3.40 Sketch within a cubic unit cell the following planes: (a) (01 1 ) (b) (112 ) (c) (102 ) (d) (13 1) Solution

Transcription

1 3.40 Sketch within a cubic unit cell the following planes: (a) (01 1 ) (b) (11 ) (c) (10 ) (d) (13 1) The planes called for are plotted in the cubic unit cells shown below.

2 3.41 Determine the Miller indices for the planes shown in the following unit cell: For plane A we will leave the origin at the unit cell as shown; this is a (403) plane, as summarized below. Intercepts Intercepts in terms of a, b, and c x y z Reciprocals of intercepts 0 Reduction Enclosure (403) a 1 b c For plane B we will move the origin of the unit cell one unit cell distance to the right along the y axis, and one unit cell distance parallel to the x axis; thus, this is a (1 1 ) plane, as summarized below. x y z Intercepts a b c Intercepts in terms of a, b, and c Reciprocals of intercepts 1 1 Reduction (not necessary) Enclosure (1 1)

3 4.4 Below, atomic radius, crystal structure, electronegativity, and the most common valence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated. Element Atomic Radius (nm) Crystal Structure Electronegativity Valence Cu FCC C H O Ag FCC Al FCC Co HCP Cr BCC Fe BCC Ni FCC Pd FCC. + Pt FCC. + Zn HCP Which of these elements would you expect to form the following with copper: (a) A substitutional solid solution having complete solubility (b) A substitutional solid solution of incomplete solubility (c) An interstitial solid solution 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%, ) 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 + C 44 H 64 O 53 Ag +13 FCC 0 1+

4 Al +1 FCC Co - HCP Cr - 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. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal structure, and thus display complete solid solubility at these temperatures. (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 +. (c) C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of Cu. 4.9 Calculate the composition, in weight percent, of an alloy that contains 18.0 kg titanium, 14.6 kg of aluminum, and 9.7 kg of vanadium. The concentration, in weight percent, of an element in an alloy may be computed using a modified form of Equation 4.3. For this alloy, the concentration of titanium (C Ti ) is just C Ti m Ti m Ti + m Al + m V kg 18 kg kg kg wt% Similarly, for aluminum C Al 14.6 kg 18 kg kg kg wt% And for vanadium

5 C V 9.7 kg 18 kg kg kg wt%

6 4.10 What is the composition, in atom percent, of an alloy that contains 98 g tin and 65 g of lead? The concentration of an element in an alloy, in atom percent, may be computed using Equation 4.5. However, it first becomes necessary to compute the number of moles of both Sn and Pb, using Equation 4.4. Thus, the number of moles of Sn is just n msn m Sn ' A Sn 98 g g /mol 0.86 mol Likewise, for Pb n mpb 65 g 07. g /mol mol Now, use of Equation 4.5 yields C ' Sn n msn n msn + n mpb mol 0.86 mol mol at% Also, C ' Pb mol 0.86 mol mol at% Additional questions: What directions have the greatest linear density in FCC and in BCC? Which of these is higher linear density? What planes of the greatest planar density in FCC and BCC? Which of these has higher planar density? Both FCC and BCC, and in fact all crystal structures, contain a direction in which atoms are touching. This is the highest possible linear direction. So both FCC and BCC have the highest possible. For FCC this is the <110> family of directions and for BCC it is the <111> family Any plane that contains the hex arrangement (both HCP and FCC do) is the highest possible planer density. BCC does not have this sort of arrangement. Therefore, the greatest planer density in FCC is greater than that in BCC. In FCC it is the {111} family of planes. The greatest planer density in BCC is the {110} family.