Chapter 18: Problem 47
We note in Section \(12.5\) (Figure 12.20) that in FeO (wüstite), the iron ions can exist in both \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) states. The number of each of these ion types depends on temperature and the ambient oxygen pressure. Furthermore, we also note that in order to retain electroneutrality, one \(\mathrm{Fe}^{2+}\) vacancy is created for every two \(\mathrm{Fe}^{3+}\) ions that are formed; consequently, in order to reflect the existence of these vacancies, the formula for wüstite is often represented as \(\mathrm{Fe}_{(1-x)} \mathrm{O}\), where \(x\) is some small fraction less than unity. In this nonstoichiometric \(\mathrm{Fe}_{(1-x)} \mathrm{O}\) material, conduction is electronic and, in fact, it behaves as a \(p\)-type semiconductor - that is, the \(\mathrm{Fe}^{3+}\) ions act as electron acceptors, and it is relatively easy to excite an electron from the valence band into an \(\mathrm{Fe}^{3+}\) acceptor state with the formation of a hole. Determine the electrical conductivity of a specimen of wüstite with a hole mobility of \(1.0 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{V} \cdot \mathrm{s}\), and for which the value of \(x\) is \(0.040\). Assume that the acceptor states are saturated (i.e., one hole exists for every \(\mathrm{Fe}^{3+}\) ion). Wüstite has the sodium chloride crystal structure with a unit cell edge length of \(0.437 \mathrm{~nm}\).