Upon examination of equation 41 all parameters

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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 Q v are given except N , the total number of atomic sites. However, N is related to the density, ( ρ ), Avogadro's number ( N A ), and the atomic weight ( A ) according to Equation (4.2) as N = N A ρ Pb A Pb = (6.023 x 10 23 atoms/mol)(9.5 g/cm 3 ) 107.9 g/mol = 5.30 x 10 22 atoms/cm 3 = 5.30 x 10 28 atoms/m 3
49 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 10 23 m -3 5.30 x 10 28 m -3 = 1.10 eV/atom 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 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) the valences should be the same, or nearly the same. Below are tabulated, for the various elements, these criteria. Crystal Electro- Element R% Structure negativity Valence Cu FCC 2+ C -44 H -64 O -53 Ag +13 FCC 0 1+ Al +12 FCC -0.4 3+ Co -2 HCP -0.1 2+ Cr -2 BCC -0.3 3+ Fe -3 BCC -0.1 2+ Ni -3 FCC -0.1 2+ Pd +8 FCC +0.3 2+ Pt +9 FCC +0.3 2+ Zn +4 HCP -0.3 2+
50 (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 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 are significantly 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 site is at the center of the edge. R R 2r a The diameter of an atom that will just fit into this site (2 r ) is just the difference between that unit cell edge length ( a ) and the radii of the two host atoms that are located on either side of the site ( R ); that is 2r = a - 2R However, for FCC a is related to R according to Equation (3.1) as a = 2 R f8e5 2; therefore, solving for r gives r = a - 2R 2 = 2R f8e5 2 - 2R 2 = 0.41R A (100) face of a BCC unit cell is shown below.
51 R R + r a/4 a/2 The interstitial atom that just fits into this interstitial site is shown by the small circle. It is situated in the plane of this (100) face, midway between the two vertical unit cell edges, and one quarter of the distance between the bottom and top cell edges. From the right triangle that is defined by the three arrows we may write ( ) a 2 2 + ( ) a 4 2 = ( ) R + r 2 However, from Equation (3.3), a = 4 R f8e5 3 , and, therefore, the above equation takes the form 4R 2 f8e5 3 2 + 4R 4 f8e5 3 2 = R 2 + 2Rr + r 2 After rearrangement the following quadratic equation results: r 2 + 2Rr - 0.667R 2 = 0 And upon solving for r , r = 0.291 R .

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