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1、【精品文档】如有侵权，请联系网站删除，仅供学习与交流工程材料科学与设计(原书第2版)课后习题答案(48)章.精品文档.Solutions to Chapter 41. FIND: What material has a property that is hugely affected by a small impurity level?SOLUTION: Electrical conductivity spans a wide range. Incorporation of a few parts per million impurities can change electrical con
2、ductivity orders of magnitude. Small cracks in brittle materials decrease their tensile strength by orders of magnitude. Small additions of impurity can change the color of gems. COMMENTS: These are but a few examples.2.COMPUTE: The temperature at which the vacancy concentration is one half that of
3、25oC.GIVEN: EQUATION:whereCv = vacancy concentrationQfv = activation energy for vacancy informationR = gas constant 8.314 J/mole-KT = absolute temperatureIn the present problemand T1 = 35 + 273 = 308KT2 = 25 + 273 = 298KalsoCv(35oC) = 2Cv(25oC)Thus,Solving for Qfv we get Qfv = 52893.5 J/mole.Using t
4、his value of Qfv, the Cv(25oC) can be calculatedThe problem requires us to calculate the temperature at which the vacancy concentration is Cv(25oC). Cv(25oC) = 2.675 x 10-10Thusfor solving T, we get:T = 288.63K or 15.63oC.3.COMPUTE:GIVEN:EQUATION:Dividing (1) by (2) we get:Solving for Q, we get:Q =
5、22033.56 J/mole= exp(-7.511)= 5.46 x 10-4The problem requires computing a temperature at which Cv = 3Cv(80oC).3Cv(80oC) = 3 x 5.46 x 10-4= 1.63 x 10-3solving for T, we get: T = 413.05K or 140.05oC4.5.FIND: Are Al and Zn completely soluble in solid solution?If Al-Zn system obeys all the Hume-Rothery
6、rules. Then it is expected to show complete solubility.(i)The atomic radii of Al and Zn are 0.143nm and 0.133 nm respectively. The difference in their radii is 7.5% which is less than 15%.(ii)The electronegativities of Al and Zn are 1.61 an 1.65 respectively which are also very similar.(iii)The most
7、 common valence of Al is +3 and +2 for Zn.(iv)Al has an FCC structure where Zn has a HCP structure.It appears that Al-Zn system obeys 3 out of 4 Hume-Rothery rules. In this case they are not expected to be completely soluble.6.SHOW: The extent of solid solution formation in the following systems usi
8、ng Hume-Rothery Rules.(a) Al in NiSize: r(Ni) = 0.125nm; r(Al) = 0.143nm difference = 14.4%Electronegativity: Al = 1.61; Ni = 1.91Most Common Valence: Al3+; Ni2+Crystal Structure: Al: FCC; Ni:FCCThe crystal structure of Al and Ni are the same and the most common valencies are also comparable. Howeve
9、r, the size difference is close to 15% and the difference is electronegativities is rather significant.Based on this, it appears that Ni and Al would not form a solid solution over the entire compositional range.(b) Ti in NiSize: r(Ti) = 0.147 nm, r(Ni) = 0.125nm difference = 17.6%Electronegativity:
10、 Ti: 1.54; Ni: 1.91Valence:Ti4+; Ni2+Crystal Structure:Ti:HCP; Ni FCCTi in Ni would not exhibit extensive solid solubility(c)Zn in FeSizer(Zn) = 0.133nm; r(Fe) - 0.124nm difference = 7.25%Electronegativity:Zn = 1.65; Fe = 1.83Most Common Valence: Zn2+; Fe2+Crystal Structure:An: HCP; Fe: BCCSince ele
11、ctronegativities and crystal structures are very different, Zn - Fe will not exhibit extensive solid solubility.(d)Si in AlSize r(Si) = 0.117 nm; r(Al) = 0.143nm; difference = 22.2%Electronegativity: (Si) = 1.90; Al = 1.61Valence: Si4+; A;3+Crystal Structure: Si: Diamond Cubic; Al: FCCSince the size
12、 difference is greater than 15%, and the crystal structures are different, Si-Al would not exhibit extensive solid solubility.(e) Li in AlSize r(li): 0.152, r(Al): 0.143; difference - 6.29%Electronegativity: Li: 0.98; Al: 1.61Most Common Valence: Li1+; Al3+Crystal Structure: Li:BCC; Al: FCCSince ele
