Materiales

ME 2105: Suggested Problem for practice

Chapters 2 & 3

Ch 2:

13, 18, 19, 20

Ch 3:

4, 5, 6, 10, 31, 34, 35, 39, 42, 49, 50, 52, 55

KEY:

2.13 Calculate the force of attraction between a K+ and an O2- ion the centers of which are separated by a distance of 1.5 nm.

Solution

The attractive force between two ions FA is just the derivative with respect to the interatomic separation of the attractive energy expression, Equation 2.8, which is just

The constant A in this expression is defined in footnote 3. Since the valences of the K+ and O2- ions (Z1 and Z2) are +1 and -2, respectively, Z1 = 1 and Z2 = 2, then

= 2.05  10-10 N

2.18 (a) Briefly cite the main differences between ionic, covalent, and metallic bonding.

(b) State the Pauli exclusion principle.

Solution

(a) The main differences between the various forms of primary bonding are:

Ionic--there is electrostatic attraction between oppositely charged ions.

Covalent--there is electron sharing between two adjacent atoms such that each atom assumes a stable electron configuration.

Metallic--the positively charged ion cores are shielded from one another, and also "glued" together by the sea of valence electrons.

(b) The Pauli exclusion principle states that each electron state can hold no more than two electrons, which must have opposite spins.

2.19 Compute the percents ionic character of the interatomic bonds for the following compounds: TiO2, ZnTe, CsCl, InSb, and MgCl2.

Solution

The percent ionic character is a function of the electron negativities of the ions XA and XB according to Equation 2.10. The electronegativities of the elements are found in Figure 2.7.

For TiO2, XTi = 1.5 and XO = 3.5, and therefore,

For ZnTe, XZn = 1.6 and XTe = 2.1, and therefore,

For CsCl, XCs = 0.7 and XCl = 3.0, and therefore,

For InSb, XIn = 1.7 and XSb = 1.9, and therefore,

For MgCl2, XMg = 1.2 and XCl = 3.0, and therefore,

2.20 Make a plot of bonding energy versus melting temperature for the metals listed in Table 2.3. Using this plot, approximate the bonding energy for copper, which has a melting temperature of 1084C.

Solution

Below is plotted the bonding energy versus melting temperature for these four metals. From this plot, the bonding energy for copper (melting temperature of 1084C) should be approximately 3.6 eV. The experimental value is 3.5 eV.

CHAPTER 3:

3.4 For the HCP crystal structure, show that the ideal c/a ratio is 1.633.

Solution

A sketch of one-third of an HCP unit cell is shown below.

Consider the tetrahedron labeled as JKLM, which is reconstructed as

The atom at point M is midway between the top and bottom faces of the unit cell--that is = c/2. And, since atoms at points J, K, and M, all touch one another,

where R is the atomic radius. Furthermore, from triangle JHM,

or

Now, we can determine the length by consideration of triangle JKL, which is an equilateral triangle,

and

Substituting this value for in the above expression yields

and, solving for c/a

3.5 Show that the atomic packing factor for BCC is 0.68.

Solution

The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or

Since there are two spheres associated with each unit cell for BCC

Also, the unit cell has cubic symmetry, that is VC = a3. But a depends on R according to Equation 3.3, and

Thus,

3.6 Show that the atomic packing factor for HCP is 0.74.

Solution

The APF is just the total sphere volume-unit cell volume ratio. For HCP, there are the equivalent of six spheres per unit cell, and thus

Now, the unit cell volume is just the product of the base area times the cell height, c. This base area is just three times the area of the parallelepiped ACDE shown below.

The area of ACDE is just the length of times the height . But is just a or 2R, and

Thus, the base area is just

and since c = 1.633a = 2R(1.633)

(3.S1)

Thus,

3.10 Some hypothetical metal has the simple cubic crystal structure shown in Figure 3.24. If its atomic weight is 70.4 g/mol and the atomic radius is 0.126 nm, compute its density.

Solution

For the simple cubic crystal structure, the value of n in Equation 3.5 is unity since there is only a single atom associated with each unit cell. Furthermore, for the unit cell edge length, a = 2R (Figure 3.24). Therefore, employment of Equation 3.5 yields

and incorporating values of the other parameters provided in the problem statement leads to

7.31 g/cm3

3.31 Determine the indices for the directions shown in the following cubic unit cell:

Solution

Direction A is a direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system

x y z

Projections 0a –b –c

Projections in terms of a, b, and c 0 –1 –1

Reduction to integers not necessary

Enclosure

Direction B is a direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system

x y z

Projections –a 0c

Projections in terms of a, b, and c –1 0

Reduction to integers –2 1 0

Enclosure

Direction C is a [112] direction, which determination is summarized as follows.

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