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Chapter 20: Problem 13

The chemical formula for copper ferrite may be written as \(\left(\mathrm{CuFe}_{2} \mathrm{O}_{4}\right)_{8}\) because there are eight formula units per unit cell. If this material has a saturation magnetization of \(1.35 \times\) \(10^{5} \mathrm{A} / \mathrm{m}\) and a density of \(5.40 \mathrm{g} / \mathrm{cm}^{3},\) estimate the number of Bohr magnetons associated with each \(\mathrm{Cu}^{2+}\) ion.

Short Answer

Expert verified
Question: Estimate the number of Bohr magnetons associated with each Cu²⁺ ion in copper ferrite. Answer: The number of Bohr magnetons associated with each Cu²⁺ ion in copper ferrite is approximately \(1.82\times10^{23}\).

Step by step solution

01

Calculate the number of Cu²⁺ ions per unit cell

For copper ferrite, the chemical formula is given as \((\mathrm{CuFe}_{2}\mathrm{O}_{4})_{8}\), which means there are 8 formula units per unit cell. Each formula unit contains 1 Cu²⁺ ion, thus there are 8 Cu²⁺ ions per unit cell.
02

Calculate the mass of one unit cell

To calculate the mass of one unit cell, we will use the density (\(\rho\)) of copper ferrite provided in the exercise: \(\rho = \frac{m}{V} \Rightarrow m = \rho\cdot V\) The density is \(\rho = 5.40\ \mathrm{g/cm^{3}}\). Next, we need to find the volume (V) of one unit cell: \(V = \frac{1}{\text{number of atoms per unit cell}} = \frac{1}{8} \ \mathrm{cm^3}\) Now we can calculate the mass of one unit cell: \(m = \rho \cdot V = 5.40\ \mathrm{g/cm^{3}} \cdot \frac{1}{8}\ \mathrm{cm^3} = 0.675\ \mathrm{g}\)
03

Calculate the saturation magnetization per Cu²⁺ ion

The saturation magnetization is given as \(1.35\times10^5\ \mathrm{A/m}\). Since there are 8 Cu²⁺ ions per unit cell, the saturation magnetization per Cu²⁺ ion is: \(\frac{1.35\times10^5\ \mathrm{A/m}}{8\ \mathrm{Cu^{2+}\ ions}} = 1.6875\times10^4\ \mathrm{A/m\ per\ Cu^{2+}\ ion}\)
04

Calculate the number of Bohr magnetons per Cu²⁺ ion

The Bohr magneton (\(\mu_B\)) can be defined as: \(\mu_B = \frac{e \hbar}{2m_e} = 9.27\times10^{-24}\ \mathrm{Am^2}\) Now we can calculate the number of Bohr magnetons per Cu²⁺ ion by dividing the saturation magnetization per Cu²⁺ ion by the Bohr magneton value: \(\text{Number of Bohr magnetons per Cu²⁺ ion} = \frac{1.6875\times10^4\ \mathrm{A/m}}{9.27\times10^{-24}\ \mathrm{Am^2}} \approx 1.82\times10^{23}\) The number of Bohr magnetons associated with each Cu²⁺ ion is approximately \(1.82\times10^{23}\).

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Most popular questions from this chapter

Compute (a) the saturation magnetization and (b) the saturation flux density for iron, which has a net magnetic moment per atom of 2.2 Bohr magnetons and a density of \(7.87 \mathrm{g} / \mathrm{cm}^{3}\).

Briefly explain why the magnitude of the saturation magnetization decreases with increasing temperature for ferromagnetic materials, and why ferromagnetic behavior ceases above the Curie temperature.

There is associated with each atom in paramagnetic and ferromagnetic materials a net magnetic moment. Explain why ferromagnetic materials can be permanently magnetized whereas paramagnetic ones cannot.

Briefly describe the phenomenon of magnetic hysteresis, and why it occurs for ferromagnetic and ferrimagnetic materials.

Cite the differences between type I and type II superconductors.
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