13、ctronegativity and crystal structures are very different, Li-Al will not exhibit extensive solid solubility.(f)Cu in AuSize r(Cu) = 0.125nm; r(au) = 0.144nm; difference = 12.5%Electronegativity: Cu = 1.90; Au = 1.93Most Common Valence: Cu+; Au+Crystal Structure: Cu:FCC; Au:FCCCu-Au will exhibit exte
14、nsive solid solubility.(g)Mn in FeSize r(Mn) = 0.112, r(Fe) = 0.124 difference = 10.71%Electronegativity: Mn 1.55; Fe 1.83Most Common Valence: Mn2+; Fe2+Crystal Structure: Mn:BCC; Fe BCCThe difference in electronegativity is high but Mn-Fe does obey the other 3 Hume-Rothery rules. Therefore, it will
15、 form solid solutions but not over the entire compositional range.(h)Cr in FeSize r(Cr) = 0.125nm, Fe = 0.144nm difference = 12.5%Electronegativity: Cr = 1.66; Fe = 1.83Most Common Valence: Cr3+; Fe2+Crystal Structure: Cr:BCC; Fe:BCCCr in Fe will exhibit extensive solid solubility but not over the e
16、ntire compositional range since it obeys only 3 of 4 Hume-Rothery rules.(i)Ni in FeSize r(Ni) = 0.125nm, r(Fe) = 0.124nm difference = 0.8%Electronegativity: Ni: 1.91; Fe 1.83Most Common Valence: Ni3+; Fe3+Crystal Structure: Ni:FCC; Fe: BCCNi and Fe obeys 3 of the 4 Hume-Rothery rules therefore, exte
17、nsive solid solution will be exhibited but not over the entire compositional range.7.(a) When one attempts to add a small amount of Ni to Cu, Ni is the solute and Cu is the solvent.(b) Based on the relative sizes of Ni and Cu, radius of Ni = 0.128nm, radius of Cu = 0.125nm, these two are expected to
18、 form substitutional solid solutions.(c) Ni and Cu will be completely soluble in each other because they obey all four Hume-Rothery rules.8. FIND: Predict how Cu dissolves in Al. DATA: Cu Alatomic radius (A)1.281.43electronegativity1.901.61valence1+,2+3+crystal structureFCCFCCSOLUTION: All of Hume-R
19、otherys rules must be followed for a substitutional solution. In this case, the valences do not match. Cu will not go into substitutional positions in Al to a large extent. COMMENTS: This principle is often used to precipitation harden Al using Cu.9.What type of solid solution is expected to form wh
20、en C is added to Fe?The radius of carbon atom is 0.077nm and that of an Fe atom is 0.124nm. The size difference between these two is 61% which is much grater than 15%. Thus, these two are not expected to form substitutional solid solution.If we compare the size ratio of C to Fe atoms with the size o
21、f tetrahedral and octahedral interstitial sites in BCC iron, we find that C does not easily fit into either type of interstitial position. C, however, forms an interstitial solid solution with Fe but the solubility is limited.10. FIND: Calculate the activation for vacancy formation in Fe. GIVEN: The
22、 vacancy concentration at 727C = 1000K is 0.00022. SOLUTION: We use equation 4.2-2 to solve this problem:Cv = exp (-Qfv/RT) Solving for Qfv: Qfv = -RT ln Cv = -(8.31 J/mole-K)(1000K) ln 0.00022 = 7.0 x 104 J/mole11.SHOW: A Schottky and Frenkel defect in MgF2 structuresA 2-D representation of the MgF
23、2 structure containing a Schottky defect and a Frenkel defect is shown below.12.Explain why the following statement is incorrect: In ionic solids the number of cation vacancies is equal to the number of anion vacancies.In ionic crystals, even in the presence of vacancies, the charge neutrality must
24、be maintained. Therefore, single vacancies do not occur in ionic crystals since removal of a single ion would lead to charge imbalance. Instead the vacancies occur in a manner such that the anion: cation vacancy ratio render the solid electrically neutral. This, however, does not mean that the anion
