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 conductivity 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 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 this 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 = 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 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 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 using 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. However, 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: 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 electronegativities 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 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 electronegativity 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 extensive 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 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 entire compositional range since it obeys only 3 of 4 Hume-Rothery rules.(i)Ni in FeSize r(Ni) = 0.125nm, r(Fe) = 0.124nm differ