine the number of atoms per cubic centimeter of one element in a solid solution, N1, given the concentration of that element specified in weight percent, C1. This computation is possible using the following expression: N 1 · N A C 1 C 1 A 1 ρ 1 + A 1 ρ 2 100−C 1 ( ) (4.18) where NA = Avogadro’s number ρ1 and ρ2 = densities of the two elements A1 = the atomic weight of element 1 Derive Equation 4.18 using Equation 4.2 and expressions contained in Section 4.4. Solution This problem asks that we derive Equation 4.18, using other equations given in the chapter. The concentration of component 1 in atom percent (C 1 ' ) is just 100 c 1 ' where c 1 ' is the atom fraction of component 1. Furthermore, c 1 ' is defined as c 1 ' = N 1 /N where N 1 and N are, respectively, the number of atoms of component 1 and total number of atoms per cubic centimeter. Thus, from the above discussion the following holds: N 1 = C 1 ' N 100 Substitution into this expression of the appropriate form of N from Equation 4.2 yields N 1 = C 1 ' N A ρ ave 100 A ave And, finally, substitution into this equation expressions for C 1 ' (Equation 4.6a), ρ ave (Equation 4.10a), A ave (Equation 4.11a), and realizing that C 2 = (C 1 – 100), and after some algebraic manipulation we obtain the desired expression: 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. N 1 = N A C 1 C 1 A 1 ρ 1 + A 1 ρ 2 (100 − C 1 ) 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. 4.20 Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold alloy that contains 10 wt% Au and 90 wt% Ag. The densities of pure gold and silver are 19.32 and 10.49 g/cm 3 , respectively. Solution To solve this problem, employment of Equation 4.18 is necessary, using the following values: C 1 = C Au = 10 wt% ρ 1 = ρ Au = 19.32 g/cm 3 ρ 2 = ρ Ag = 10.49 g/cm 3 A 1 = A Au = 196.97 g/mol Thus N Au = N A C Au C Au A Au ρ Au + A Au ρ Ag (100 − C Au ) = (6.022 × 10 23 atoms /mol )(10 wt%) (10 wt %)(196.97 g/mol ) 19.32 g/cm 3 + 196.97 g/mol 10.49 g/cm 3 (100 − 10 wt%) = 3.36 × 10 21 atoms/cm 3 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. 4.21 Germanium forms a substitutional solid solution with silicon. Compute the number of germanium atoms per cubic centimeter for a germanium-silicon alloy that contains 15 wt% Ge and 85 wt% Si. The densities of pure germanium and silicon are 5.32 and 2.33 g/cm 3 , respectively. Solution To solve this problem, employment of Equation 4.18 is necessary, using the following values: C 1 = C Ge = 15 wt% ρ 1 = ρ Ge = 5.32 g/cm 3 ρ 2 = ρ Si = 2.33 g/cm 3 A 1 = A Ge = 72.64 g/mol Thus N Ge = N A C Ge C Ge A Ge ρ Ge + A Ge ρ Si (100 − C Ge ) = (6.022 × 10 23 atoms /mol )(15 wt%) (15 wt%)(72.64 g/mol ) 5.32 g/cm 3 + 72.64 g/mol 2.33 g/cm 3 (100 − 15 wt%) = 3.16 × 10 21 atoms/cm 3 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. 4.22 Sometimes it is desirable to be able to determine the weight percent of one element, C1, that will produce a specified concentration in terms of the number of atoms per cubic centimeter, N1, for an alloy composed of two types of atoms. This computation is possible using the following expression: C 1 · 100 1 + N A ρ 2 N 1 A 1 − ρ 2 ρ 1 (4.19) where NA = Avogadro’s number ρ1 and ρ2 = densities of the two elements A1 and A2 = the atomic weights of the two elements Derive Equation 4.19 using Equation 4.2 and expressions contained in Section 4.4. Solution The number of atoms of component 1 per cubic centimeter is just equal to the atom fraction of component 1 (c 1 ' ) times the total number of atoms per cubic centimeter in the alloy (N). Thus, using the equivalent of Equation 4.2, we may write N 1 = c 1 ' N = c 1 ' N A ρ ave A ave Realizing that c 1 ' = C 1 ' 100 and C 2 ' = 100 −C 1 ' and substitution of the expressions for ρ ave and A ave , Equations 4.10b and 4.11b, respectively, leads to N 1 = c 1 ' N A ρ ave A ave 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. = N A C 1 ' ρ 1 ρ 2 C 1 ' ρ 2 A 1 + (100 − C 1 ' )ρ 1 A 2 And, solving for C 1 ' C 1 ' = 100 N 1 ρ 1 A 2 N A ρ 1 ρ 2 − N 1 ρ 2 A 1 + N 1 ρ 1 A 2 Substitution of this expression for C 1 ' into Equation 4.7a, which may be written in the following form C 1 = C 1 ' A 1 C 1 ' A 1 + C 2 ' A 2 × 100 = C 1 ' A 1 C 1 ' A 1 + (100− C 1 ' )A 2 × 100 yields C 1 = 100 1 + N A ρ 2 N 1 A 1 − ρ 2 ρ 1 the desired expression. 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. 4.23 Molybdenum forms a substitutional solid solution with tungsten. Compute the weight percent of molybdenum that must be added to tungsten to yield an alloy that contains 1.0 × 10 22 Mo atoms per cubic centimeter. The densities of pure Mo and W are 10.22 and 19.30 g/cm 3 , respectively. Solution To solve this problem, employment of Equation 4.19 is necessary, using the following values: N 1 = N Mo = 10 22 atoms/cm 3 ρ 1 = ρ Mo = 10.22 g/cm 3 ρ 2 = ρ W = 19.30 g/cm 3 A 1 = A Mo = 95.94 g/mol A 2 = A W = 183.84 g/mol Thus C Mo = 100 1 + N A ρ W N Mo A Mo − ρ W ρ Mo = 100 1 + (6.022 × 10 23 atoms /mol )(19.30 g/cm 3 ) (10 22 atoms /cm 3 )(95.94 g/mol ) − 19.30 g/cm 3 10.22 g/cm 3 | . ` · = 8.91 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. 4.24 Niobium forms a substitutional solid solution with vanadium. Compute the weight percent of niobium that must be added to vanadium to yield an alloy that contains 1.55 × 10 22 Nb atoms per cubic centimeter. The densities of pure Nb and V are 8.57 and 6.10 g/cm 3 , respectively. Solution To solve this problem, employment of Equation 4.19 is necessary, using the following values: N 1 = N Nb = 1.55 × 10 22 atoms/cm 3 ρ 1 = ρ Nb = 8.57 g/cm 3 ρ 2 = ρ V = 6.10 g/cm 3 A 1 = A Nb = 92.91 g/mol A 2 = A V = 50.94 g/mol Thus C Nb = 100 1 + N A ρ V N Nb A Nb − ρ V ρ Nb = 100 1 + (6.022 × 10 23 atoms /mol )(6.10 g/cm 3 ) (1.55 × 10 22 atoms /cm 3 )(92.91 g/mol ) − 6.10 g/cm 3 8.57 g/cm 3 | . ` · = 35.2 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. 4.25 Silver and palladium both have the FCC crystal structure, and Pd forms a substitutional solid solution for all concentrations at room temperature. Compute the unit cell edge length for a 75 wt% Ag–25 wt% Pd alloy. The room-temperature density of Pd is 12.02 g/cm 3 , and its atomic weight and atomic radius are 106.4 g/mol and 0.138 nm, respectively. Solution First of all, the atomic radii for Ag (using the table inside the front cover) and Pd are 0.144 and 0.138 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 silver and palladium are 10.49 g/cm 3 (as taken from inside the front cover) and 12.02 g/cm 3 , respectively, the average density is just ρ ave = 100 C Ag ρ Ag + C Pd ρ Pd = 100 75 wt% 10.49 g/cm 3 + 25 wt% 12.02 g/cm 3 = 10.83 g/cm 3 And for the average atomic weight A ave = 100 C Ag A Ag + C Pd A Pd = 100 75 wt % 107.9 g/mol + 25 wt % 106.4 g/mol = 107.5 g/mol Now, V C is determined from Equation 3.5 as 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. V C = nA ave ρ ave N A = (4 atoms/unit cell)(107.5 g/mol) (10.83 g/cm 3 )(6.022 ×10 23 atoms/mol) = 6.59 × 10 -23 cm 3 /unit cell And, finally a = (V C ) 1/3 =(6.59×10 −23 cm 3 /unit cell) 1/3 = 4.04 × 10 -8 cm = 0.404 nm 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. Dislocations—Linear Defects 4.26 Cite the relative Burgers vector–dislocation line orientations for edge, screw, and mixed dislocations. Solution The Burgers vector and dislocation line are perpendicular for edge dislocations, parallel for screw dislocations, and neither perpendicular nor parallel for mixed dislocations. 