Imperfections in Solids Problem Solutions Part 9

DESIGN PROBLEMS Specification of Composition 4.D1 Aluminum–lithium (Al-Li) 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 2.47 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 = 100 100 = CLi CAl CLi 100 − C Li + + ρ Li ρ Al ρLi ρ Al inasmuch as CLi + CAl = 100. According to the table inside the front cover, the respective densities of Li and Al are 0.534 and 2.71 g/cm3. Upon algebraic manipulation of the above equation and solving for CLi leads to C Li = 100 ρ Li (ρ Al − ρave ) ρave ( ρAl − ρ Li ) Incorporation of values for ρave, ρ Al, and ρLi yields the following value for CLi: C Li= (100) ( 0.534 g/cm3)( 2.71 g/cm 3 − 2.47 g/cm 3) ( 2.47 g/cm3)(2.71 g/cm3 − 0.534 g/cm3 ) = 2.38 wt% 4.D2 Copper (Cu) and platinum (Pt) both have the FCC crystal structure, and Cu forms a substitutional solid solution for concentrations up to approximately 6 wt% Cu at room temperature. Determine the concentration in weight percent of Cu that must be added to Pt to yield a unit cell edge length of 0.390 nm. Solution To begin, it is necessary to employ Equation 3.8, and solve for the unit cell volume, VC, as VC = nAave ρ aveN A where Aave and ρave are the atomic weight and density, respectively, of the Pt-Cu alloy. Inasmuch as both of these materials have the FCC crystal structure, which has cubic symmetry, VC is just the cube of the unit cell length, a. That is VC = a 3 = (0.390 nm)3 = ( 3.90 × 10 −8 cm )3 = 5.932 × 10 −23 cm 3 It is now necessary to construct expressions for Aave and ρave in terms of the concentration of copper, CCu, using Equations 4.11a and 4.10a. For Aave we have Aave = = 100 CCu (100 − C Cu) + ACu APt 100 CCu 63.55 g/mol + (100 − CCu ) 195.08 g/mol since the atomic weights for Cu and Pt are, respectively, 63.55 and 195.08 g/mol (as noted inside the front cover). Now, the expression for ρ is as follows: ave ρ ave = 100 CCu ρ Cu + (100 − C Cu) ρ Pt = 100 (100 − C Cu ) + 8.94 g/cm3 21.45 g/cm3 CCu given the densities of 8.94 and 21.45 g/cm3 for the respective metals. Within the FCC unit cell there are 4 equivalent atoms, and thus, the value of n in Equation 3.8 is 4; hence, the expression for VC may be written in terms of the concentration of Cu in weight percent as follows: VC = 5.932 × 10 −23 cm3 = nAave ρave NA     100   (4 atoms/unit cell) CCu (100 − CCu )   +  63.55 g/mol 195.08 g/mol    =     100   (6.022 × 1023 atoms/mol)  CCu ( 100 − CCu )  +    8.94 g/cm 3 21.45 g/cm3  And solving this expression for CCu leads to CCu = 2.825 wt%. FUNDAMENTALS OF ENGINEERING QUESTIONS AND PROBLEMS 4.1FE Calculate the number of vacancies per cubic meter at 1000°C for a metal that has an energy for vacancy formation of 1.22 eV/atom, a density of 6.25 g/cm3, and an atomic weight of 37.4 g/mol. (A) 1.49 × 1018 m−3 (B) 7.18 × 1022 m−3 (C) 1.49 × 1024 m− 3 (D) 2.57 × 1024 m −3 Solution Determination of the number of vacancies per cubic meter at 1000 °C (1273 K) requires the utilization of Equations 4.1 and 4.2 as follows:  Q  N ρ  Q  Nv = N exp  − v÷ = A exp  − v ÷  kT   kT  A Incorporation into this expression values for Qv, ρ , and A provided in the problem statement yields the following for Nv: Nv =   1.22 eV/atom (6.022 × 1023atoms/mol)(6.25 g/cm3 ) exp −  −5 37.4 g/mol eV/atom-K)(1273 K) (8.62 × 10   = 1.49 × 1018 cm−3 = 1.49 × 1024 m−3 which is answer C. 4.2FE What is the composition, in atom percent, of an alloy that consists of 4.5 wt% Pb and 95.5 wt% Sn? The atomic weights for Pb and Sn are 207.19 g/mol and 118.71 g/mol, respectively. (A) 2.6 at% Pb and 97.4 at% Sn (B) 7.6 at% Pb and 92.4 at% Sn (C) 97.4 at% Pb and 2.6 at% Sn (D) 92.4 at% Pb and 7.6 at% Sn Solution We are asked to compute the composition of an alloy in atom percent. Employment of Equation 4.6a leads to C ′Pb= = CPb ASn C PbASn + CSn APb × 100 (4.5)(118.71 g/mol) × 100 (4.5)(118.71 g/mol) + (95.5)(207.19 g/mol) = 2.6 at% C Sn ′ = = CSn APb × 100 CSn APb + CPb ASn (95.5)(207.19 g/mol) ×100 (95.5)(207.19 g/mol) + (4.5)(118.71 g/mol) = 97.4 at% which is answer A (2.6 at% Pb and 97.4 at% Sn). 4.3FE What is the composition, in weight percent, of an alloy that consists of 94.1 at% Ag and 5.9 at% Cu? The atomic weights for Ag and Cu are 107.87 g/mol and 63.55 g/mol, respectively. (A) 9.6 wt% Ag and 90.4 wt% Cu (B) 3.6 wt% Ag and 96.4 wt% Cu (C) 90.4 wt% Ag and 9.6 wt% Cu (D) 96.4 wt% Ag and 3.6 wt% Cu Solution To compute composition, in weight percent, of a 94.1 at% Ag-5.9 at% Cu alloy, we employ Equation 4.7 as follows: C Ag = = C′AgAAg × 100 C′Ag AAg + CCu ′ ACu (94.1)(107.87 g/mol) × 100 (94.1)(107.87 g/mol) + (5.9)(63.55 g/mol) = 96.4 wt% C Cu = = CCu ′ ACu × 100 ′ AAg + CCu ′ ACu CAg (5.9)(63.55 g/mol) × 100 (94.1)(107.87 g/mol) + (5.9)(63.55 g/mol) = 3.6 wt% which is answer D (96.4 wt% Ag and 3.6 wt% Cu).