Imperfections in Solids Problem Solutions Part 6

4.33 Sometimes it is desirable to determine the weight percent of one element, C 1, that will produce a specified concentration in terms of the number of atoms per cubic centimeter, N 1, for an alloy composed of two types of atoms. This computation is possible using the following expression: C1 = 100 NA ρ2 ρ 2 1+ − N1A1 ρ 1 (4.22) where NA is Avogadro’s number, ρ 1 and ρ2 are the densities of the two elements, and A 1 is the atomic weight of element 1. Derive Equation 4.22 using Equation 4.2 and expressions contained in Section 4.4. Solution This problem asks that we derive Equation 4.22, 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 (c1′ ) times the total number of atoms per cubic centimeter in the alloy ( N). Thus, using the equivalent of Equation 4.2 for N, we may write N1 = c1′ N = Realizing that c1′ = c′NA ρave Aave C1′ 100 and C 2′ = 100 − C1′ and substitution of the expressions for ρave and Aave, Equations 4.10b and 4.11b, respectively, leads to N1 = c1′ NA ρ ave Aave    C′ A + C′2A2 ÷ (c1′ NA )  1 1 C A C′ A ÷  1′ 1 + 2 2 ÷ ρ2   ρ1 = C′A1 + C2′ A2 100 = And, solving for C1′ N A C1′ ρ1 ρ 2 C1′ ρ2 A1 + (100 − C1′ ) ρ1 A2 leads to C1′ = Substitution of this expression for C1′ 100 N1 ρ1 A2 N Aρ1 ρ2 − N1ρ 2 A1 + N1 ρ1 A2 into Equation 4.7a, which may be written in the following form   100 N1 ρ 1 A2 ÷ A1   NA ρ 1 ρ2 − N1 ρ2 A1 + N1 ρ1 A2  C1 = × 100      100 N1 ρ 1A2 100 N1ρ 1 A2   ÷ A1 + 100 − ÷ A2    N Aρ1 ρ 2 − N1 ρ2 A1 + N1 ρ1 A2   NAρ1 ρ2 − N1 ρ 2 A1+ N1 ρ1 A2   which simplifies to C1 = the desired expression. 100 NA ρ 2 ρ 2 − 1+ N1 A1 ρ1 4.34 Gold (Au) forms a substitutional solid solution with silver (Ag). Compute the weight percent of gold that must be added to silver to yield an alloy that contains 5.5 × 10 21 Au atoms per cubic centimeter. The densities of pure Au and Ag are 19.32 and 10.49 g/cm3, respectively. Solution To solve this problem, employment of Equation 4.22 is necessary, using the following values: N1 = NAu = 5.5 × 1021 atoms/cm3 ρ1 = ρ Au = 19.32 g/cm3 ρ2 = ρ Ag = 10.49 g/cm 3 A1 = AAu = 196.97 g/mol (found inside front cover) A2 = AAg = 107.87 g/mol (found inside from cover) Thus CAu = 1+ = 100 N A ρAg N Au AAu − ρ Ag ρ Au 100 ( 6.022 × 1+ 1023 atoms/mol)(10.49 g/cm 3) ( 5.5 × 1021 atoms/cm 3 )(196.97 g/mol) = 15.9 wt% 10.49 g/cm 3  − ÷ 19.32 g/cm 3  4.35 Germanium (Ge) forms a substitutional solid solution with silicon (Si). Compute the weight percent of germanium that must be added to silicon to yield an alloy that contains 2.43 × 10 21 Ge atoms per cubic centimeter. The densities of pure Ge and Si are 5.32 and 2.33 g/cm3, respectively. Solution To solve this problem, employment of Equation 4.22 is necessary, using the following values: N1 = NGe = 2.43 × 1021 atoms/cm3 ρ1 = ρ Ge = 5.32 g/cm3 ρ2 = ρ Si = 2.33 g/cm3 A1 = AGe = 72.64 g/mol (found inside front cover) A2 = ASi = 28.09 g/mol (found inside front cover) Thus C Ge = = 100 NA ρSi ρ − Si 1+ NGe AGe ρ Ge 100 1023 atoms/mol )(2.33 g/cm3 )  2.33 g/cm 3  − ÷ (2.43 × 10 21 atoms/cm 3) (72.64 g/mol)  5.32 g/cm 3  ( 6.022 × 1+ = 11.7 wt% 4.36 Electronic devices found in integrated circuits are composed of very high purity silicon to which has been added small and very controlled concentrations of elements found in Groups IIIA and VA of the periodic table. For Si to which has had added 6.5 × 1021 atoms per cubic meter of phosphorus, compute (a) the weight percent and (b) the atom percent of P present. Solution (a) This part of the problem is solved using Equation 4.22, in which C1 = concentration of P in wt% N1 = number of P atoms per cubic centimeter = (6.5 × 1021 atoms/m3)(1 m3/106 cm3) = 6.5 × 1015 atoms/cm3 A1 = atomic weight of P = 30.97 g/mol (inside front cover) ρ 1 = density of P = 1.82 g/cm3 (inside front cover) ρ 2 = density of silicon = 2.33 g/cm3 (inside front cover) And, from Equation 4.22 C1 = 100 NA ρ 2 ρ 2 − 1+ N1 A1 ρ 1 100 = 1+ (6.022 × 1023 atoms/mol)(2.33 g/cm 3) (6.5 × 1015 atoms/cm 3)(30.97 g/mol) − 2.33 g/cm 3 1.82 g/cm3 = 1.43 × 10−5 wt% (b) To convert from weight percent to atom percent we use Equation 4.6a where C ′1 = atom percent of P C1 = weight percent of P = 1.43 × 10−5 wt% C2 = weight percent of Si = 100 wt% − 1.43 × 10−5 wt% = 99.9999857 A1 = atomic weight of P = 30.97 g/mol A2 = atomic weight of Si = 28.09 g/mol And from Equation 4.6a we obtain the following: C1′ = = (1.43 × 10 −5 C1A2 × 100 C1A2 + C2 A1 (1.43 × 10 −5 wt%)(28.09 g/mol) wt%)(28.09 g/mol) + (99.9999857 wt%)(30.97 g/mol) = 1.30 × 10−5 at% × 100 4.37 Iron and vanadium both have the BCC crystal structure, and V forms a substitutional solid solution for concentrations up to approximately 20 wt% V at room temperature. Compute the unit cell edge length for a 90 wt% Fe–10 wt% V alloy. Solution This problems asks that we compute the unit cell edge length for a 90 wt% Fe-10 wt% V alloy. First of all, the atomic radii for Fe and V (using the table inside the front cover) are 0.124 and 0.132 nm, respectively. Also, using Equation 3.8 it is possible to compute the unit cell volume, and inasmuch as the unit cell is cubic, the unit cell edge length is just the cube root of the volume. However, it is first necessary to calculate the density and average atomic weight of this alloy using Equations 4.10a and 4.11a. Inasmuch as the densities of iron and vanadium are 7.87g/cm3 and 6.10 g/cm3, respectively, (as taken from inside the front cover), the average density is just ρ ave = = 100 C + Fe ρ V ρFe CV 100 10 wt% 6.10 g/cm3 + 90 wt% 7.87 g/cm3 = 7.65 g/cm3 And for the average atomic weight Aave = = 100 CV CFe + AV AFe 100 10 wt% 90 wt% + 50.94 g/mol 55.85 g/mol = 55.32 g/mol Now, VC is determined from Equation 3.8 as (taking the number of atoms per unit cell, n = 2, since the crystal structure of the alloy is BCC) VC = = nAave ρ aveN A (2 atoms/unit cell)(55.32 g/mol) (7.65 g/cm3 )(6.022 × 10 23 atoms/mol) − = 2.40 × 10 23 cm3/unit cell And, finally, because the unit cell is of cubic symmetry a = (VC )1/3 = (2.40 × 10 −23 cm 3 /unit cell) 1/3 − = 2.89 × 10 8 cm = 0.289 nm