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. Interfacial Defects 4.27 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 3.54 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 3.54, planar densities for FCC (100) and (111) planes are 1 4R 2 and 1 2R 2 3 , respectively —that is 0.25 R 2 and 0.29 R 2 (where R is the atomic radius). Thus, since the planar density for (111) is greater, it will have the lower surface energy. 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. 4.28 For a BCC single crystal, would you expect the surface energy for a (100) plane to be greater or less than that for a (110) plane? Why? (Note: You may want to consult the solution to Problem 3.55 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 3.55, the planar densities for BCC (100) and (110) are 3 16R 2 and 3 8R 2 2 , respectively —that is 0.19 R 2 and 0.27 R 2 . Thus, since the planar density for (110) is greater, it will have the lower surface energy. 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. 4.29 (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. 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. 4.30 (a) Briefly describe a twin and a twin boundary. (b) Cite the difference between mechanical and annealing twins. Solution (a) A twin boundary is an interface such that atoms on one side are located at mirror image positions of those atoms situated on the other boundary side. The region on one side of this boundary is called a twin. (b) Mechanical twins are produced as a result of mechanical deformation and generally occur in BCC and HCP metals. Annealing twins form during annealing heat treatments, most often in FCC metals. 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. 4.31 For each of the following stacking sequences found in FCC metals, cite the type of planar defect that exists: (a) . . . A B C A B C B A C B A . . . (b) . . . A B C A B C B C A B C . . . Now, copy the stacking sequences and indicate the position(s) of planar defect(s) with a vertical dashed line. Solution (a) The interfacial defect that exists for this stacking sequence is a twin boundary, which occurs at the indicated position. The stacking sequence on one side of this position is mirrored on the other side. (b) The interfacial defect that exists within this FCC stacking sequence is a stacking fault, which occurs between the two lines. Within this region, the stacking sequence is HCP. 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. Grain Size Determination 4.32 (a) Using the intercept method, determine the average grain size, in millimeters, of the specimen whose microstructure is shown in Figure 4.14(b); use at least seven straight-line segments. (b) Estimate the ASTM grain size number for this material. Solution (a) Below is shown the photomicrograph of Figure 4.14(b), on which seven straight line segments, each of which is 60 mm long has been constructed; these lines are labeled “1” through “7”. In order to determine the average grain diameter, it is necessary to count the number of grains intersected by each of these line segments. These data are tabulated below. Line Number No. Grains Intersected 1 11 2 10 3 9 4 8.5 5 7 6 10 7 8 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. The average number of grain boundary intersections for these lines was 9.1. Therefore, the average line length intersected is just 60 mm 9.1 = 6.59 mm Hence, the average grain diameter, d, is d = ave. line length intersected magnification = 6.59 mm 100 =6.59 ×10 −2 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 4.16. Inasmuch as the magnification is 100× , the value of N is measured directly from the micrograph. The photomicrograph on which has been constructed a square 1 in. on a side is shown below. The total number of complete grains within this square is approximately 10 (taking into account grain fractions). Now, in order to solve for n in Equation 4.16, it is first necessary to take logarithms as logN ·(n −1)log2 From which n equals 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. n · log N log 2 +1 · log10 log 2 +1·4.3 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. 4.33 (a) Employing the intercept technique, determine the average grain size for the steel specimen whose microstructure is shown in Figure 9.25(a); use at least seven straight-line segments. (b) Estimate the ASTM grain size number for this material. Solution (a) Below is shown the photomicrograph of Figure 9.25(a), on which seven straight line segments, each of which is 60 mm long has been constructed; these lines are labeled “1” through “7”. In order to determine the average grain diameter, it is necessary to count the number of grains intersected by each of these line segments. These data are tabulated below. Line Number No. Grains Intersected 1 7 2 7 3 7 4 8 5 10 6 7 7 8 The average number of grain boundary intersections for these lines was 8.7. Therefore, the average line length intersected is just 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. 60 mm 8.7 = 6.9mm Hence, the average grain diameter, d, is d = ave. linelengthintersected magnification = 6.9mm 90 =0.077mm (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 4.16. However, the magnification of this micrograph is not 100× , but rather 90× . Consequently, it is necessary to use Equation 4.17 N M M 100 | . ` · 2 ·2 n−1 where N M = 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: logN M + 2 l o g M 1 0 0 | . ` · ·(n − 1 ) l o g 2 Solving this expression for n gives n · log N M +2 log M 100 | . ` · log 2 +1 The photomicrograph on which has been constructed a square 1 in. on a side is shown below. 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. From Figure 9.25(a), N M is measured to be approximately 7, which leads to n · log 7+2log 90 100 | . ` · log 2 +1 = 3.5 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. 4.34 For an ASTM grain size of 8, approximately how many grains would there be per square inch at (a) a magnification of 100, and (b) without any magnification? Solution (a) This part of problem asks that we compute the number of grains per square inch for an ASTM grain size of 8 at a magnification of 100× . All we need do is solve for the parameter N in Equation 4.16, inasmuch as n = 8. Thus N ·2 n−1 = 2 8−1 = 128 grains/in. 2 (b) Now it is necessary to compute the value of N for no magnification. In order to solve this problem it is necessary to use Equation 4.17: N M M 100 | . ` · 2 ·2 n−1 where N M = the number of grains per square inch at magnification M, and n is the ASTM grain size number. Without any magnification, M in the above equation is 1, and therefore, N 1 1 100 | . ` · 2 ·2 8−1 ·128 And, solving for N 1 , N 1 = 1,280,000 grains/in. 2 . 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. 4.35 Determine the ASTM grain size number if 25 grains per square inch are measured at a magnification of 600. Solution This problem asks that we determine the ASTM grain size number if 8 grains per square inch are measured at a magnification of 600. In order to solve this problem we make use of Equation 4.17: N M M 100 | . ` · 2 · 2 n −1 where N M = the number of grains per square inch at magnification M, and n is the ASTM grain size number. Solving the above equation for n, and realizing that N M = 8, while M = 600, we have n · log N M +2 log M 100 | . ` · log 2 +1 · log 8 +2 log 600 100 | . ` · log 2 +1 ·9.2 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. 