MECH 340 Lecture, Fall 2004

4-1 CHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS Vacancies and Self-Interstitials 4.1 In order to compute the fraction of atom sites that are vacant in copper at 1357 K, we must employ Equation 4.1. As stated in the problem, Qv = 0.90 eV/atom. Thus, N v N = exp − Qv kT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = exp − 0.90 eV/atom (8.62 × 10−5 eV/atom- K)(1357 K) ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 4.56 x 10-4 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. 4-2 4.2 Determination of the number of vacancies per cubic meter in gold at 900°C (1173 K) requires the utilization of Equations 4.1 and 4.2 as follows: Nv = N exp − Qv kT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = NAρAu AAu exp − Qv kT ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = (6.023 × 1023atoms /mol)(18.63 g/cm3) 196.9 g/mol exp − 0.98 eV/atom (8.62 × 10−5 eV/atom− K) (1173 K) ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 3.52 x 1018 cm-3 = 3.52 x 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 4.3 This problem calls for the computation of the activation energy for vacancy formation in silver. 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, (ρ), Avogadro's number (NA), and the atomic weight (A) according to Equation 4.2 as N = N AρPb APb = (6.023 × 1023 atoms /mol)(9.5 g /cm3) 107.9 g/mol = 5.30 x 1022 atoms/cm3 = 5.30 x 1028 atoms/m3 Now, taking natural logarithms of both sides of Equation 4.1, ln Nv = ln N − Qv kT and, after some algebraic manipulation Qv = − kT ln Nv N ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − (8.62 × 10-5 eV/atom- K)(800°C + 273 K) ln 3.60 × 1023 m−3 5.30 × 1028 m−3 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 1.10 eV/atom Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4-4 Impurities in Solids 4.4 In this problem we are asked to cite which of the elements listed form with Ni the three possible solid solution types. For complete substitutional solubility the following criteria must be met: 1) the difference in atomic radii between Ni and the other element (∆R%) must be less than ±15%, 2) the crystal structures must be the same, 3) the electronegativities must be similar, and 4) the valences should be the same, or nearly the same. Below are tabulated, for the various elements, these criteria. Crystal ∆Electro- Element ∆R% Structure negativity Valence Ni FCC 2+ C –43 H –63 O –52 Ag +16 FCC +0.1 1+ Al +15 FCC -0.3 3+ Co +0.6 HCP 0 2+ Cr +0.2 BCC -0.2 3+ Fe -0.4 BCC 0 2+ Pt +11 FCC +0.4 2+ Zn +7 HCP -0.2 2+ (a) Pt is the only element that meets all of the criteria and thus forms a substitutional solid solution having complete solubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal structure, and thus display complete solid solubility at these temperatures. (b) Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Ni are greater than ±15%, and/or have a valence different than 2+. (c) C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of Ni. 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 4.5 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 = 2 R 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. 4-6 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 = R2 + 2Rr + r2 After rearrangement the following quadratic equation results: r2 + 2Rr − 0.667R 2 = 0 And upon solving for r: r = −(2R) ± (2R)2 − (4)(1)(−0.667R2) 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. 4-7 = −2R± 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. 4-8 Specification of Composition 4.6 (a) This problem asks that we derive Equation 4.7a. To begin, C1 is defined according to Equation 4.3 as C1 = m1 m1 + m2 × 100 or, equivalently C1 = m1 ' m1 ' + m2 ' × 100 where the primed m's indicate masses in grams. From Equation 4.4 we may write m1 ' = nm1 A1 m 2 ' = nm2 A2 And, substitution into the C1 expression above C1 = nm1 A1 nm1 A1 + nm2 A2 × 100 From Equation 4.5 it is the case that nm1 = C1 ' (nm1 + nm2) 100 nm2 = C2 ' (nm1 + nm2) 100 And substitution of these expressions into the above equation leads to 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 C1 = C1 ' A1 C1 ' A1 + C2 ' A2 × 100 which is just Equation 4.7a. (b) This problem asks that we derive Equation 4.9a. To begin, is defined as the mass of component 1 per unit volume of alloy, or