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 \(0.437 \mathrm{~nm}\).

Short Answer

Expert verified
Using the given information about the nonstoichiometric Fe(1-x)O material, we determined the concentration of holes (Fe³⁺ ions) and then calculated the number of ions per unit volume. Finally, we used the formula for electrical conductivity to calculate the conductivity of the material, which is approximately 7.69 × 10⁻³ S/m.

Step by step solution

01

Determine the concentration of Fe³⁺ ions

Given the value of x as 0.040, for every 1 mole of Fe atoms, there are 0.040 moles of Fe³⁺ ions. Since we are also given that the acceptor states are saturated, this means that there is one hole for every Fe³⁺ ion. So, the concentration of holes is the same as the concentration of Fe³⁺ ions.
02

Calculate the number of ions per unit volume

To calculate the number of ions per unit volume, we first need to find how many Fe and O atoms are there in one unit cell. We know that wüstite has the sodium chloride (NaCl) crystal structure, which has 4 formula units per unit cell, meaning that there are 4 Fe atoms and 4 O atoms within one unit cell. With a unit cell edge length of 0.437 nm, the volume of one unit cell is: Volume = (0.437 x 10⁻⁹ m)³ = 8.33 x 10⁻²⁸ m³ So, the number of Fe³⁺ ions per unit volume can be calculated as follows: (0.040 moles of Fe³⁺ ions / (1 mole of Fe atoms + 1 mole of O atoms)) x (4 Fe atoms + 4 O atoms in a unit cell / (8.33 x 10⁻²⁸ m³)) The 4 Fe atoms and 4 O atoms cancel out, leaving: (0.040 Fe³⁺ ions / (8.33 x 10⁻²⁸ m³))
03

Calculate the electrical conductivity

Now that we have the concentration of holes and the hole mobility, we can calculate the electrical conductivity using the following formula: Electrical Conductivity (σ) = (q * hole concentration * hole mobility) Where q is the elementary charge of 1.6 x 10⁻¹⁹ C, the hole concentration is 0.040 Fe³⁺ ions per unit volume as calculated in Step 2, and hole mobility is given as 1.0 x 10⁻⁵ m²/V⋅s. σ = (1.6 x 10⁻¹⁹ C) x (0.040 Fe³⁺ ions / (8.33 x 10⁻²⁸ m³)) x (1.0 x 10⁻⁵ m²/V⋅s) The electrical conductivity of the wüstite specimen is approximately 7.69 × 10⁻³ S/m.

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