4.36 Determine the ASTM grain size number if 20 grains per square inch are measured at a magnification of 50. Solution This problem asks that we determine the ASTM grain size number if 20 grains per square inch are measured at a magnification of 50. In order to solve this problem we make use of Equation 4.17—viz. N M M 100 | . ` · 2 ·2 n−1 where N M = the number of grains per square inch at magnification M, and n is the ASTM grain size number. Solving the above equation for n, and realizing that N M = 20, while M = 50, we have n · log N M +2 log M 100 | . ` · log 2 +1 · log 20 +2 log 50 100 | . ` · log 2 +1 ·3.3 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. DESIGN PROBLEMS Specification of Composition 4.D1 Aluminum–lithium alloys have been developed by the aircraft industry to reduce the weight and improve the performance of its aircraft. A commercial aircraft skin material having a density of 2.55 g/cm 3 is desired. Compute the concentration of Li (in wt%) that is required. Solution Solution of this problem requires the use of Equation 4.10a, which takes the form ρ ave = 100 C Li ρ Li + 100 − C Li ρ Al inasmuch as C Li + C Al = 100. According to the table inside the front cover, the respective densities of Li and Al are 0.534 and 2.71 g/cm 3 . Upon solving for C Li from the above equation, we get C Li = 100 ρ Li (ρ Al − ρ ave ) ρ ave (ρ Al − ρ Li ) And incorporating specified values into the above equation leads to C Li = (100)(0.534 g/cm 3 )(2.71 g/cm 3 − 2.55 g/cm 3 ) (2.55 g/cm 3 )(2.71 g/cm 3 − 0.534 g/cm 3 ) = 1.540 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. 4.D2 Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to approximately 20 wt% V at room temperature. Determine the concentration in weight percent of V that must be added to iron to yield a unit cell edge length of 0.289 nm. Solution To begin, it is necessary to employ Equation 3.5, and solve for the unit cell volume, V C , as V C = nA ave ρ ave N A where A ave and ρ ave are the atomic weight and density, respectively, of the Fe-V alloy. Inasmuch as both of these materials have the BCC crystal structure, which has cubic symmetry, V C is just the cube of the unit cell length, a. That is V C =a 3 =(0.289nm) 3 ·(2.89×10 −8 cm) 3 ·2.414 ×10 −23 cm 3 It is now necessary to construct expressions for A ave and ρ ave in terms of the concentration of vanadium, C V , using Equations 4.11a and 4.10a. For A ave we have A ave = 100 C V A V + (100 − C V ) A Fe = 100 C V 50.94g/mol + (100 − C V ) 55.85 g/mol whereas for ρ ave ρ ave = 100 C V ρ V + (100 − C V ) ρ Fe 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. = 100 C V 6.10 g/cm 3 + (100 − C V ) 7.87 g/cm 3 Within the BCC unit cell there are 2 equivalent atoms, and thus, the value of n in Equation 3.5 is 2; hence, this expression may be written in terms of the concentration of V in weight percent as follows: V C = 2.414 × 10 -23 cm 3 = nA ave ρ ave N A = (2 atoms /unit cell ) 100 C V 50.94 g/mol + (100 − C V ) 55.85 g/mol ] ] ] ] ] ] 100 C V 6.10 g/cm 3 + (100 − C V ) 7.87 g/cm 3 ] ] ] ] ] ] (6.022 × 10 23 atoms /mol ) And solving this expression for C V leads to C V = 12.9 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. 4.2 Calculate the number of vacancies per cubic meter in iron at 850 °C. The energy for vacancy formation is 1.08 eV/atom. Furthermore, the density and atomic weight for Fe are 7.65 g/cm 3 and 55.85 g/mol, respectively. Solution Determination of the number of vacancies per cubic meter in iron at 850°C (1123 K) requires the utilization of Equations 4.1 and 4.2 as follows: Q N ρe Q N v =N e v = A F e v x − p x − p kT kT AF e And incorporation of values of the parameters provided in the problem statement into the above equation leads to N= v (.2 ×12as /m)75 g m)x 62 0 03 t o m o (. l 6 /c 3 e − p 5.5 g o 58 /m l 18 e /a . 0 Vt o m 8 6 05 Vt o K8 C 2 K m− ) 5 ( 0° +7 ) 3 ( .2 ×1− e /a = 1.18 × 1018 cm-3 = 1.18 × 1024 m-3 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. 4.3 Calculate the activation energy for vacancy formation in aluminum, given that the equilibrium number of vacancies at 500°C (773 K) is 7.57 × 1023 m-3. The atomic weight and density (at 500°C) for aluminum are, respectively, 26.98 g/mol and 2.62 g/cm3. Solution Upon examination of Equation 4.1, all parameters besides Qv are given except N, the total number of atomic sites. However, N is related to the density, (ρAl), Avogadro's number (NA), and the atomic weight (AAl) according to Equation 4.2 as N = N AρA l AA l = ( .2 ×12 a s /m) 22 g m) 62 0 03 tm o ( . o l 6 /c 3 2.8 g o 69 /m l = 5.85 × 1022 atoms/cm 3 = 5.85 × 1028 atoms/m 3 Now, taking natural logarithms of both sides of Equation 4.1, Q l v = l nN nN − v kT and, after some algebraic manipulation N Qv = − kT ln v N 3 . 7 ×1 2 m 7 5 03 − l n 3 . 8 08 − 55 ×1 2 m = −(.2 × 0- ea 8 6 1 5V /tm -K(0 ° +7) o )5 C 2K 0 3 = 0.75 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 4.4 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 Radius (nm) 0.1278 0.071 0.046 0.060 0.1445 0.1431 0.1253 0.1249 0.1241 0.1246 0.1376 0.1387 0.1332 Element Cu C H O Ag Al Co Cr Fe Ni Pd Pt Zn Crystal Structure FCC Electronegativity 1.9 Valence +2 FCC FCC HCP BCC BCC FCC FCC FCC HCP 1.9 1.5 1.8 1.6 1.8 1.8 2.2 2.2 1.6 +1 +3 +2 +3 +2 +2 +2 +2 +2 Which of these elements would you expect to form the following with copper: (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 Cu the three possible solid solution types. For complete substitutional solubility the following criteria must be met: 1) the difference in atomic radii between Cu 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 Cu C H O Ag Al Co Cr ∆R% –44 –64 –53 +13 +12 -2 -2 Crystal Structure FCC ∆Electronegativity Valence 2+ FCC FCC HCP BCC 0 -0.4 -0.1 -0.3 1+ 3+ 2+ 3+ 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. Fe Ni Pd Pt Zn -3 -3 +8 +9 +4 BCC FCC FCC FCC HCP -0.1 -0.1 +0.3 +0.3 -0.3 2+ 2+ 2+ 2+ 2+ (a) Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions 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 Cu 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 Cu. 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. solving for r from the However. compute the radius r of an impurity atom that will just fit into one of these sites in terms of the atomic radius R of the host atom. and is normally occupied by impurity atoms. lying on {100} faces. the interstitial site is at the center of the edge.1 as a = above equation gives r = a− R 2 2R 2 − R 2 = = 0. therefore.5 For both FCC and BCC crystal structures. this larger one is located at the center of each edge of the unit cell.41R 2 2 A (100) face of a BCC unit cell is shown below. The diameter of an atom that will just fit into this site (2r) is just the difference between that unit cell edge length (a) and the radii of the two host atoms that are located on either side of the site (R). for FCC a is related to R according to Equation 3.4. there are two different types of interstitial sites. For FCC. In each case. it is termed a tetrahedral interstitial site. For both FCC and BCC crystal structures. it is termed an octahedral interstitial site. Solution In the drawing below is shown the atoms on the (100) face of an FCC unit cell. 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. . that is 2r = a – 2R 2R 2 . with BCC the larger site type is found at 0 1 2 1 4 positions—that is. one site is larger than the other. 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. On the other hand. and situated midway between two unit cell edges on this face and one-quarter of the distance between the other two unit cell edges. midway between the two vertical unit cell edges. 