Homework Solutions (Assignment 3) PDF

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Homework Solutions (Assignment 3)...

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BG1005 Materials Sciences_ Homework #3_Solutions Question 1. (Callister 10.4) (a) For the solidification of nickel, calculate the critical radius r* and the activation free energy ∆G* if nucleation is homogeneous. Values for the latent heat of fusion and surface free energy are -2.53 x 109 J/m3 and 0.255 J/m2, respectively. Use the supercooling value found in Table 10.1. Solution. This portion of the problem asks that we compute r* and ΔG* for the homogeneous nucleation of the solidification of Ni. First of all, Equation 10.6 is used to compute the critical radius. The melting temperature for nickel, found inside the front cover is 1455°C; also values of ΔHf (–2.53 x 109 J/m3) and γ (0.255 J/m2) are given in the problem statement, and the supercooling value found in Table 10.1 is 319 °C (or 319 K). Thus, from Equation 10.6 we have

For computation of the activation free energy, Equation 10.7 is employed. Thus

(b) Now calculate the number of atoms found in a nucleus of critical size. Assume a lattice parameter of 0.360 nm for solid nickel at its melting temperature. Solution: In order to compute the number of atoms in a nucleus of critical size (assuming a spherical nucleus of radius r*), it is first necessary to determine the number of unit cells, which we then multiply by the number of atoms per unit cell. The number of unit cells found in this critical nucleus is just the ratio of critical nucleus and unit cell volumes. Inasmuch as nickel has the FCC crystal structure, its unit cell volume is just a3 where a is the unit cell length (i.e., the lattice parameter); this value is 0.360 nm, as cited in the problem statement. Therefore, the number of unit cells found in a radius of critical size is just

Inasmuch as 4 atoms are associated with each FCC unit cell, the total number of atoms per critical nucleus is just (116 unitcells/criticalnucleus)(4 atoms/unitcell) = 464 atoms/criticalnucleus

Question 2. (Callister 10.5) (a) Assume for the solidification of nickel (Problem 10.4) that nucleation is homogeneous, and the number of stable nuclei is 106 nuclei per cubic meter. Calculate the critical radius and the number of stable nuclei that exist at the following degrees of supercooling: 200 K and 300 K. Solution: For this part of the problem we are asked to calculate the critical radius for the solidification of nickel (per Question 1), for 200 K and 300 K degrees of supercooling, and assuming that the there are 106 nuclei per meter cubed for homogeneous nucleation. In order to calculate the critical radii, we replace the (Tm – T) term in Equation 10.6 by the degree of supercooling (denoted as ΔT) cited in the problem. For 200 K supercooling,

For 300 K supercooling,

In order to compute the number of stable nuclei that exist at 200 K and 300 K degrees of supercooling, it is necessary to use Equation 10.8. However, we must first determine the value of K1 in Equation 10.8, which in turn requires that we calculate ΔG* at the homogeneous nucleation temperature using Equation 10.7; this was done in Problem 10.4, and yielded a value of ΔG* = 1.27 x 10-18 J. Now for the computation of K1, using the value of n* for at the homogenous nucleation temperature (106 nuclei/m3):

1728 K

2.32 x 1034

Now for 200 K supercooling, it is first necessary to recalculate the value ΔG* of using Equation 10.7, where, again, the (Tm – T) term is replaced by the number of degrees of supercooling, denoted as ΔT, which in this case is 200 K. Thus

And, from Equation 10.8, the value of n* is

2.32 x 1034 1728 K

4.31 x 10-33

Now, for 300 K supercooling the value of ΔG* is equal to

from which we compute the number of stable nuclei at 300 K of supercooling as

2.32 x 1034

= 427 stable nuclei 1728 K

(b) What is significant about the magnitudes of these critical radii and the numbers of stable nuclei? Solution: Relative to critical radius, r* for 300 K supercooling is slightly smaller that for 200 K (1.16 nm versus 1.74 nm). [From Problem 10.4, the value of r* at the homogeneous nucleation temperature (319 K) was 1.09 nm.] More significant, however, are the values of n* at these two degrees of supercooling, which are dramatically different— 4.31 x 10-33 stable nuclei at ΔT = 200 K, versus 427 stable nuclei at ΔT = 300 K!

Question 3. (Callister 12.4) Demonstrate that the minimum cation-to-anion radius ratio for a coordination number of 8 is 0.732. From the cubic unit cell shown below

the unit cell edge length is 2rA, and from the base of the unit cell x2 = (2rA)2 + (2rA)2 = 8rA2 Or x = 2 2 rA Now from the triangle that involves x, y, and the unit cell edge x2 + (2rA)2 = y2 = (2rA+2rC)2 (2 2 rA)2 + 4rA2 = (2rA + 2rC)2 Which reduces to 2rA( 3 − 1) = 2rC Or

rC = 3 − 1 = 0.732 rA

Question 4. (Callister 12.2) Show that the minimum cation-to-anion radius ratio for a coordination number of 4 is 0.225. (hint: this case is a bit difficult. Let the tetrahedron be ABCD, O is the center of the tetrahedron, O’ is the center of equilateral triangle BCD, you should be able to show that 6 AO=(3/4)AO’= a , where a is the edge of the tetrahedron. The other way to visualize 4 the tetrahedron is to put the four apex as 4 corners of a cube. Then the center of tetrahedron must also be the center of the cube.) Solution: If lines are drawn from the centers of the anions, then a tetrahedron is formed. The tetrahedron may be inscribed within a cube as shown below.

The spheres at the apexes of the tetrahedron are drawn at the corners of the cube, and designated as positions A, B, C, and D. (These are reduced in size for the sake of clarity.) The cation resides at the center of the cube, which is designated as point E. Let us now express the cation and anion radii in terms of the cube edge length, designated as a. The spheres located at positions A and B touch each other along the bottom face diagonal. Thus,

But

or

and Note that from symmetry, E is the center of the cube and the tetrahedron, therefore the cube diagonal AEF will be related to the ionic radii as

(By writing this way, it doesn’t mean there will also be an anion at F, which is in touch with the cation E. )

From the triangle ABF

From above, thus,

Solving for the rC/rA ratio leads to

Question 5. (Callister 12.16)

This problem asks that we compute the theoretical density of ZnS given that the Zn—S distance and bond angle are 0.234 nm and 109.5°, respectively. The first thing we need do is to determine the unit cell volume from the given Zn—S distance. The unit cell volume VC is just a3, a being the unit cell edge length, and

Now we must utilize Equation 12.1 with n' = 4 formula units, and AZn and AS being 65.39 and 32.06 g/mol, respectively. Thus

ρ=

n ′(ΣAC + ΣAA ) VC N A

= 4.11 g/cm3 The measured value of the density is 4.10 g/cm3.

Question 6. (Callister 12.21)

This problem asks us to compute the atomic packing factor for Fe3O4 given its density and unit cell edge length. It is first necessary to determine the number of formula units in the unit cell in order to calculate the sphere volume. Solving for n' from Equation 12.1 leads to

= 8.0 formula units/unit cell Thus, in each unit cell there are 8 Fe2+, 16 Fe3+, and 32 O2- ions (FeO-Fe2O3). From Table 12.3, rFe2+ = 0.077 nm, rFe3+ = 0.069 nm, and rO2- = 0.140 nm. Thus, the total sphere volume in Equation 3.2 (which we denote as VS), is just

= 4.05 x 10-22 cm3 Now, the unit cell volume (VC) is just VC = a3= (8.39 x 10-8 cm)3 = 5.90x10−22 cm3 Finally, the atomic packing factor (APF) from Equation 3.2 is just

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