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.
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.
1)
Unit Cell 
Number of Lattice Points per Unit Cell 

Primitive (P) 

Body Centered (I) 

Base Centered (B) 

Face Centered (F) 

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) 

Body Centered (I) 

Base Centered (B) 

Face Centered (F) 

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 twoatom basis set. The complete description of this crystal, then, is:
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 12: [0 1/2 1/2]
Atoms 13: [1/2 0 1/2]
Atoms 14: [1/2 1/2 1]
These three vectors are the three vectors needed for the facecentered type cell. The complete description of this crystal might look like:
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:
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 Itype 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 Itype translation vector, so orthorhombicI 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:
The volume of the unit cell is given by the cube of the lattice parameter (Appendix 10) as V = (0.5431nm)^3 = (5.431e10 m)^3 = 1.6e28 m^3.
The calculated density of Si is thus DENSITY = M/V = (3.73e22 g/u.c.)/(1.6e28 m^3/u.c.) = 2.33 Mg/m^3, which compares pretty well to the experimentally measured value of 2..329 Mg/m^3.
Taking the density, etc. from appendix 5, we have these values:
Element 
Density [g/cm^3] 
Z 
A 

Li 



Si 



Ti 



W 



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 0100%).