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. from Equation 3.3. . a = form 4R . and. and one quarter of the distance between the bottom and top cell edges.The interstitial atom that just fits into this interstitial site is shown by the small circle. From the right triangle that is defined by the three arrows we may write 2 2 a a 2 + = ( R +r ) 2 4 However. making this substitution.667 R 2 = 0 And upon solving for r: 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. the above equation takes the 3 2 2 4R 4R + = R 2 + 2Rr + r 2 2 3 4 3 After rearrangement the following quadratic equation results: r 2 + 2R r − 0. It is situated in the plane of this (100) face. therefore. Thus.r = − ( 2R) ± ( 2 − 4)1)− 667 R 2 ) 2R) ( ( ( 0. only the r(+) root is possible. 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. = 0. 2R 582R 2 And. r = 0. 291R 2 − − 582R 2R 2. r( ) = − =− 291R 2. finally − + 582R 2R 2. therefore. 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. for a host atom of radius R.4 times that for BCC. 2 = − ± 2. .291R. 2 r( ) = + Of course. the size of an interstitial site for FCC is approximately 1. and. substitution into the C1 expression above C1 = n m1 A1 + n m2 A2 n m1 A1 × 100 From Equation 4. C1 is defined according to Equation 4. From Equation 4. equivalently ' m1 ' ' m1 + m2 C1 = × 100 where the primed m's indicate masses in grams.4 we may write ' m1 = n m1 A1 m 2' = m2n A2 And.6 Derive the following equations: (a) Equation 4.7a. 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.3 as C1 = m1 + m2 m1 × 100 or.10a (d) Equation 4.7a (b) Equation 4. To begin.9a (c) Equation 4.11b Solution (a) This problem asks that we derive Equation 4.Specification of Composition 4.5 it is the case that 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. . " C1 is defined as the mass of component 1 per unit volume of alloy.e. the volume of each constituent is related to its density and mass as 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.. or "= C1 m1 V m1 If we assume that the total alloy volume V is equal to the sum of the volumes of the two constituents--i.9a. 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. (b) This problem asks that we derive Equation 4. To begin.nm1 = ' C1 (nm1 + nm2 ) 100 nm2 = ' n C 2 ( m1 + nm2 ) 100 And substitution of these expressions into the above equation leads to ' C1 A1 ' ' C1 A1 + C 2 A2 C1 = × 100 which is just Equation 4.7a. V = V1 + V2--then C1 " = V1 +2 V Furthermore. . inasmuch as 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. m1 and m2 may be expressed as follows: m1 = C1 (m1 + m2 ) 100 C 2 ( 1 + m2 ) m 100 m2 = Substitution of these equations into the preceding expression yields C 1 = C 1 = C1 C1 C 2 + ρ1 ρ2 If the densities ρ1 and ρ2 are given in units of g/cm 3. . 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.V1 = m1 ρ1 V2 = m2 ρ2 This leads to C1 " = m m 1 + ρ 1 From Equation 4.3. then conversion to units of kg/m3 requires that we multiply this equation by 103. 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.1 g/cm 3 = 103 kg/m3 Therefore. normally V will not be exactly equal to (V1 + V2)]. the previous equation takes the form C1 C1 " = C1 C + ρ ρ 1 which is the desired expression. The density of an alloy ρave is just the total alloy mass M divided by its volume V ρave = Or. V = 1 m1 ρ1 V2 = m2 ρ2 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. which is only an approximation.10a. . Each of V1 and V2 may be expressed in terms of its mass and density as. (c) Now we are asked to derive Equation 4. in terms of the component elements 1 and 2 M V ρave = m1 + m2 V1 + V2 [Note: here it is assumed that the total alloy volume is equal to the separate volumes of the individual components. 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. we get ρave = m1 + m2 m1 m + 2 ρ1 ρ2 Furthermore. this equation reduces to = C1 100 C + 2 ρ1 ρ2 (d) And. the derivation of Equation 4. The alloy average molecular weight is just the ratio of total alloy mass in grams M’ and the total number of moles in the alloy Nm. . when substituted into the above ρ expression yields ave ρav e = m1 + m2 C1 (m1 + m2 ) C 2 (m1 + m2 ) 10 0 100 + ρ1 ρ2 And.11b for Aave is requested. from Equation 4. finally.4 we may write ' m1 = nm1 A1 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. finally. That is ' ' m1 + m2 MÕ = Nm nm1 + nm2 Aave = But using Equation 4.When these expressions are substituted into the above equation.3 m1 = C1 ( 1 + m2 ) m 100 C 2 ( 1 + m2 ) m 100 m2 = Which. 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. when substituted into the above Aave expression yields Aave = n A + nm2 A2 M' = m1 1 Nm nm1 + nm2 Furthermore. from Equation 4.' m2 = nm2 A2 Which. substitution of these expressions into the above equation for Aave yields ' C1 A1 (nm1 + nm2 ) + 100 Aave = nm1 + ' C 2 A2 (nm1 + nm2 ) 100 nm2 = ' ' C1 A1 + C2 A2 100 which is the desired result. .5 ' n C1 ( m1 + nm2 ) 100 nm1 = nm2 = ' n C 2 ( m1 + nm2 ) 100 Thus. 6 as ' CZ = n CZ AC n u ×1 0 0 CZ AC + C AZ n u Cu n = ( )6. in atom percent.6 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.4.5gm + 7)6. .5gm + 7)6.1gm 3(35 / o ) (0 54 / o ) 0 l ( l = 70. 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.1gm 3(35 / o ) (0 54 / o ) 0 l ( l = 29. of a 30 wt% Zn-70 wt% Cu alloy. in atom percent.5gm 3(35 / o ) 0 l ×0 1 0 ( )6.1gm 7(54 / o ) 0 l ×0 1 0 ( )6.7 What is the composition. we employ Equation 4. of an alloy that consists of 30 wt% Zn and 70 wt% Cu? Solution In order to compute composition.4 at% ' CC = u CC AZ u n ×1 0 0 CZ AC + C AZ n u Cu n = ( )6. 0 wt% CS = n ' CS AS n n ' ' CP AP +CS AS b b n n × 10 0 = ( )1 7 gm 9( 8. l = 10.1 / o ) ( 7.0 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.4. l ×0 1 0 ( )2 2 / o ) + 9)1 7 gm 6 0 gm (4(1 1 / o ) ( 7.7 as ' CP AP b b ' ' CP AP +CS AS b b n n CP = b × 10 0 = ( )2 2 / o ) 6 0 gm ( 7. we employ Equation 4. . l 8. of an alloy that consists of 6 at% Pb and 94 at% Sn? Solution In order to compute composition. l 41 l = 90. 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. in weight percent.1 / o ) 41 l ×0 1 0 ( )2 2 / o ) ( )1 7 gm 6 0 gm +9( 8.8 What is the composition. of a 6 at% Pb-94 at% Sn alloy. in weight percent. k 1 g 1 6g 9 g 8 4 7 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.0 kg titanium. k +.4. k 46g ×0 =. k 1 g 1 6g 9 g 8 4 7 Similarly. in weight percent. k 1 g 1 6g 9 g 8 4 7 And for vanadium C= V 9 k . of an element in an alloy may be computed using a modified form of Equation 4. of an alloy that contains 218. For this alloy. k +. % 9 7 2 k + . Solution The concentration. for aluminum C= A l 1. g 7 ×0 =. 14.7 kg of vanadium. in weight percent. w 1 0 6t 0% 3 2 k + .3. and 9. the concentration of titanium (CTi) is just CT = i mT i × 10 0 mT +mA +mV i l = 2 k 1 g 8 ×0 =9 t 1 0 8w . . w 1 0 4t 0% 0 2 k + .9 Calculate the composition.6 kg of aluminum. k +. in atom percent.% 5 02 m + 01 m . for Pb nm P = b 6g 5 =. ' CP = = b 01 m .4 o 3 l × 0 = 7t 1 0 2a .1o 0m 3l 4 2 2 /m 0 go 7. l Now.10 What is the composition. 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.% 5 02 m + 01 m . may be computed using Equation 4. it first becomes necessary to compute the number of moles of both Sn and Pb. . of an alloy that contains 98 g tin and 65 g of lead? Solution The concentration of an element in an alloy.6 o 8 l . use of Equation 4.4 o 3 l 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. using Equation 4.4. Thus.4.7 g/m 1 1 o l Likewise. in atom percent.5 yields ' CSn = nmSn + nm Pb nmSn × 10 0 = 02 m .6 o 8 l ×0 =2t 1 0 7a . However.4 o 3 l Also. the number of moles of Sn is just nmS = n ' mS n AS n = 9 g 8 =02m . 