We noted in Section 12.5 (Figure 12.20) that in FeO (wüstite), the iron ions can exist in both Fe2+ and Fe3+ states. The number of each of these ion types depends on temperature and the ambient oxygen pressure. Furthermore, we also noted that in order to retain electroneutrality, one Fe2+ vacancy will be created for every two Fe3+ ions that are formed; consequently, in order to reflect the existence of these vacancies the formula for wüstite is often represented as Fe(1 – x)O, where x is some small fraction less than unity. In this nonstoichiometric Fe(1 – x)O material, conduction is electronic, and, in fact, it behaves as a p-type semiconductor—that is, the Fe3+ ions act as electron acceptors, and it is relatively easy to excite an electron from the valence band into an Fe3+ acceptor state, with the formation of a hole. Determine the electrical conductivity of a specimen of wüstite that has a hole mobility of 1.0 × 10–5 m2/V.s and for which the value of x is 0.040. Assume that the acceptor states are saturated (i.e., one hole exists for every Fe3+ ion). Wüstite has the sodium chloride crystal structure with a unit cell edge length of 0.437 nm.

To determine the electrical conductivity of the wüstite specimen, we need to follow these steps:

1. Calculate the number of Fe atoms per unit cell: Wüstite has the sodium chloride (NaCl) crystal structure. In this structure, there are 4 Fe atoms per unit cell.

2. Determine the volume of the unit cell: The unit cell edge length is given as 0.437 nm. $\text{Volume of unit cell} = (0.437 \times 10^{-9} \text{ m})^3 = 8.36 \times 10^{-29} \text{ m}^3$

3. Calculate the number of unit cells per cubic meter: $\text{Number of unit cells per cubic meter} = \frac{1}{8.36 \times 10^{-29} \text{ m}^3} = 1.20 \times 10^{28} \text{ unit cells/m}^3$

4. Calculate the number of Fe atoms per cubic meter: $\text{Number of Fe atoms per cubic meter} = 4 \times 1.20 \times 10^{28} = 4.80 \times 10^{28} \text{ Fe atoms/m}^3$

5. Determine the number of Fe3+ ions per cubic meter: Given that $x = 0.040$, the fraction of Fe3+ ions is $2x$ (since one Fe2+ vacancy is created for every two Fe3+ ions). $\text{Number of Fe3+ ions per cubic meter} = 2x \times \text{Number of Fe atoms per cubic meter} = 2 \times 0.040 \times 4.80 \times 10^{28} = 3.84 \times 10^{27} \text{ Fe3+ ions/m}^3$

6. Determine the hole concentration: Since each Fe3+ ion corresponds to one hole (acceptor state is saturated), $\text{Hole concentration} = 3.84 \times 10^{27} \text{ holes/m}^3$

7. Calculate the electrical conductivity: The electrical conductivity $\sigma$ is given by: $\sigma = q \cdot p \cdot \mu_h$ where:

   • $q$ is the charge of an electron ($1.6 \times 10^{-19} \text{ C}$)
   • $p$ is the hole concentration ($3.84 \times 10^{27} \text{ holes/m}^3$)
   • $\mu_h$ is the hole mobility ($1.0 \times 10^{-5} \text{ m}^2/\text{V.s}$)

   Substituting the values: $\sigma = (1.6 \times 10^{-19} \text{ C}) \times (3.84 \times 10^{27} \text{ holes/m}^3) \times (1.0 \times 10^{-5} \text{ m}^2/\text{V.s})$ $\sigma = 6.14 \times 10^{3} \text{ S/m}$

Therefore, the electrical conductivity of the wüstite specimen is $6.14 \times 10^{3} \text{ S/m}$.