C1 " C1 " = m1 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 = V1 + V2--then C1 " = m1 V1 + V2 Furthermore, the volume of each constituent is related to its density and mass as V1 = m1 ρ1 V2 = m2 ρ2 This leads to C1 " = m1 m1 ρ1 + m2 ρ2 From Equation 4.3, m1 and m2 may be expressed as follows: m1 = C1(m1 + m2) 100 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 m2 = C2 (m1 + m2) 100 Substitution of these equations into the preceding expression yields C1 " = C1(m1 + m2) 100 C1(m1 + m2) 100 ρ1 + C2 (m1 + m2) 100 ρ2 = C1 C1 ρ1 + C2 ρ2 If the densities ρ1 and ρ2 are given in units of g/cm3, then conversion to units of kg/m3 requires that we multiply this equation by 103, inasmuch as 1 g/cm3 = 103 kg/m3 Therefore, the previous equation takes the form C1 " = C1 C1 ρ1 + C2 ρ2 x 103 which is the desired expression. (c) Now we are asked to derive Equation 4.10a. The density of an alloy ρave is just the total alloy mass M divided by its volume V ρave = M V Or, in terms of the component elements 1 and 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. 4-11 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, which is only an approximation; normally V will not be exactly equal to (V1 + V2)]. Each of V1 and V2 may be expressed in terms of its mass and density as, V1 = m1 ρ1 V2 = m2 ρ2 When these expressions are substituted into the above equation, we get ρave = m1 + m2 m1 ρ1 + m2 ρ2 Furthermore, from Equation 4.3 m1 = C1(m1 + m2) 100 m2 = C2 (m1 + m2) 100 Which, when substituted into the above ρave expression yields ρave = m1 + m2 C1(m1 + m2) 100 ρ1 + C2 (m1 + m2) 100 ρ2 And, finally, this equation reduces to = 100 C1 ρ1 + C2 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. 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 4.7 In order to compute composition, in atom percent, of a 92.5 wt% Ag-7.5 wt% Cu alloy, we employ Equation 4.6 as CAg ' = CAgACu CAgACu + CCu AAg × 100 = (92.5)(63.55 g/mol) (92.5)(63.55 g/mol) + (7.5)(107.87g/mol) × 100 = 87.9 at% CCu ' = CCu AAg CAgACu + CCu AAg × 100 = (7.5)(107.87 g/mol) (92.5)(63.55 g/mol) + (7.5)(107.87g/mol) × 100 = 12.1 at% 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 4.8 In order to compute composition, in weight percent, of a 5 at% Cu-95 at% Pt alloy, we employ Equation 4.7 as CCu = CCu ' ACu CCu ' ACu + CPt ' APt × 100 = (5)(63.55 g /mol) (5)(63.55 g /mol) + (95)(195.08 g /mol) × 100 = 1.68 wt% CPt = CPt ' APt CCu ' ACu + CPt ' APt × 100 = (95)(195.08 g /mol) (5)(63.55 g/mol) + (95)(195.08 g/mol) x 100 = 98.32 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. 4-15 4.9 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 iron (CFe) is just CFe = mFe mFe + mC + mCr × 100 = 105 kg 105 kg + 0.2 kg + 1.0 kg × 100 = 98.87 wt% Similarly, for carbon CC = 0.2 kg 105 kg + 0.2 kg + 1.0 kg × 100 = 0.19 wt% And for chromium CCr = 1.0 kg 105 kg + 0.2 kg + 1.0 kg × 100 = 0.94 wt% 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 4.10 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 Cu and Zn, using Equation 4.4. Thus, the number of moles of Cu is just nmCu = mCu ' ACu = 33 g 63.55 g/mol = 0.519 mol Likewise, for Zn nmZn = 47 g 65.39 g /mol = 0.719 mol Now, use of Equation 4.5 yields CCu ' = nmCu nmCu + nmZn × 100 = 0.519 mol 0.519 mol + 0.719 mol × 100 = 41.9 at% Also, CZn ' = 0.719 mol 0.519 mol + 0.719 mol × 100 = 58.1 at% Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 4-17 4.11 In this problem we are asked to determine the concentrations, in atom percent, of the Ag-Au-Cu alloy. It is first necessary to convert the amounts of Ag, Au, and Cu into grams. mAg ' = (44.5 lbm)(453.6 g/lbm) = 20,185 g mAu ' = (83.7 lbm)(453.6 g/lbm) = 37,966 g mCu ' = (5.3 lbm)(453.6 g/lbm) = 2, 404 g These masses must next be converted into moles (Equation 4.4), as nmAg = mAg ' AAg = 20,185 g 107.87 g/mol = 187.1 mol nmAu = 37,966 g 196.97 g /mol = 192.8 mol nmCu = 2,404 g 63.55 g /mol = 37.8 mol Now, employment of a modified form of Equation 4.5, gives CAg ' = nmAg nmAg + nmAu + nmCu × 100 = 187.1 mol 187.1 mol + 192.8 mol + 37.8 mol × 100 = 44.8 at% CAu ' = 192.8 mol 187.1 mol + 192.8 mol + 37.8 mol × 100 = 46.2 at% CCu ' = 37.8 mol 187.1 mol + 192.8 mol + 37.8 mol × 100 = 9.0 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. 