6o 8 l 18.6 o 8 l . 3 6. and 2. o 6l ' CZ = n 7 3o 0 m 7. 1 b (b These masses must next be converted into moles (Equation 4. g m= 2 ) 9 5 3 m=g ' mP=l 2 . 5 /m l u nm C u = nm Z n = 4.2 g 66 7 =0m 7 o 7l . l ×0 =9t 1 0 4a . in atom percent./ (g 4l 5b 3 u (l ) 42 54 . of the Cu-Zn-Pb alloy.1 lbm lead? Solution In this problem we are asked to determine the concentrations.5. l . as ' mC 52 4 u = 4.m 4o 6l 2 2 /m 0 g o 7.1g o 54 /m l nm P b = 9 g 5 3 =.11 What is the composition. l 0 m 7. of an alloy that contains 99. l ×0 =0t 1 0 5a . It is first necessary to convert the amounts of Cu. m 16o 1 l AC 65g o 3. employment of a modified form of Equation 4. Zn.% 0 7 6 o +7 3 o +4 m 1 m 1.7 lb m copper. l 0 m 7. 9 9 7 m6 b ). .% 7 7 6 o +7 3 o +4 m 1 m 1./ (g 4b 5 3 l m6 ) 46 67 . l . 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. gives ' CC = u nmCu +nm Zn + nm Pb nm Cu × 100 = 7 6o 1 m 1./ (g 4l 5b 3 m6 ). ' mC=.4. and Pb into grams. g m= 2 ' mZ=2 1 0 b n (l ).4). o 6l 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. l Now.2 g =7. in atom percent. 102 lbm zinc. 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. . l . o 6l 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. l 0 m 7.a 1 0 0t 3 % 7 6 o + 7 3 o +4 m 1 m 1.' CP = b 4 m . o 6l ×0 =. 4.5gm) (58 / o l ×0 1 0 35. in atom percent.12 What is the composition.9gm) +35. 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.6 leads to ' CF = e CF AS e i ×10 0 CF AS + CS AF e i i e = 9 2.9gm (58 / o l 7(80 / o ) l = 5. .5gm) +9 2. Employment of Equation 4. of an alloy that consists of 97 wt% Fe and 3 wt% Si? Solution We are asked to compute the composition of an Fe-Si alloy in atom percent.9gm 7(80 / o ) l ×0 1 0 9 2.2 at% ' CS = i CSAF i e ×10 0 CSAF + CF AS i e e i = 35.8 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.5gm 7(80 / o l (58 / o ) l = 94. 4.3 at% Pb. l . l . (7. gm 0 ( . l = ×0 1 0 ( .0 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. l = ×0 1 0 ( . )6 5gm 50 3 5 /o) 0 ( . 49. Modification of Equation 4. l 30 l = 50. Solution The composition in atom percent for Problem 4. )6 5gm +47 64gm +0)2 2 /o) 50 35 /o) (9 ) 5 1 /o) ( . 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.13 Convert the atom percent composition in Problem 4.0 at% Cu. l 30 l = 1. )2 2 /o) 0 (7. l 30 l = 49. ( .0 wt% CZ = n ' C Z AZ n n C' C AC u u + C Z AZ + CP AP n n b b ' ' × 10 0 ( .7 at% Zn.11 is 50.11 to weight percent. ( . gm 30 l = ×0 1 0 ( .1 wt% CP = b ' CP AP b b ' ' ' CC AC + C Z AZ + CP AP u u n n b b × 10 0 (. (7. )6 4gm 47 5 1 /o) 9 ( . . gm 0 ( . )6 5gm +47 64gm +0)2 2 /o) 50 3 5 /o) (9 ) 5 1 /o) ( .7 to take into account a three-component alloy leads to the following ' CC AC u u ' ' ' CC AC + C Z AZ + CP AP u u n n b b CC = u × 10 0 ( . and 0. )6 5gm +47 64gm +0)2 2 /o) 50 3 5 /o) (9 ) 5 1 /o) ( . ( . l . (7. gm 0 ( . 2.05 × 1028 atoms/m 3 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.4.14 Calculate the number of atoms per cubic meter in aluminum. Solution In order to solve this problem. N = ( .98 g/mol. l 7 /c 3 2. Thus.71 g/cm 3. while its atomic weight is 26. N = N AρA l AA l The density of Al (from the table inside of the front cover) is 2. . 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. one must employ Equation 4.05 × 1022 atoms/cm 3 = 6.2 ×12 as /m) 21g m) 62 0 03 t o m o (.8g o 69 /m l = 6. densities for carbon and iron are 2. . 1 5 0 .9 as CC " CC = CC + ρ C From inside the front cover.25 and 7.4.87 g/cm 3.85 wt% Fe alloy we must employ Equation 4. therefore " CC = 0 .15 The concentration of carbon in an iron-carbon alloy is 0. and. 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. 8 7 = 11.8 kg/m3 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. respectively. 1 5 9 9 + 3 2g .15 wt% C-99.15 wt%. / 2 5 c m 7 . What is the concentration in kilograms of carbon per cubic meter of alloy? Solution In order to compute the concentration in kg/m 3 of C in a 0. and Pb—i. . 33.94 g/cm 3.5 t 2 w% .0 t + + 8 4 g/c 3 .5 wt% Zn..13 g/cm 3.4.0 wt% Pb.3 m = 8. the density is computed as follows: ρv = ae 10 0 6 w% 4.35 g/cm 3—(as taken from inside the front cover of the text).9 m 7 3 g/c 3 1 5 g/c 3 .5 wt% Cu.16 Determine the approximate density of a high-leaded brass that has a composition of 64.10a is modified to take the following form: ρave = 100 CCu C Zn CPb + + ρCu ρZn ρPb And.5 t 3 w% 3. 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. and 2. Zn.27 g/cm 3 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. and 11.e. 8. 7.1 m 1. Solution In order to solve this problem. Equation 4. using the density values for Cu. 11a and 4. then nAave VC N A ρave = a3N A nAave And solving this equation for the unit cell edge length.17 Calculate the unit cell edge length for an 85 wt% Fe-15 wt% V alloy.94 g/mol. respectively (Figure 2. when incorporated into the above expression yields 1/ 3 100 n CFe + CV AFe AV a = 100 N CFe CV A + ρV ρFe Since the crystal structure is BCC. then VC = a3. All of the vanadium is in solid solution. . leads to 1/3 nA ave a = ρ N ave A Expressions for Aave and ρave are found in Equations 4. which. respectively. Solution In order to solve this problem it is necessary to employ Equation 3. in this expression density and atomic weight will be averages for the alloy—that is ρ ave = Inasmuch as the unit cell is cubic. and.85 and 50. whereas the densities for the Fe 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. The atomic weights for Fe and V are 55.10a.6).4. 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. the value of n in the above expression is 2 atoms per unit cell. at room temperature the crystal structure for this alloy is BCC. 5/ o a = 10 0 62 .m 2 8 9 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. 2 × 1 2 am o 0 0 3 t s l o/ m 5t % 1w 5t % 8w + 3 . g 8 c 60/m 3 .87 g/cm 3 and 6. 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.and V are 7.10 g/cm 3 (from inside the front cover). 4/ o 0 g l 9 m 5. g 1 c 77/m 1 /3 ( ) 8 = × -c =n 2 . into the above equation gives 10 0 ( a mncl) 2t s i e o/ t l u 5t % 1w 5t % 8w + 5 g l 8 m 5. . as well as the concentration values stipulated in the problem statement. Substitution of these. 8 1 m0 9 0 . 18 Some hypothetical alloy is composed of 12. then VC = a3. the crystal structure will be either BCC or FCC. determine whether the crystal structure for this alloy is simple cubic. When we solve the above expression for n the result is as follows: n = ρ a3N A ave Aave Expressions for Aave and ρave are found in Equations 4.7 g/mol. the unit cell is cubic. For n =1. 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. in order to determine the crystal structure it is necessary to solve for n.5. . respectively. the crystal structure is simple cubic.4 and 125. when incorporated into the above expression yields 100 a 3 N A CA CB + ρB ρA n = 100 CA CB + AB AA 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. respectively.4. Assume a unit cell edge length of 0.395 nm. If the densities of metals A and B are 4. whereas for n values of 2 and 4. in this expression density and atomic weight will be averages for the alloy—that is ρ ave = nAave VC N A Inasmuch as for each of the possible crystal structures. whereas their respective atomic weights are 61. or ρave = a3N A nAave And.27 and 6.35 g/cm3. face-centered cubic.5 wt% of metal A and 87.11a and 4.5 wt% of metal B. or body-centered cubic. respectively. Solution In order to solve this problem it is necessary to employ Equation 3. which. the number of atoms per unit cell.10a. 5 w t% 12.Substitution of the concentration values (i.4 g/m × 10 23 atom ol s/m ) n = = 2. into the above equation gives 100 3. 