HW 4 Solution - Spring 2018 Homework 4 Answers, Professor: Dr. Cogan PDF

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Spring 2018 Homework 4 Answers, Professor: Dr. Cogan...

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Impurities in Solids 4.6 Atomic radius, crystal structure, electronegativity, and the most common valence are given in the following table for several elements; for those that are nonmetals, only atomic radii are indicated. Element

Atomic Radius (nm)

Crystal Structure

Electronegativity

Valence

Ni C H O Ag Al Co Cr Fe Pt Zn

0.1246 0.071 0.046 0.060 0.1445 0.1431 0.1253 0.1249 0.1241 0.1387 0.1332

FCC

1.8

+2

FCC FCC HCP BCC BCC FCC HCP

1.4 1.5 1.7 1.6 1.7 1.5 1.7

+1 +3 +2 +3 +2 +2 +2

Which of these elements would you expect to form the following with nickel: (a) A substitutional solid solution having complete solubility (b) A substitutional solid solution of incomplete solubility (c) An interstitial solid solution

Solution For complete substitutional solubility the four Hume-Rothery rules must be satisfied: (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.

Element

R%

Ni C H O Ag Al Co Cr Fe Pt Zn

–43 –63 –52 +16 +15 +0.6 +0.2 -0.4 +11 +7

Crystal Structure

Electronegativity

FCC

FCC FCC HCP BCC BCC FCC HCP

Valence 2+

-0.4 -0.3 -0.1 -0.2 -0.1 -0.3 -0.1

1+ 3+ 2+ 3+ 2+ 2+ 2+

(a) Pt is the only element that meets all of the criteria and thus forms a substitutional solid solution 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 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 Ni.

4.7 Which of the following systems (i.e., pair of metals) would you expect to exhibit complete solid solubility? Explain your answers. (a) Cr-V (b) Mg-Zn (c) Al-Zr (d) Ag-Au (e) Pb-Pt

Solution In order for there to be complete solubility (substitutional) for each pair of metals, the four Hume-Rothery rules must be satisfied: (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.

(a) A comparison of these four criteria for the Cr-V system is given below: Metal

Atomic Radius (nm)

Crystal Structure

Electronegativity

Valence

Cr

0.125

BCC

1.6

+3

V

0.132

BCC

1.5

+5 (+3)

For chromium and vanadium, the percent difference in atomic radii is approximately 6%, the crystal structures are the same (BCC), and there is very little difference in their electronegativities. The most common valence for Cr is +3; although the most common valence of V is +5, it can also exist as +3. Therefore, chromium and vanadium are completely soluble in one another.

(b) A comparison of these four criteria for the Mg-Zn system is given below: Metal

Atomic Radius (nm)

Crystal Structure

Electronegativity

Valence

Mg

0.160

HCP

1.3

+2

Zn

0.133

HCP

1.7

+2

For magnesium and zinc, the percent difference in atomic radii is approximately 17%, the crystal structures are the same (HCP), and there is some difference in their electronegativities (1.3 vs. 1.7). The most common valence for both Mg and Zn is +2. Magnesium and zinc are not completely soluble in one another, primarily because of the difference in atomic radii.

(c) A comparison of these four criteria for the Al-Zr system is given below: Metal

Atomic Radius (nm)

Crystal Structure

Electronegativity

Valence

Al

0.143

FCC

1.5

+3

Zr

0.159

HCP

1.2

+4

For aluminum and zirconium, the percent difference in atomic radii is approximately 11%, the crystal structures are different (FCC and HCP), there is some difference in their electronegativities (1.5 vs. 1.2). The most common valences for Al and Zr are +3 and +4, respectively. Aluminum and zirconium are not completely soluble in one another, primarily because of the difference in crystal structures.

(d) A comparison of these four criteria for the Ag-Au system is given below:

Metal

Atomic Radius (nm)

Crystal Structure

Electronegativity

Valence

Ag

0.144

FCC

1.4

+1

Au

0.144

FCC

1.4

+1

For silver and gold, the atomic radii are the same, the crystal structures are the same (FCC), their electronegativities are the same (1.4), and their common valences are +1. Silver and gold are completely soluble in one another because all four criteria are satisfied.

(d) A comparison of these four criteria for the Pb-Pt system is given below: Metal

Atomic Radius (nm)

Crystal Structure

Electronegativity

Valence

Pb

0.175

FCC

1.6

+2

Pt

0.139

FCC

1.5

+2

For lead and platinum, the percent difference in atomic radii is approximately 20%, the crystal structures are the same (FCC), their electronegativities are nearly the same (1.6 vs. 1.5), and the most common valence for both of them is +2. Lead and platinum are not completely soluble in one another, primarily because of the difference in atomic radii.

4.8 (a) Compute the radius r of an impurity atom that will just fit into an FCC octahedral site in terms of the atomic radius R of the host atom (without introducing lattice strains). (b) Repeat part (a) for the FCC tetrahedral site. (Note: You may want to consult Figure 4.3a.)

Solution (a) In the drawing below is shown the atoms on the (100) face of an FCC unit cell; the small circle represents an impurity atom that just fits within the octahedral interstitial site that is located at the center of the unit cell edge.

The diameter of an atom that will just fit into this site (2r) 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

; therefore, solving for r from the above

equation gives

(b) Drawing (a) below shows one quadrant of an FCC unit cell, which is a cube; corners of the tetrahedron correspond to atoms that are labeled A, B, C, and D. These corresponding atom positions are noted in the FCC unit cell in drawing (b). In both of these drawings, atoms have been reduced from their normal sizes for clarity. The interstitial atom resides at the center of the tetrahedron, which is designated as point E, in both (a) and (b).

Let us now express the host and interstitial atom radii in terms of the cube edge length, designated as a. From Figure (a), the spheres located at positions A and B touch each other along the bottom face diagonal. Thus,

But

or

And

There will also be an anion located at the corner, point F (not drawn), and the cube diagonal

will be related to

the atomic radii as

(The line AEF has not been drawn to avoid confusion.) From the triangle ABF (P4.8a) But,

and

from above. Substitution of the parameters involving r and R noted above into Equation P4.8a leads to the following:

Solving for r leads to...