Crystal Structure and Band Structure of Vanadium: A Detailed Analysis

This week's questions are based on Week 2 and 3 lecture notes and covers the topics of crystal structure (e.g. FCC, BCC, diamond), lattice and basis, the bandstructure of metals, insulators and semiconductors, energy band diagrams, Fermi level and workfunction.

1. Vanadium forms the body-centred cubic (BCC) crystal structure. The radius of a vanadium atom is 153 pm.

a) Draw the unit cell for the BCC. How many atoms are there per unit cell? b) If R is the radius of one vanadium atom, show that the lattice parameter, a = 4 * R / √3. c) What is the atomic concentration of the vanadium crystal?

a) The unit cell for the body-centred cubic (BCC) crystal structure is a cube with an atom at each of its eight corners and one atom at the centre. Therefore, there are 8 atoms per unit cell.

b) In a BCC crystal structure, the relationship between the atomic radius (R) and the lattice parameter (a) is given by the equation: a = 4 * R / √3.

To show this, we can substitute the given radius (153 pm) into the equation: a = 4 * (153 pm) / √3 a ≈ 287.95 pm

c) The atomic concentration of the vanadium crystal can be calculated by dividing the volume occupied by the vanadium atoms by the total volume of the unit cell. In a BCC crystal structure, each atom contributes a volume equal to (4/3) * π * R^3 to the unit cell.

The volume of the unit cell (V_uc) can be calculated using the lattice parameter (a): V_uc = a^3

The volume occupied by the vanadium atoms (V_atom) is equal to the volume contributed by one atom (4/3) * π * R^3 multiplied by the number of atoms per unit cell (8): V_atom = (4/3) * π * R^3 * 8

The atomic concentration (C) is given by: C = V_atom / V_uc

Substituting the values we have: C = [(4/3) * π * (153 pm)^3 * 8] / (287.95 pm)^3

After performing the calculations, the atomic concentration of the vanadium crystal can be determined.