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. on the basis of this value.5 wt% + 6. .5 w t% 12.00 atoms/unit cell Therefore.022 ( 87.5 wt%) as well as values for the other parameters given in the problem statement.. 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.e.35 g/cm 3 4.5 wt% and CB = 87.95 × 10 -8 nm )3 (6.27 g/cm 3 100 t% 87.7 g/m ol 61. CA = 12.5 w + ol 125. the crystal structure is body-centered cubic. N1.11a).19 For a solid solution consisting of two elements (designated as 1 and 2). C1. 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.10a). from the above discussion the following holds: N1 = C1' N 100 Substitution into this expression of the appropriate form of N from Equation 4.4.6a). and after some algebraic manipulation we obtain the desired expression: 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. the number of atoms 1 1 of component 1 and total number of atoms per cubic centimeter. Thus. Aave 1 (Equation 4. finally. given the concentration of that element specified in weight percent. . ρave (Equation 4. This computation is possible using the following expression: N1 = NAC1 C1A1 A + 1( 100−C1) ρ1 ρ2 (4. Solution This problem asks that we derive Equation 4.4. respectively.18. concentration of component 1 in atom percent (C ' ) 1 The is just 100 c ' where c ' is the atom fraction of 1 1 component 1.2 yields N1 = C1' N A ρ ave 100 Aave And. using other equations given in the chapter. Furthermore.18) where NA = Avogadro’s number ρ1 and ρ2 = densities of the two elements A1 = the atomic weight of element 1 Derive Equation 4. and realizing that C2 = (C1 – 100).2 and expressions contained in Section 4. c ' is defined as c ' = N1/N where N1 and N are. sometimes it is desirable to determine the number of atoms per cubic centimeter of one element in a solid solution.18 using Equation 4. substitution into this equation expressions for C ' (Equation 4. . 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.N1 = C1 A1 ρ1 N AC1 + ρ2 A1 (100 − C1) 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. 49 g/cm 3 A1 = AAu = 196.02 0 3 tm o 1 t ) l (0 w% 9 7 g/m ) 16. respectively.9 (1 o l 9 7 o l + (0 − 1 w% 10 0 t ) 3 1 2 g/c 9.18 is necessary. using the following values: C1 = CAu = 10 wt% ρ1 = ρAu = 19. 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. . Compute the number of gold atoms per cubic centimeter for a silver-gold alloy that contains 10 wt% Au and 90 wt% Ag.49 g/cm3.4.32 g/cm 3 ρ2 = ρAg = 10.97 g/mol Thus NA = u CA AA u u ρA u N ACA u Au + A ( 0 − CA ) 10 u ρA g = (6 2 × 1 2 aos /m ) (0 w% .20 Gold forms a substitutional solid solution with silver.36 × 1021 atoms/cm 3 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.32 and 10. The densities of pure gold and silver are 19.4 m = 3. employment of Equation 4.3 m 1 9 g/c 3 0. Solution To solve this problem.9 g/m 1 t ) 6. .64 g/mol Thus N Ge = CGe AGe ρGe N ACGe A + Ge ( − CGe) 100 ρSi = ( . Compute the number of germanium atoms per cubic centimeter for a germanium-silicon alloy that contains 15 wt% Ge and 85 wt% Si.0 ×1 2 am /m )(5w 62 2 03 t s o o 1 t l %) (5w (7 4 g o ) 1 t %) 2.6 /m l 7 4g o 2. The densities of pure germanium and silicon are 5.33 g/cm3. employment of Equation 4. Solution To solve this problem.3 /c 3 = 3.21 Germanium forms a substitutional solid solution with silicon. using the following values: C1 = CGe = 15 wt% ρ1 = ρGe = 5. 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.3 /c 2 3g m .32 g/cm 3 ρ2 = ρSi = 2.4.6 /m l + (0 − 1 w 1 0 5t %) 3 5 2g m .16 × 1021 atoms/cm 3 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.18 is necessary.32 and 2.33 g/cm 3 A1 = AGe = 72. respectively. Solution The number of atoms of component 1 per cubic centimeter is just equal to the atom fraction of component 1 (c ' ) times the total number of atoms per cubic centimeter in the alloy (N). This computation is possible using the following expression: C1 = 100 N ρ ρ 1+ A 2 − 2 N1A1 ρ1 (4. Equations 4.11b. we may write ' c1 N A ρ ave ' N 1 = c1 N = Aave Realizing that ' C1 ' c1 = 100 and ' C 2 = 100 −C1' and substitution of the expressions for ρ and A . N1.4. 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. that will produce a specified concentration in terms of the number of atoms per cubic centimeter.19) where NA = Avogadro’s number ρ1 and ρ2 = densities of the two elements A1 and A2 = the atomic weights of the two elements Derive Equation 4.2. leads to ave ave N1 = c1' N Aρ ave Aave 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. C 1.4. .22 Sometimes it is desirable to be able to determine the weight percent of one element.10b and 4.19 using Equation 4. using the 1 equivalent of Equation 4. Thus.2 and expressions contained in Section 4. respectively. for an alloy composed of two types of atoms. solving for C ' 1 100 N 1 ρ A2 1 N Aρ ρ − N 1 ρ A1 + N 1 ρ A2 1 2 2 1 C1' = Substitution of this expression for C ' into Equation 4. 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. which may be written in the following form 1 ' C1 A1 ' ' C1 A1 + C 2 A2 C1 = × 10 0 = C1' A1 C1' A1 + (100− C1' ) A2 × 100 yields C1 = 1+ 100 N A ρ2 ρ − 2 N 1 A1 ρ1 the desired expression.7a.= N AC1' ρ1 ρ2 ' C1 ρ2 A1 + ( 100 − C1' )ρ1 A2 And. . respectively. Compute the weight percent of molybdenum that must be added to tungsten to yield an alloy that contains 1.0 × 1022 Mo atoms per cubic centimeter.91 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. The densities of pure Mo and W are 10.22 and 19.22 g/cm 3 ρ2 = ρW = 19. 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. employment of Equation 4.19 is necessary. Solution To solve this problem. .30 g/cm 3 A1 = AMo = 95.23 Molybdenum forms a substitutional solid solution with tungsten.94 g/mol A2 = AW = 183.3 g/c 3 o 1 0 m ) 1 0 g/c 3 l 9.4.84 g/mol Thus CMo = 100 N AρW ρ 1+ − W N Mo AMo ρMo = 1+ 10 0 ( .3 m − (1 2 am /c 3) 9 4 g/m ) 0 2 t s m ( 5.2 m 1 2 g/c 3 × 1 2 am 03 t s o = 8. using the following values: N1 = NMo = 1022 atoms/cm 3 ρ1 = ρMo = 10.9 o o l 0.30 g/cm3.02 62 /m )(9. 19 is necessary. Solution To solve this problem.94 g/mol Thus CNb = 100 N AρV ρ 1+ − V N Nb ANb ρNb = 1+ 10 0 (6 2 .57 g/cm 3 ρ2 = ρV = 6.02 a m /m )( .1 m − ( .24 Niobium forms a substitutional solid solution with vanadium.5 × 1 2 aos /c 3)(9 1 g/m ) 8 7 g/c 3 1 5 0 2 t m m 2. .5 m ×1 2 03 = 35.9 o l .57 and 6.55 × 1022 atoms/cm 3 ρ1 = ρNb = 8. Compute the weight percent of niobium that must be added to vanadium to yield an alloy that contains 1.1 g/c 3) 6 0 g/c 3 t s o 60 m o l . employment of Equation 4. 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.4.55 × 1022 Nb atoms per cubic centimeter.91 g/mol A2 = AV = 50. using the following values: N1 = NNb = 1.2 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.10 g/cm3. respectively. The densities of pure Nb and V are 8.10 g/cm 3 A1 = ANb = 92. 