Collister Chap 04 Solns

Collister Chap 04 Solns - CHAPTER 4 IMPERFECTIONS IN SOLIDS...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS 4.1 In order to compute the fraction of atom sites that are vacant in lead at 600 K, we must employ Equation (4.1). As stated in the problem, Q v = 0.55 eV/atom. Thus, N V N = exp - Q V kT = exp - 0.55 eV / atom 8.62 x 10- 5 eV / atom-K ( 29 (600 K) = 2.41 x 10-5 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 A Au exp - Q V kT = 6.023 x 10 23 atoms/ mol ( 29 19.32 g / cm 3 ( 29 196.9 g / mol exp - 0.98 eV / atom 8.62 x 10- 5 eV / atom- K ( 29 (1173 K) = 3.65 x 10 18 cm-3 = 3.65 x 10 24 m-3 4.3 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 ), Avogadro's number ( N A ), and the atomic weight ( A ) according to Equation (4.2) as N = N A ρ Ag A Ag = 6.023 x 10 23 atoms/ mol ( 29 10.49 g / cm 3 ( 29 107.87 g / mol 57 = 5.86 x 10 22 atoms/cm 3 = 5.86 x 10 28 atoms/m 3 Now, taking natural logarithms of both sides of Equation (4.1), and, after some algebraic manipulation - N N V ln kT = Q V = - 8.62 x 10-5 eV/atom- K ( 29 (1073 K) ln 3.60 x 10 23 m- 3 5.86 x 10 28 m- 3 = 1.11 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 Cu 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 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+ 58 (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+....
View Full Document

This homework help was uploaded on 04/17/2008 for the course ENGR 106 taught by Professor Vanantwerp during the Spring '06 term at Calvin.

Page1 / 28

Collister Chap 04 Solns - CHAPTER 4 IMPERFECTIONS IN SOLIDS...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online