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This problem calls for the computation of the activation energy for vacancy formation in silver.Upon examination of Equation (4.1), all parameters besides Qvare given except N, the totalnumber of atomic sites. However, Nis related to the density, (ρ), Avogadro's number (NA), andthe atomic weight (A) according to Equation (4.2) asN = NAρPbAPb= (6.023 x 1023atoms/mol)(9.5 g/cm3)107.9 g/mol= 5.30 x 1022atoms/cm3= 5.30 x 1028atoms/m3
49Now, taking natural logarithms of both sides of Equation (4.1), and, after some algebraicmanipulationQV= - RT ln NVN= - (8.62 x 10-5eV/atom-K)(1073 K) ln 3.60 x 1023m-35.30 x 1028m-3= 1.10 eV/atom4.4 In this problem we are asked to cite which of the elements listed form with Cu the three possiblesolid solution types. For complete substitutional solubility the following criteria must be met: 1)the difference in atomic radii between Ni and the other element (∆R%) must be less than ±15%,2) the crystal structures must be the same, 3) the electronegativities must be similar, and 4) thevalences should be the same, or nearly the same. Below are tabulated, for the variouselements, these criteria.Crystal∆Electro-Element∆R%StructurenegativityValenceCuFCC2+C-44H-64O-53Ag+13FCC01+Al+12FCC-0.43+Co-2HCP-0.12+Cr-2BCC-0.33+Fe-3BCC-0.12+Ni-3FCC-0.12+Pd+8FCC+0.32+Pt+9FCC+0.32+Zn+4HCP-0.32+
50(a) Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions havingcomplete solubility.(b) Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. Allthese metals have either BCC or HCP crystal structures, and/or the difference between theiratomic radii and that for Ni 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 aresignificantly smaller than the atomic radius of Cu.4.5 In the drawing below is shown the atoms on the (100) face of a FCC unit cell; the interstitial siteis at the center of the edge.RR2raThe diameter of an atom that will just fit into this site (2r) is just the difference between that unitcell edge length (a) and the radii of the two host atoms that are located on either side of the site(R); that is2r = a - 2RHowever, for FCC ais related to Raccording to Equation (3.1) as a= 2R√f8e52; therefore, solvingfor rgivesr = a - 2R2= 2R√f8e52 - 2R2= 0.41RA (100) face of a BCC unit cell is shown below.
51RR + ra/4a/2The interstitial atom that just fits into this interstitial site is shown by the small circle. It is situatedin the plane of this (100) face, midway between the two vertical unit cell edges, and one quarterof the distance between the bottom and top cell edges. From the right triangle that is definedby the three arrows we may write( )a22+ ( )a42= ()R + r2However, from Equation (3.3), a= 4R√f8e53, and, therefore, the above equation takes the form4R2√f8e532+ 4R4√f8e532= R2+ 2Rr + r2After rearrangement the following quadratic equation results:r2+ 2Rr - 0.667R 2= 0And upon solving for r, r= 0.291R.