25、 vacancies are equal to cation vacancies. For example, a Schottky defect in MgCl2 or MgF2 involves two Cl- or F- cation vacancies for every Mg2+ anion vacancy to maintain electrical neutrality.The number of cation vacancies equals the number of anion vacancies only for the limiting case where the ch
26、emical formula of the compound is MX.13.Calculate the number of defects created when 2 moles of NiO are added to 98 moles of SiO2. Also, determine the type of defect created.GIVEN: Neglect interstial vacanciesWe have 2 moles of NiO and 98 moles of SiO2. Since NiO is a 1:1 compound there are 2 moles
27、of Ni2+ ions and 2 moles of O2- ions present. SiO2 on the other hand is a 1:2 compound; therefore, there are 98 moles of Si4+ and 196 moles of O2-. The total number of each type of ion isNNi = 2 molesNSi = 98 molesNO2 = 196 molesThe total number of moles of ions in the system is NT = NNi + NSi + NO
28、= 2 + 98 + 196 = 196 molesEach substitution of an Ni2+ for Si4+ results in a loss of 2 positive charges. If no interstitials are created, this loss of positive charge is balanced by the creation of anion vacancies. Charge neutrality requires one oxygen vacancy created for every Ni2+ ion. Therefore,
29、the number of oxygen vacancies isNOv = NNi = 2 molesThere are 2 moles of oxygen ion vacancies created with the addition of 2 moles of NiO to 98 moles of SiO2.14.Calculate the number of defects created when 1 mole of MgO is added to 99 moles of Al2O3.MgO is a 1:1 compound, therefore there is 1 mole o
30、f Mg2+ ions and 1 mole of O2- ions in the system.From Al2O3, there are 198 moles of Al3+ ions and 297 moles of O2- ions in the system.Each substitution of an Mg2+ ion for Al3+ ion results in a loss of one positive charge. This loss of positive charge is balanced by oxygen vacancy. Charge neutrality
31、requires one oxygen vacancy to be created for every two Mg2+ ion3. Therefore the number of oxygen ion vacancies created is0.5 moles of oxygen ion vacancies are created by the addition of 1 mole of MgO to 99 moles of Al2O3.15.COMPUTE: Relative concentration of cation vacancies, anion vacancies and ca
32、tion interstitials.GIVEN:QCv = 20kJ/moleQAv = 40kJ/moleQCI = 30kJ/moleASSUMPTION: assume room temperatureT = 298KConcentration of cation vacancies, CCv is given bySimilarly for anion vacanciesand for cation interstitials16.(a) Describe a Schottky defect in U2(b) Would you expect to find more cation
33、or anion Frenkel defects in this compound? Why?UO2 has a fluorite structure with U4+ ions occupying FCC lattice sites and O2- occupying tetrahedral interstitial sites.(a) A Schottky defect in UO2 will involve one U4+ cation vacancy and 2 O2- anion vacancies.(b) In general cation Frenkel defects are
34、more common than anion Frenkel defects because cations are usually smaller. In this case, the radii of U4+ is 0.106nm and that of O2- is 0.132nm. The U4+ cation is smaller than the O2- anion. However, the size difference is not very high. Still, cation Frenkel defects are expected to be more.17.Ioni
35、c compound Li2O(a) Describe a Schottky defect(b) Describe a Frenkel defectLi2O has an antifluorite structure. O2- ions occupy FCC lattice sites and Li+ occupies tetrahedral interstitial sites.(a) A Schottky defect in Li2O involves 2 Li2+ cation vacancies and one O2- anion vacancy(b) The ionic radii
36、of Li+ and O2- are 0.078nm and 0.132nm respectively. This material is most likely to exhibit cation Frenkel defect since the size of the cation is much smaller than the anion.18.DETERMINE:(a) Interstitial Na+ ions(b) Interstitial O2- ions(c) Vacant Na+ sites(d) Vacant O2- sites in Na2OGIVEN: r(Na+)