144 and 0. it is first necessary to calculate the density and average atomic weight of this alloy using Equations 4.25 Silver and palladium both have the FCC crystal structure.02 g/cm 3.5 as 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.02 g /cm3 = 10. respectively.11a.138 nm. respectively.49 g/cm 3 (as taken from inside the front cover) and 12.9 g/m 0 o l 1 0 6. and its atomic weight and atomic radius are 106. respectively. Inasmuch as the densities of silver and palladium are 10.83 g/cm 3 And for the average atomic weight Aave = 100 C Ag AAg + CPd APd = 10 0 7 w% 5 t 2 w% 5 t + 17.10a and 4. the unit cell edge length is just the cube root of the volume.4 g/m o l = 107.49 g /cm3 12. However. Also.4 g/mol and 0. VC is determined from Equation 3. Compute the unit cell edge length for a 75 wt% Ag–25 wt% Pd alloy.02 g/cm3. 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.4. the average density is just ρave = 100 C Ag ρAg + C Pd ρPd = 100 75 wt% 25 wt% + 10. Solution First of all.5 g/mol Now. the atomic radii for Ag (using the table inside the front cover) and Pd are 0. using Equation 3. . and inasmuch as the unit cell is cubic.5 it is possible to compute the unit cell volume.138 nm. The room-temperature density of Pd is 12. and Pd forms a substitutional solid solution for all concentrations at room temperature. g /mol) 4 ( 5 ( 83 g /cm3 )6.404 nm 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. .04 × 10-8 cm = 0.VC = ρ NA ave nAave = ( atoms/unit cell)107. ( = 6. × − c3 ( 6 5 1 2 / c) 9 03 u e1 m n l / i t = 4. 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. finally a = (VC )1/3 3 =.022 ×1023 atoms/mol) 10.59 × 10-23 cm3/unit cell And. . Cite the relative Burgers vector–dislocation line orientations for edge. and neither perpendicular nor parallel for mixed dislocations. parallel for screw dislocations. and mixed 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. screw. 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.26 dislocations.Dislocations—Linear Defects 4. Solution The Burgers vector and dislocation line are perpendicular for edge dislocations. e. 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. respectively 2R 2 3 1 (where R is the atomic radius). 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 3.Interfacial Defects 4.54. the higher the packing density.29 R2 4R 2 and . Thus. since the planar density for (111) is greater.27 For an FCC single crystal. .. From the solution to Problem 3.11)]—that is.) Solution The surface energy for a crystallographic plane will depend on its packing density [i. the greater the number of nearest-neighbor atoms. the planar density (Section 3.25 R2 and 0. planar densities for FCC (100) and (111) planes are —that is 1 0. consequently.54 at the end of Chapter 3. and the more atomic bonds in that plane that are satisfied. the lower the surface energy. and. it will have the lower surface energy. it will have the lower surface 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. . Thus.19 R2 and 0.e. and the more atomic bonds in that plane that are satisfied. would you expect the surface energy for a (100) plane to be greater or less than that for a (110) plane? Why? (Note: You may want to consult the solution to Problem 3.) Solution The surface energy for a crystallographic plane will depend on its packing density [i. consequently.55. the planar densities for BCC (100) and (110) are —that is energy. the higher the packing density.28 For a BCC single crystal. the lower the surface energy. the planar density (Section 3. 3 0. respectively 8R 2 2 3 . 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.27 R2 16 R 2 and .11)]—that is. and..55 at the end of Chapter 3.4. the greater the number of nearest-neighbor atoms. since the planar density for (110) is greater. From the solution to Problem 3. there will be fewer unsatisfied bonds along a grain boundary. 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. (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. would you expect the surface energy to be greater than. the same as. . thus. some atoms on one side of a boundary will bond to atoms on the other side. Why is this so? Solution (a) The surface energy will be greater than the grain boundary energy. and.29 (a) For a given material.4. Therefore. such is not the case for surface atoms. 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. there are fewer unsatisfied bonds. 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. For grain boundaries. (b) Cite the difference between mechanical and annealing twins. Annealing twins form during annealing heat treatments. 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. most often in FCC metals. .4. (b) Mechanical twins are produced as a result of mechanical deformation and generally occur in BCC and HCP metals. 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. Solution (a) A twin boundary is an interface such that atoms on one side are located at mirror image positions of those atoms situated on the other boundary side.30 (a) Briefly describe a twin and a twin boundary. The region on one side of this boundary is called a twin. 31 For each of the following stacking sequences found in FCC metals.4. Solution (a) The interfacial defect that exists for this stacking sequence is a twin boundary. Now. 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. . . . . Within this region. 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. . which occurs between the two lines. A B C A B C B C A B C . . cite the type of planar defect that exists: (a) . The stacking sequence on one side of this position is mirrored on the other side. A B C A B C B A C B A . (b) . . (b) The interfacial defect that exists within this FCC stacking sequence is a stacking fault. the stacking sequence is HCP. . . copy the stacking sequences and indicate the position(s) of planar defect(s) with a vertical dashed line. which occurs at the indicated position. of the specimen whose microstructure is shown in Figure 4.32 (a) Using the intercept method. use at least seven straight-line segments. Grains Intersected 11 10 9 8.14(b). determine the average grain size. it is necessary to count the number of grains intersected by each of these line segments.14(b). in millimeters. In order to determine the average grain diameter.Grain Size Determination 4.5 7 10 8 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. (b) Estimate the ASTM grain size number for this material. each of which is 60 mm long has been constructed. Solution (a) Below is shown the photomicrograph of Figure 4. . on which seven straight line segments. Line Number 1 2 3 4 5 6 7 No. These data are tabulated below. these lines are labeled “1” through “7”. 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. the average line length intersected is just 6m 0 m =6 m . is related to the number of grains per square inch. . d. it is first necessary to take logarithms as l N =n − l 2 o g ( 1) o g From which n equals 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. Hence.16.9 m 5 91 . is d = a le lg v i e e. Therefore. The average grain size number. at a magnification of 100× according to Equation 4.1. the average grain diameter. in order to solve for n in Equation 4. Now. 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. Inasmuch as the magnification is 100× . n n t h mtn aii ga no i f c iet nc tsd re e = 69m . m 5 =. on a side is shown below. 6 ×0 2m 5 9 1− m 1 0 0 (b) This portion of the problem calls for us to estimate the ASTM grain size number for this same material.The average number of grain boundary intersections for these lines was 9. the value of N is measured directly from the micrograph.16. n. The total number of complete grains within this square is approximately 10 (taking into account grain fractions). The photomicrograph on which has been constructed a square 1 in. N. 3 lg 2 o 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.n= lg N o +1 lg 2 o lg 1 o 0 = + =4 1 . . 