3.40 Sketch within a cubic unit cell the following planes ...

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

The planes called for are plotted in the cubic unit cells shown below.

3.41 Determine the Miller indices for the planes shown in the following unit cell:

Solution

For plane A we will leave the origin at the unit cell as shown; this is a (403) plane, as summarized below.

xyz

Intercepts a ∞b 2c

23

Intercepts in terms of a, b, and c 1 ∞ 2

23

Reciprocals of intercepts 20 3
2

Reduction 403

Enclosure (403)

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 12) plane, as summarized below.

Intercepts xy z
–a –b c
Intercepts in terms of a, b, and c
Reciprocals of intercepts –1 –1 2
Reduction –1 –1 1
Enclosure 2
(not necessary)
(1 12) 2

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 Crystal Structure Electronegativity Valence
Cu (nm) FCC 1.9 +2
C 0.1278
H 0.071 FCC 1.9 +1
O 0.046 FCC 1.5 +3
Ag 0.060 HCP 1.8 +2
Al 0.1445 BCC 1.6 +3
Co 0.1431 BCC 1.8 +2
Cr 0.1253 FCC 1.8 +2
Fe 0.1249 FCC 2.2 +2
Ni 0.1241 FCC 2.2 +2
Pd 0.1246 HCP 1.6 +2
Pt 0.1376
Zn 0.1387
0.1332

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

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%, 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.

Element ∆R% Crystal ∆Electro- Valence
Structure negativity 2+
Cu –44
C –64 FCC
H –53
O +13 FCC 0 1+
Ag

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+

(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 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.

4.9 Calculate the composition, in weight percent, of an alloy that contains 218.0 kg titanium, 14.6 kg of aluminum,
and 9.7 kg of vanadium.

Solution

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 (CTi) is just

CTi = mTi × 100
mTi + mAl + mV

= 218 kg × 100 = 89.97 wt%
218 kg + 14.6 kg + 9.7 kg

Similarly, for aluminum

CAl = 14.6 kg × 100 = 6.03 wt%
218 kg + 14.6 kg + 9.7 kg

And for vanadium

CV = 9.7 kg × 100 = 4.00 wt%
218 kg + 14.6 kg + 9.7 kg

4.10 What is the composition, in atom percent, of an alloy that contains 98 g tin and 65 g of lead?

Solution

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

nmSn = mS' n = 98 g = 0.826 mol
ASn 118.71 g / mol

Likewise, for Pb

nmPb = 65 g = 0.314 mol
207.2 g / mol

Now, use of Equation 4.5 yields

CS'n = nmSn × 100
nmSn + nmPb

= 0.826 mol × 100 = 72.5 at%
0.826 mol + 0.314 mol

Also,

CP' b = = 0.314 mol × 100 = 27.5 at%
0.826 mol + 0.314 mol

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.