Homework #4 PROBLEM SOLUTIONS

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1 Homework #4 PROBLEM SOLUTIONS 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: N V N exp Q V kt N A ρ Au exp Q V A Au kt ( x atoms/mol)19.32 g / cm g /mol ( ) exp 0.98 ev / atom ( 8.62 x 10 5 ev /atom K) (1173 K) 3.65 x cm x m This problem calls for the computation of the energy for vacancy formation in silver. 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), Avogadros number (N A ), and the atomic weight (A) according to Equation (4.2) as N N A ρ Ag ( x atoms /mol)10.49 ( g /cm 3 ) g /mol 5.86 x atoms/cm x atoms/m 3 Now, taking natural logarithms of both sides of Equation (4.1), and, after some algebraic manipulation Q V RT ln N V N ( 8.62 x 10-5 ev/atom - K)(1073 K) ln 3.60 x m x m ev/atom 57

2 4.4 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 (ΔR%) 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 ΔElectro- Element ΔR% Structure negativity Valence Cu FCC 2+ C -44 H -64 O -53 Ag +13 FCC 0 1+ Al +12 FCC Co -2 HCP Cr -2 BCC Fe -3 BCC Ni -3 FCC Pd +8 FCC Pt +9 FCC Zn +4 HCP (a) Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions having complete solubility. (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. 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 58

3 C Ag C Ag A Cu C Ag A Cu + C Cu (92.5)(63.55 g /mol) (92.5)(63.55 g/ mol) + (7.5)(107.87g /mol) 87.9 at% C Cu C Cu C Ag A Cu + C Cu (7.5)( g /mol) (92.5)(63.55 g/ mol) + (7.5)(107.87g /mol) 12.1 at% 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 C Cu C A Cu Cu C A + C A Cu Cu Pt Pt (5)(63.55 g /mol) (5)(63.55 g /mol) + (95)( g /mol) 1.68 wt% C Pt C A Pt Pt C A + C A Cu Cu Pt Pt (95)( g /mol) (5)(63.55 g /mol) + (95)( g /mol) wt% 59

4 4.12 We are asked to compute the composition of an alloy in atom percent. Employment of Equation (4.6) leads to C Pb C Pb A Sn C Pb A Sn + C Sn A Pb 5.5( g / mol) 5.5 ( g /mol) (207.2 g /mol) 3.2 at% C Sn C Sn A Pb C Sn A Pb + C Pb A Sn 94.5 (207.2 g / mol) 94.5(207.2 g /mol) + 5.5( g /mol) 96.8 at% 4.14 This problem calls for a determination of the number of atoms per cubic meter for aluminum. 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 g/mol. Thus, N ( x atoms/mol)2.71 ( g /cm 3 ) g /mol 6.05 x atoms/cm x atoms/m 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 (4.17) is necessary, using the following values: C 1 C Mo 16.4 wt% 60

5 ρ 1 ρ Mo g/cm 3 ρ 2 ρ W 19.3 g/cm 3 A 1 A Mo g/mol Thus N Mo C Mo A Mo ρ Mo N A C Mo + A Mo ρ W ( C Mo ) ( x atoms/mol) (16.4) (16.4)(95.94 g / mol) (10.22 g / cm 3 ) g / mol g /cm3 ( ) 1.73 x atoms/cm This problem asks us to determine the weight percent of Ge that must be added to Si such that the resultant alloy will contain 2.43 x10 21 Ge atoms per cubic centimeter. To solve this problem, employment of Equation (4.18) is necessary, using the following values: N 1 N Ge 2.43 x atoms/cm 3 ρ 1 ρ Ge 5.32 g/cm 3 ρ 2 ρ Si 2.33 g/cm 3 A 1 A Ge g/mol A 2 A Si g/mol Thus C Ge N A ρ Si N Ge A Ge ρ Si ρ Ge ( x10 23 atoms/ mol)(2.33 g/cm 3 ) ( 2.43 x10 21 atoms / cm 3 )(72.59 g /mol) 2.33 g/ cm g/ cm 3 61

6 11.7 wt% 4.21 This problem asks us to determine the weight percent of Au that must be added to Ag such that the resultant alloy will contain 5.5 x Au atoms per cubic centimeter. To solve this problem, employment of Equation (4.18) is necessary, using the following values: N 1 N Au 5.5 x atoms/cm 3 ρ 1 ρ Au g/cm 3 ρ 2 ρ Ag g/cm 3 A 1 A Au g/mol A 2 Ag g/mol Thus C Au N A ρ Ag N Au A Au ρ Ag ρ Au ( x atoms /mol) (10.49 g /cm3 ) (5.5 x atoms / cm 3 )( g /mol) g / cm g / cm wt% 4.30 (a) This problem calls for a determination of the average grain size of the specimen which microstructure is shown in Figure 4.12b. Seven line segments were drawn across the micrograph, each of which was 60 mm long. The average number of grain boundary intersections for these lines was 8.7. Therefore, the average line length intersected is just 60 mm mm Hence, the average grain diameter, d, is d ave. line length int ersec ted magnification 6.9 mm 6.9 x 10 2 mm 62

7 (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 x according to Equation Inasmuch as the magnification is x, the value of N is measured directly from the micrgraph, which is approximately 12 grains. Rearranging Equation 4.16 and solving for n leads to n log N log log 12 log