7.25(a). Therefore. each of which is 60 mm long has been constructed. these lines are labeled “1” through “7”. In order to determine the average grain diameter. it is necessary to count the number of grains intersected by each of these line segments. . 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.33 (a) Employing the intercept technique.4. Solution (a) Below is shown the photomicrograph of Figure 9. These data are tabulated below. (b) Estimate the ASTM grain size number for this material. Grains Intersected 7 7 7 8 10 7 8 The average number of grain boundary intersections for these lines was 8. use at least seven straight-line segments. determine the average grain size for the steel specimen whose microstructure is shown in Figure 9. Line Number 1 2 3 4 5 6 7 No. on which seven straight line segments.25(a). the average line length intersected is just 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. m 9 =0 0m . 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. and n is the ASTM grain size number. at a magnification of 100× according to Equation 4.e t c l ned i gs d = mo ac gt nn ia fi i = 6m .7 7 m 9 0 (b) This portion of the problem calls for us to estimate the ASTM grain size number for this same material. it is necessary to use Equation 4.16. but rather 90× . on a side is shown below.6m 0 m =6m . . Taking logarithms of both sides of this equation leads to the following: M logN M + 2 lo g =(n − 1 ) lo 2 g 1 0 0 Solving this expression for n gives n= M l g N M +2 l g o o 100 lg 2 o +1 The photomicrograph on which has been constructed a square 1 in. is antnt vl he e eire . the average grain diameter. m 9 8 . n.17 2 M n− N M =2 1 0 1 0 where NM = the number of grains per square inch at magnification M. is related to the number of grains per square inch. 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. d. the magnification of this micrograph is not 100× . However. The average grain size number. N.7 Hence. Consequently. 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.25(a). NM is measured to be approximately 7.5 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.From Figure 9. . which leads to 9 0 l g 7+2 lo o g 10 0 n= +1 lo 2 g = 3. 16. 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.280. Without any magnification. All we need do is solve for the parameter N in Equation 4. In order to solve this problem it is necessary to use Equation 4. M in the above equation is 1. approximately how many grains would there be per square inch at (a) a magnification of 100. Thus N = n− 2 1 = 1 28− = 128 grains/in. inasmuch as n = 8. 2 1 1 N1 =28− =128 100 And. N1 = 1. 2. and n is the ASTM grain size number.17: 2 M n− 1 NM =2 100 where NM = the number of grains per square inch at magnification M. .4. 2 (b) Now it is necessary to compute the value of N for no magnification. and (b) without any magnification? Solution (a) This part of problem asks that we compute the number of grains per square inch for an ASTM grain size of 8 at a magnification of 100× .34 For an ASTM grain size of 8. and therefore. 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.000 grains/in. solving for N1. 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. In order to solve this problem we make use of Equation 4.2 lg 2 o 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. we have n= M l g N M +2 l g o o 100 lg 2 o +1 600 lg 8 + 2 lg o o 100 = + 1 =9.35 Determine the ASTM grain size number if 25 grains per square inch are measured at a magnification of 600.17: M N M 100 2 = 2n −1 where NM = the number of grains per square inch at magnification M. .4. and n is the ASTM grain size number. Solution This problem asks that we determine the ASTM grain size number if 8 grains per square inch are measured at a magnification of 600. Solving the above equation for n. while M = 600. and realizing that NM = 8. . while M = 50.36 Determine the ASTM grain size number if 20 grains per square inch are measured at a magnification of 50. Solution This problem asks that we determine the ASTM grain size number if 20 grains per square inch are measured at a magnification of 50.4.3 lg 2 o 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. and realizing that NM = 20. Solving the above equation for n. and n is the ASTM grain size number. we have n= M l g N M +2 l g o o 100 lg 2 o +1 50 l g 20 + 2 l g o o 100 = + 1 =3. 2 M n− 1 NM =2 100 where NM = the number of grains per square inch at magnification M.17—viz. 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. In order to solve this problem we make use of Equation 4. 540 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.DESIGN PROBLEMS Specification of Composition 4. . Compute the concentration of Li (in wt%) that is required.534 and 2. 0 5 7 /c 3 . Upon solving for CLi from the above equation. which takes the form ρave = 100 CLi 100 − CLi + ρLi ρA l inasmuch as CLi + CAl = 100.5g m)21g m −03 g m) 2 5 /c 3 .10a. the respective densities of Li and Al are 0. A commercial aircraft skin material having a density of 2. Solution Solution of this problem requires the use of Equation 4.55 g/cm 3 is desired. 4 /c 3 ( .D1 Aluminum–lithium alloys have been developed by the aircraft industry to reduce the weight and improve the performance of its aircraft. we get C Li = 100 ρ ( Al − ρ ) Li ρ ave ρ ( Al − ρ ) ρ ave Li And incorporating specified values into the above equation leads to C= L i (0) 03 g m) 21g m −25g m) 1 ( . 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.71 g/cm 3. ( 7 /c 3 . 4 /c 3 5 = 1. 5 /c 3 ( . According to the table inside the front cover. 8 0 c3 4 9 = 1 4 12m 03 3 It is now necessary to construct expressions for Aave and ρave in terms of the concentration of vanadium. a.4. VC. That is VC = 3=8 a (9 0) . Determine the concentration in weight percent of V that must be added to iron to yield a unit cell edge length of 0. .289 nm. using Equations 4. For Aave we have Aave = 100 CV (100 − CV) + AV AFe 10 0 = CV (0 − CV) 10 + 5 4g/m 0.D2 Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to approximately 20 wt% V at room temperature.9 o l 5 5 g/m 5. as VC = ρ NA ave nAave where Aave and ρave are the atomic weight and density.11a and 4.5. CV.n 2 m 3 = × −m 2 × − c ( 2 18 ) . Solution To begin. of the Fe-V alloy. . it is necessary to employ Equation 3.10a. respectively. and solve for the unit cell volume. 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. VC is just the cube of the unit cell length.8 o l whereas for ρ ave ρave = 100 CV (100 − CV) + ρV ρFe 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. Inasmuch as both of these materials have the BCC crystal structure. which has cubic symmetry. and thus. 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.= 10 0 CV (1 0 − CV) 0 + 3 6.9 = 10 0 (6 2 × 1 2 ao s /m .1 ) And solving this expression for CV leads to CV = 12.1 g/c 0 m 7.8 m . the value of n in Equation 3. this expression may be written in terms of the concentration of V in weight percent as follows: VC = 2. hence.02 0 3 tm o l CV ( 0 − CV) 10 + 6 0 g/c 3 m 7 7 g/c 3 .5 is 2. .414 × 10-23 cm3 = ρave N A nAave 10 0 (2 ao s /ui cl ) tm n el t CV ( 0 − CV) 10 + 5 4 g/m o l 5 5 g/m 5.8 o l 0.9 wt%.8 g/c 3 7 m Within the BCC unit cell there are 2 equivalent atoms. 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.