4-18 4.12 We are asked to compute the composition of a Pb-Sn alloy in atom percent. Employment of Equation 4.6 leads to CPb ' = CPb ASn CPb ASn + CSn APb × 100 = 5.5(118.69 g/mol) 5.5(118.69 g /mol) + 94.5(207.2 g/mol) × 100 = 3.2 at% CSn ' = CSn APb CSn APb + CPbASn × 100 = 94.5(207.2 g/mol) 94.5(207.2 g/mol) + 5.5(118.69 g/mol) × 100 = 96.8 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. 4-19 4.13 This problem calls for a conversion of composition in atom percent to composition in weight percent. The composition in atom percent for Problem 4.11 is 44.8 at% Ag, 46.2 at% Au, and 9.0 at% Cu. Modification of Equation 4.7 to take into account a three-component alloy leads to the following CAg = CAg ' AAg CAg ' AAg + CAu ' AAu + CCu ' ACu × 100 = (44.8) (107.87 g /mol) (44.8)(107.87 g /mol) + (46.2) (196.97 g /mol) + (9.0) (63.55 g /mol) × 100 = 33.3 wt% CAu = CAu ' AAu CAg ' AAg + CAu ' AAu + CCu ' ACu × 100 = (46.2) (196.97 g /mol) (44.8)(107.87 g /mol) + (46.2) (196.97 g /mol) + (9.0) (63.55 g /mol) × 100 = 62.7 wt% CCu = CCu ' ACu CAg ' AAg + CAu ' AAu + CCu ' ACu × 100 = (9.0) (63.55 g/mol) (44.8)(107.87 g /mol) + (46.2) (196.97 g /mol) + (9.0) (63.55 g /mol) × 100 = 4.0 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. 4-20 4.14 This problem calls for a determination of the number of atoms per cubic meter for lead. In order to solve this problem, one must employ Equation 4.2, N = N AρPb APb The density of Pb (from the table inside of the front cover) is 11.35 g/cm3, while its atomic weight is 207.2 g/mol. Thus, N = (6.023 x 1023 atoms /mol)(11.35 g/cm3) 207.2 g /mol = 3.30 x 1022 atoms/cm3 = 3.30 x 1028 atoms/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. 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 4.15 In order to compute the concentration in kg/m3 of Si in a 0.25 wt% Si-99.75 wt% Fe alloy we must employ Equation 4.9 as CSi " = CSi CSi ρSi + CFe ρFe × 103 From inside the front cover, densities for silicon and iron are 2.33 and 7.87 g/cm3, respectively; and, therefore CSi " = 0.25 0.25 2.33 g/cm3 + 99.75 7.87 g/cm3 × 103 = 19.6 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. 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 4.16 We are asked in this problem to determine the approximate density of a Ti-6Al-4V titanium alloy that has a composition of 90 wt% Ti, 6 wt% Al, and 4 wt% V. In order to solve this problem, Equation 4.10a is modified to take the following form: ρave = 100 CTi ρTi + CAl ρAl + CV ρV And, using the density values for Ti, Al, and V—i.e., 4.51 g/cm3, 2.71 g/cm3, and 6.10 g/cm3—(as taken from inside the front cover of the text), the density is computed as follows: ρave = 100 90 wt% 4.51 g /cm3 + 6 wt% 2.71 g/cm3 + 4 wt% 6.10 g/cm3 = 4.38 g/cm3 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 4.17 This problem asks that we determine the unit cell edge length for a 80 wt% Ag-20 wt% Pd alloy. 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 ρave = nAave VC NA Inasmuch as the unit cell is cubic, then VC = a3, then ρave = nAave a3N A And solving this equation for the unit cell edge length, leads to a = nAave ρaveNA ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ 1/3 Expressions for Aave and ρave are found in Equations 4.11a and 4.10a, respectively, which, when incorporated into the above expression yields a = n 100 CAg AAg + CPd APd ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ 100 CAg ρAg + CPd ρPd ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ N A ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 1/3 Since the crystal structure is FCC, the value of n in the above expression is 4 atoms per unit cell. The atomic weights for Ag and Pd are 107.9 and 106.4 g/mol, respectively (Figure 2.6), whereas the densities for the Ag and Pd are 10.49 g/cm3 (inside front cover) and 12.02 g/cm3. Substitution of these, as well as the concentration values stipulated in the problem statement, into the above equation gives 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 a = (4 atoms/unit cell) 100 80 wt% 107.9 g/mol + 20 wt% 106.4 g/mol ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ 100 80 wt% 10.49 g/cm3 + 20 wt% 12.02 g/cm3 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ 6.023 × 1023 atoms/mol ( ) ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 1/3 = 4.050 × 10-8 cm = 0.4050 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. 