ME588: Homework One
Fall 2000

This page last updated October 2, 2000

Due October 2, 2000

1) Draw the following planes and directions in an orthorhombic unit cell: (112), (1 -1 1), (002), [222], [-101], [201]. Be sure to label your cell axes.


2) Draw the following planes and directions in an hexagonal unit cell: (11-20), (11-22), (0-112), [101], [0-21], [001]. Be sure to label your cell axes.


3) The planes (-111), (1-11), and (002) all belong to the same zone. What zone is it? Name one other plane in this zone.


4) The unit cells of several orthorhombic crystals are described below. Determine the Bravais lattice of each and explain how you know. Do not change axes. HINT: In solving these kinds of problems, it is easiest to look for centering translations that connect the atoms, rather than trying to draw a picture.

a) Two atoms of the same kind per unit cell at positions of 0, 1/4, 0; and 1/4, 0, 1/4.

b) Four atoms of the same kind per unit cell at positions of 1/2, 1/4, 0; 1/2, 3/4, 1/2; 1, 1/4, -1/2; and 0, 3/4, 1.

c) Two atoms of the same kind per unit cell at positions of x, y, z; and x+1/2, -y, -z.

d) Two atoms of one kind, A, located at 0, 0, 1/2; and -1/2, 1/2, 0, and two atoms of another kind, B, located at 1/4, 1/4, 1/4; and -1/4, 3/4, -1/4.


5) Use the lattice parameter data from appendix 10 to calculate the theoretical density of silicon. Compare with the accepted value of 2329 kg/m^3.


6) Calculate and plot the probability of elastic scattering (no. of events/electron) for angles larger than 2 degrees in a foil of thickness 100 nm, as a function of electron beam energy between 10-100 keV for the following materials: Li, Si, Ti, and W. Use appendix 5 for the data on density.


SOLUTIONS


1)


2)


3) The planes (-111), (1-11), and (002) all contain (are parallel to) the direction [ 1 1 0 ]. Any other plane parallel to [110] is alos in the [110]-zone. There are a lot of possibilities: (0 0 1), (1 -1 0), (1 -1 2), (1 -1 4), ...


4) There are two pieces of information that make these problems easy to solve. The first is knowing how many lattice points there are in each type of unit cell.

Unit Cell

Number of Lattice Points per Unit Cell

Primitive (P)

1

Body Centered (I)

2

Base Centered (B)

2

Face Centered (F)

4

The second information is the types of translation vectors that move you from one lattice point to another (centering translations).

Unit Cell

Centering Translations

Primitive (P)

[001], [010], [100]

Body Centered (I)

P + [1/2 1/2 1/2]

Base Centered (B)

P + [1/2 1/2 0] OR [1/2 0 1/2] OR [0 1/2 1/2]

Face Centered (F)

P + [1/2 1/2 0] + [1/2 0 1/2] + [0 1/2 1/2]

a) Two atoms of the same kind per unit cell at positions of 0, 1/4, 0; and 1/4, 0, 1/4. With 2 atoms per unit cell, we know that the unit cell cannot be a face centered cell. To choose between P,I, and C cells, we need to examine the vector connecting the two atoms. The two atoms are separated by a vector [1/4 -1/4 1/4]. This is not a centering translation for any of the cells. This means that, if the first atom is on a lattice point, then the second cannot be on a lattice point: it must be part of a two-atom basis set. The complete description of this crystal, then, is:

Orthorhombic-P with a two-atom basis of 0,0,0 1/4, -1/4, 1/4.

b) Four atoms of the same kind per unit cell at positions of 1/2, 1/4, 0; 1/2, 3/4, 1/2; 1, 1/4, -1/2; and 0, 3/4, 1. Now we have to consider all the unit cells as possibilities, based on the number of atoms in the cell. Let's look at the vectors separating the atoms.

Atoms 1-2: [0 1/2 1/2]

Atoms 1-3: [1/2 0 -1/2]

Atoms 1-4: [-1/2 1/2 1]

These three vectors are the three vectors needed for the face-centered type cell. The complete description of this crystal might look like:

Orthorhombic-F with a one-atom basis of 0,0,0.

c) Two atoms of the same kind per unit cell at positions of x, y, z; and x+1/2, -y, -z. With 2 atoms per unit cell, we know that the unit cell cannot be a face centered cell. The two atoms are separated by a vector [1/2 -2y -2z]. Since we don't know what the values of y and z are, this is not in general a centering translation for any of the cells. The complete description of this crystal, then, is:

Orthorhombic-P with a two-atom basis of 0,0,0 1/2, -2y, -2z.

d) Two atoms of one kind, A, located at 0, 0, 1/2; and -1/2, 1/2, 0, and two atoms of another kind, B, located at 1/4, 1/4, 1/4; and -1/4, 3/4, -1/4. The first question we need to answer is whether or not the two A atoms are both on lattice points or not. The vector separating them is [-1/2 1/2 1/2], which looks a lot like an I-type cell. If the I designation is correct, then both the A and the B atoms have to have the same symmetry independently. The vecotr separating the B atoms is [-1/2 1/2 -1/2]. This is also an I-type translation vector, so orthorhombic-I looks good. (NOTE: the vector connecting the two B atoms doesn't hav e to be exactly the same as the A atoms, as long as it has the correct symmetry!) The basis will be defined by the vector between an A and a B atom: [1/4 1/4 -1/4]. The final description is:

Orthorhombic-I with a two-atom basis of and A atom at 0,0,0 and a B atom at 1/4, 1/4, -1/4.


5) The definition of the density is the mass per unit volume of the material. Use the unit cell as the unit volume. For Si, which has a diamond-cubic crystal structure, there are four lattice points per unit cell, and two Si per lattice point (see p. 33) for a total of eight Si atoms per unit cell. The atomic weight of Si is 28.0855 g/mole. This works out to a mass of M = (8 atoms/u.c.)*(28.0855 g/mole)/(No atoms/mole) = 3.73e-22 g/u.c.

The volume of the unit cell is given by the cube of the lattice parameter (Appendix 10) as V = (0.5431nm)^3 = (5.431e-10 m)^3 = 1.6e-28 m^3.

The calculated density of Si is thus DENSITY = M/V = (3.73e-22 g/u.c.)/(1.6e-28 m^3/u.c.) = 2.33 Mg/m^3, which compares pretty well to the experimentally measured value of 2..329 Mg/m^3.


6) Putting together the scattering cross section equation from the Goldstein book handout, and the probability equation, we get:

Taking the density, etc. from appendix 5, we have these values:

Element

Density [g/cm^3]

Z

A

Li

0.533

3

6.941

Si

2.33

14

28.0855

Ti

4.51

22

47.88

W

19.25

74

183.85

I put the equation for the probability of an elastic scattering event into a spreadsheet (EXCEL) and plotted the data. The result is shown below.

NOTE: The "probability" that is plotted is really the number of elastic scattering events that would occur for an electron traversing the entire 100 nm of metal foil. It is not a true probability (which would have values between 0-100%).


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