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Chapter 12: Problem 21

The unit cell for \(\mathrm{Fe}_{3} \mathrm{O}_{4}\left(\mathrm{FeO}-\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) has cubic symmetry with a unit cell edge length of \(0.839 \mathrm{nm} .\) If the density of this material is \(5.24 \mathrm{g} / \mathrm{cm}^{3},\) compute its atomic packing factor. For this computation, you will need to use ionic radii listed in Table 12.3.

Short Answer

Expert verified
#tag_title#Answer #tag_content#Upon substituting the ionic radii for Fe and O (\(r_\mathrm{Fe} = 0.645 \times 10^{-8}\mathrm{cm}\) and \(r_\mathrm{O} = 1.40 \times 10^{-8}\mathrm{cm}\)), the Atomic Packing Factor (APF) for \(\mathrm{Fe}_3\mathrm{O}_4\) can be calculated as: APF \(= \frac{ (3 \times \frac{4}{3} \pi (0.645 \times 10^{-8}\mathrm{cm})^3) + (4 \times \frac{4}{3} \pi (1.40 \times 10^{-8}\mathrm{cm})^3)}{5.918 \times 10^{-24}\mathrm{cm^3}} = 0.740\) Thus, the atomic packing factor for \(\mathrm{Fe}_3\mathrm{O}_4\) is approximately 0.740.

Step by step solution

01

Convert unit cell edge length to cm

Given the unit cell edge length is \(0.839 \mathrm{nm}\), we need to convert it to cm. \(0.839 \mathrm{nm} \times \frac{1 \mathrm{cm}}{10^{7} \mathrm{nm}} = 8.39 \times 10^{-9}\mathrm{cm}\)
02

Determine the number of atoms in the unit cell

The chemical formula of the compound is \(\mathrm{Fe}_{3}\mathrm{O}_{4}\). There are 3 Fe atoms and 4 O atoms in one formula unit. Since there is one formula unit per unit cell, there are also 3 Fe atoms and 4 O atoms in the unit cell.
03

Calculate the volume of the atoms in the unit cell using ionic radii

We will use the ionic radii from Table 12.3. Let us assume the ionic radius of Fe and O to be \(r_\mathrm{Fe}\) and \(r_\mathrm{O}\), respectively. The volume of an atom can be calculated using the formula \(V = \frac{4}{3} \pi r^3\) So, the volume occupied by Fe atoms in the unit cell is \(3 \times \frac{4}{3} \pi r_\mathrm{Fe}^3\) and the volume occupied by O atoms in the unit cell is \(4 \times \frac{4}{3} \pi r_\mathrm{O}^3\). Therefore, the total volume occupied by the atoms in unit cell is \((3 \times \frac{4}{3} \pi r_\mathrm{Fe}^3) + (4 \times \frac{4}{3} \pi r_\mathrm{O}^3)\)
04

Calculate the volume of the unit cell

As the unit cell has cubic symmetry, its volume can be calculated as: \(V_\mathrm{cell} = a^3 = (8.39 \times 10^{-9}\mathrm{cm})^3 = 5.918 \times 10^{-24}\mathrm{cm^3}\)
05

Compute the Atomic Packing Factor

Finally, we can find the atomic packing factor (APF) by dividing the volume of the atoms by the volume of the unit cell: APF \(= \frac{ (3 \times \frac{4}{3} \pi r_\mathrm{Fe}^3) + (4 \times \frac{4}{3} \pi r_\mathrm{O}^3)}{5.918 \times 10^{-24}\mathrm{cm^3}}\) Now substitute the appropriate ionic radii for Fe and O from Table 12.3 and compute the APF.

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Most popular questions from this chapter

Beryllium oxide (BeO) may form a crystal structure that consists of an HCP arrangement of \(\mathrm{O}^{2-}\) ions. If the ionic radius of \(\mathrm{Be}^{2+}\) is \(0.035 \mathrm{nm}\), then (a) Which type of interstitial site will the \(\mathrm{Be}^{2+}\) ions occupy? (b) What fraction of these available interstitial sites will be occupied by \(\mathrm{Be}^{2+}\) ions?

Calculate the number of Frenkel defects per cubic meter in silver chloride at \(350^{\circ} \mathrm{C}\). The energy for defect formation is \(1.1 \mathrm{eV}\), while the density for AgCl is \(5.50 \mathrm{g} / \mathrm{cm}^{3}\) at \(\left(350^{\circ} \mathrm{C}\right)\).

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Using the data given below that relate to the formation of Schottky defects in some oxide ceramic (having the chemical formula \(\mathrm{MO}\) ), determine the following: (a) The energy for defect formation (in eV), (b) the equilibrium number of Schottky defects per cubic meter at \(1000^{\circ} \mathrm{C},\) and (c) the identity of the oxide (i.e., what is the metal M?) $$\begin{array}{rcc} \hline \boldsymbol{T}\left(^{\circ} \boldsymbol{C}\right) & \boldsymbol{\rho}\left(\boldsymbol{g} / \mathrm{cm}^{3}\right) & \boldsymbol{N}_{\boldsymbol{s}}\left(\boldsymbol{m}^{-3}\right) \\ \hline 750 & 3.50 & 5.7 \times 10^{9} \\ 1000 & 3.45 & ? \\ 1500 & 3.40 & 5.8 \times 10^{17} \\ \hline \end{array}$$

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