ENGR 321: Homework Set 3
Fall 2001

Prof. W. H. Warnes
Office: Rogers Hall 308
Phone and Voice Mail: (541) 737-7016
FAX: (541) 737-2600


This page last updated February 10, 2002

email: warnesw@engr.orst.edu


DUE IN CLASS Wednesday, February 6, 2002


MILLER INDICES OF PLANES AND PLANAR DENSITIES

1) Sketch within a unit cell the following planes: ( 2 0 1 ), ( 1 -1 1 ), ( 3 2 -1 ), ( 0 -1 0 )

2) Determine the planar density of the ( 0 0 1 ) plane in HCP.

3) Callister problem 3.48.

COORDINATION NUMBER AND IONICALLY BONDED CRYSTALS

4) Using the ionic charge and radii, determine the crystal structures for the following solids:

  1. NaI
  2. CuS
  3. CsBr
  4. BaCl

5) Calculate the atomic packing factor for a solid with the NaCl crystal structure as a function of the ionic radius ratio, r/R, and plot this function for values of 0.414<r/R<0.732.

POINT DEFECTS AND COMPOSITION CALCULATIONS

6) Assuming the equilibrium concentration of vacancy defects at the melting point of most metals is 1E-4, determine the energy needed to form a single vacancy in W, Ni, and Ga.

7) Using the information from problem 6 to find the vacancy concentration in W, Ni and Ga at 0 deg C.


SOLUTIONS


1) Sketch within a unit cell the following planes: ( 2 0 1 ), ( 1 -1 1 ), ( 3 2 -1 ), ( 0 -1 0 )


2) Determine the planar density of the ( 0 0 1 ) plane in HCP.

The (001) plane is the basal plane (perpendicular to the c-axis) in HCP. Refering to a drawing of the crystal structure (such as figure 3.3 in Callister), we can see that the (001) plane is a close packed plane made up of atoms touching in equilateral triangles. The planar density can be found from the geometry of the equilateral triangles (see figure to the right).

The height of the triangle is h = R sqrt(3), and the area of the triangle is Rh. Each triangle contains a 60 degree slice of each atom, for a total of 1/2 an atom. The planar density is therefore:


3) Callister problem 3.48.

The first thing to do is draw out the three planes given in the problem on a unit cell:

Based on the positions of the planes in the unit cell, and the positions of the atoms on the planes, this looks like a body-centered cell. The unit cell axes are a = 0.30nm, b = 0.40nm, and c = 0.35nm. Since all three unit cell axes are of different lengths, and the angles are all 90 degrees,

a) this is an ORTHORHOMBIC unit cell (see Table 3.2, p. 39).

b) The atom positions show an atom in the center of the unit cell, so we would call this a Body-centered Orthorhombic crystal.

c) We can find the density using the unit cell dimensions, the fact that there are 2 atoms per unit cell, and the density equation to find:

(NOTE: This data does not match any known element...)


4) First, it is helpful to accumulate the data needed for the problem:

Ion
Ionic Radius [pm]
Ionic Charge
Na
102
+1
I
220
-1
Cu
96
+1
S
184
-2
Cs
170
+1
Br
196
-1
Ba
136
+2
Cl
181
-1

Now, looking at the pairs of ions, we can find the coordination number of the smallest ion in the pair by taking the ratio of the ionic radii and using Table 13.2. Then, we have to look at the relative charge to finally determine the crystal structure:

a) NaI: The ionic radius ratio is r/R = 102pm/220pm = 0.464, which gives this pair a CN = 6. The two ions have the same charge, so there will be an equal number of them in each unit cell. The crystal structure that matches these requirements is the ROCK SALT (or NaCl) structure.

b) CuS: This one is the most difficult, and your answer for this won't be graded (extra credit if you get it right). In fact, I haven't been able to find an answer for this problem. That's what I get for making it up myself, I guess. There is not a crystal structure matching the requirements of this ion pair in the book.

c) CsBr: The ionic radius ratio is r/R = 170pm/196pm = 0.867, which gives this pair a CN = 8. The two ions have the same charge, so there will be an equal number of them in each unit cell. The crystal structure that matches these requirements is the CESIUM CHLORIDE structure.

d) BaCl: The ionic radius ratio is r/R = 136pm/181pm = 0.751, which gives this pair a CN = 8. The two ions have the same charge, so there will be an equal number of them in each unit cell. The crystal structure that matches these requirements is the FLUORITE (or CaF2) structure.


5) Calculate the atomic packing factor for a solid with the NaCl crystal structure as a function of the ionic radius ratio, r/R, and plot this function for values of 0.414<r/R<0.732.

For the NaCl structure (see Figure 13.2, p. 386), the atoms touch along the [100], [010], and [001] directions, and along these directions we get the relation between the lattice parameter and the ionic radii: a = 2(r + R). In the unit cell there are 4 large atoms, and 4 small atoms. With this information, we can write out the equation foor the APF:

With this equation, I have plotted the function in EXCEL as shown below. The APF increases as the radius ratio decreases.


6) Assuming the equilibrium concentration of vacancy defects at the melting point of most metals is 1E-4, determine the energy needed to form a single vacancy in W, Ni, and Ga.

From the melting temperatures of the three elements (see table inside front cover of Callister) we can get the information we need to plug into the vacancy concentration equation:


7) Using the information from problem 6 to find the vacancy concentration in W, Ni and Ga at 0 deg C.

Work the equation the other direction:


End of File.