4-25 4.18 This problem asks that we determine, for a hypothetical alloy that is composed of 25 wt% of metal A and 75 wt% of metal B, whether the crystal structure is simple cubic, face-centered cubic, or body-centered cubic. We are given the densities of the these metals (ρA = 6.17 g/cm3 and ρB = 8.00 g/cm3 for B), their atomic weights (AA = 171.3 g/mol and AB = 162.0 g/mol), and that the unit cell edge length is 0.332 nm (i.e., 3.32 x 10-8 cm). 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 ρave = nAave VC NA Inasmuch as for each of the possible crystal structures, the unit cell is cubic, then VC = a3, or ρave = nAave a3N 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 = ρavea3NA Aave Expressions for Aave and ρave are found in Equations 4.11a and 4.10a, respectively, which, when incorporated into the above expression yields n = 100 CA ρA + CB ρB ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ a3N A 100 CA AA + CB AB ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ Substitution of the concentration values (i.e., CA = 25 wt% and CB = 75 wt%) as well as values for the other parameters given in the problem statement, into the above equation gives 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. 4-26 n = 100 25 wt% 6.17 g/cm3 + 75 wt% 8.00 g/cm3 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ (3.32 × 10-8 nm)3(6.023 × 1023 atoms/mol) 100 25 wt% 171.3 g/mol + 75 wt% 162.0 g/mol ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ = 1.00 atom/unit cell Therefore, on the basis of this value, the crystal structure is simple 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. 4-27 4.19 This problem asks that we derive Equation 4.18, using other equations given in the chapter. The concentration of component 1 in atom percent is just 100 where is the atom fraction of component 1. Furthermore, is defined as = N1/N where N1 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: (C1 ') c1 ' c1 ' c1 ' c1 ' N1 = C1 ' N 100 Substitution into this expression of the appropriate form of N from Equation 4.2 yields N1 = C1 ' NA ρave 100 Aave And, finally, substitution into this equation expressions for (Equation 4.6a), ρave (Equation 4.10a), Aave (Equation 4.11a), and realizing that C2 = (C1 – 100), and after some algebraic manipulation we obtain the desired expression: C1 ' N1 = NAC1 C1 A1 ρ1 + A1 ρ2 (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. 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 4.20 This problem asks us to determine the number of molybdenum atoms per cubic centimeter for a 16.4 wt% Mo-83.6 wt% W solid solution. To solve this problem, employment of Equation 4.18 is necessary, using the following values: C1 = CMo = 16.4 wt% ρ1 = ρMo = 10.22 g/cm3 ρ2 = ρW = 19.3 g/cm3 A1 = AMo = 95.94 g/mol Thus NMo = N ACMo CMo AMo ρMo + AMo ρW (100 − CMo) = (6.023 × 1023 atoms /mol) (16.4 wt%) (16.4 wt%)(95.94 g/mol) 10.22 g/cm3 + 95.94 g /mol 19.3 g/cm3 (100 − 16.4 wt%) = 1.73 x 1022 atoms/cm3 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 4.21 This problem asks us to determine the number of niobium atoms per cubic centimeter for a 24 wt% Nb-76 wt% V solid solution. To solve this problem, employment of Equation 4.18 is necessary, using the following values: C1 = CNb = 24 wt% ρ1 = ρNb = 8.57 g/cm3 ρ2 = ρV = 6.10 g/cm3 A1 = ANb = 92.91 g/mol Thus NNb = NACNb CNb ANb ρNb + ANb ρV (100 − CNb) = (6.023 × 1023 atoms /mol) (24 wt%) (24 wt%)(92.91 g/mol) 8.57 g/cm3 + 92.91 g/mol 6.10 g/cm3 (100 − 24 wt%) = 1.02 x 1022 atoms/cm3 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 4.22 This problem asks that we derive Equation 4.19, using other equations given in the chapter. The number of atoms of component 1 per cubic centimeter is just equal to the atom fraction of component 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 (c1 ') N1 = c1 ' N = c1 ' N A ρave Aave Realizing that c1 ' = C1 ' 100 and C2 ' = 100 − C1 ' and substitution of the expressions for ρave and Aave, Equations 4.10b and 4.11b, respectively, leads to N1 = c1 ' N Aρave Aave = NAC1 ' ρ1ρ2 C1 ' ρ2 A1 + (100 − C1 ' )ρ1A2 And, solving for C1 ' C1 ' = 100 N1ρ1 A2 NAρ1ρ2 − N1ρ2 A1 + N1ρ1 A2 Substitution of this expression for C1 into Equation 4.7a, which may be written in the following form ' 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 C1 = C1 '