37、= 0.098nm r(O2-) = 0.132nmNa2O structure is similar to antifluorite structure. Na+ ions occupy tetrahedral interstitial sites and O2- ions occupy FCC lattice sites.Since the ratio of Na:0 is 2:1 for this materials, a Schottky defect results in 2 cation vacancies for every one anion vacancy.no. of va
38、cant Na+ sites = 2 x no. of vacant O2- sitesA cation Frenkel defect is more likely to occur in this material(a) Interstitial Na+ ions = 1(b) Interstitial O2- ions = 0(c) Vacant Na+ sites = 2(d) Vacant O2- sites = 119.SOLVENT: AuSOLUTE: N, Ag or CsDETERMINE: (a) which element is most likely to form a
39、n interstitial solid solution. (b) which element is most likely to form a substitutional solid solution.r(Au) = 0.144nmr(N) = 0.071nmr(Ag) = 0.144nmr(Cs) = 0.265nm(a) Based on atomic radii N is most likely to form are interstitial solid solution with Au as solvent.(b) Ag is most likely to form a sub
40、stitutional solid solution because the size difference between Au & N and Au & Cs is more than 15%.In addition, Au and Ag have similar valence, and crystal structure. The electronegativities are not quite similar, but since Ag-Au system obeys 3 out of 4 of the Hume-Rothery rules, Ag is the most like
41、ly element with which Au forms a substitutional solid solution.Section 4.4 Diffusion20. Under what condition can Ficks first law be used to solve diffusion problems.The Ficks first law can be used to solve diffusion problems provided the concentration gradient does not change with time.21.GIVEN: 1 w
42、t% B is added to Fe.FIND: (a) if B would be present as an interstitial impurity or substitutional impurity, (b) fraction of sites occupied by B atoms, (c) if Fe containing B were to be gas carburized, would the process be faster or slower than for Fe which has no B? Explain.r(B) = 0.097nmr(Fe) = 0.1
43、24nm(a) Based on the atomic radii B would be present as an interstitial impurity(b) amount of B present = 1 wt%As a basis of calculation assume 100gms of material.Determine the no. of moles of Fe and B present.Total no. of moles of Fe and B = 1.773 + 0.092 = 1.865 moles.Fraction of sites occupied by
44、 B atoms = = mole fraction of B = 0.049Thus, B roughly occupies 5% of the sites.(c) If Fe containing B were to be gas carburized the process would be slower than for Fe which has no B simply because the presence of B atoms already in interstitial sites leave fewer sites for interstitial C to diffuse
45、 through.22.Determine which type of diffusion would be easier(a) C in HCP Ti(b) N in BCC Ti(c) Ti in BCC Tir(C) r(Ti) so we can predict that diffusion occurs via an interstitial mechanism r(N)r(Ti). In this case the diffusion also occurs via interstitial mechanism.Ti in BCC Ti is a case of self-diff
46、usion and self-diffusion occurs via a vacancy mechanism. In general the activation energy for self diffusion is higher than interstitial mechanism because vacancy mechanism involves two steps. One is to create a vacancy and second is to promote a vacancy/atom exchange. Thus Ti in BCC Ti will be the
47、slowest.The activation energy for diffusion via interstitial mechanism is just the energy necessary to move an atom into a neighboring interstitial site. An open crystal structure, as opposed to a dense structure, should have a lower activation energy. Between BCC Ti and HCP Ti, BCC Ti has a more open structure (lower APF) than HCP Ti.Thus, N in BCC Ti diffusion would be the easiest by virtue of its lowest activation energy.23.GIVEN:C1 = 0.19 at % at surfaceC2