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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
Based on the given information and calculations, the estimated number of Bohr magnetons associated with each Cu²⁺ ion in copper ferrite is approximately 1.819 x 10²⁷.

Step by step solution

01

Calculate the number of copper ions in a unit cell

Given that the chemical formula for copper ferrite is \(\left(\mathrm{CuFe}_{2} \mathrm{O}_{4}\right)_{8}\), we can see that there are eight formula units per unit cell. Since there is one \(\mathrm{Cu}^{2+}\) ion per formula unit, there are a total of 8 \(\mathrm{Cu}^{2+}\) ions in a unit cell.
02

Calculate the magnetization contributed by the copper ions in a unit cell

We are given the saturation magnetization of the material as \(1.35 \times 10^{5} \mathrm{~A} / \mathrm{m}\). We know that the magnetization comes from the \(\mathrm{Cu}^{2+}\) ions in the unit cell, so the saturation magnetization corresponds to the magnetization contributed by all 8 copper ions in a unit cell. Therefore, the total magnetization contributed by the 8 copper ions in a unit cell is \(1.35 \times 10^{5} \mathrm{~A} / \mathrm{m}\).
03

Calculate the individual magnetization of each copper ion

To find the individual magnetization of each \(\mathrm{Cu}^{2+}\) ion, we need to divide the total magnetization by the number of copper ions in a unit cell. Individual magnetization of a single \(\mathrm{Cu}^{2+}\) ion = (Total magnetization) / (Number of copper ions) = \((1.35 \times 10^{5} \mathrm{~A} / \mathrm{m}) / 8 = 1.6875 \times 10^{4} \mathrm{~A} / \mathrm{m}\)
04

Divide the individual magnetization by the magnetization of a single Bohr magneton

We know that the magnetization of a single Bohr magneton is \(9.27 \times 10^{-24} \mathrm{~A} \cdot \mathrm{m}^{2}\). To calculate the number of Bohr magnetons per copper ion, we need to divide the individual magnetization of a \(\mathrm{Cu}^{2+}\) ion by the magnetization of a single Bohr magneton: Number of Bohr magnetons per \(\mathrm{Cu}^{2+}\) ion = (Individual magnetization of a single \(\mathrm{Cu}^{2+}\) ion) / (Magnetization of a single Bohr magneton) = \((1.6875 \times 10^{4} \mathrm{~A} / \mathrm{m}) / (9.27 \times 10^{-24} \mathrm{~A} \cdot \mathrm{m}^{2})\)
05

Estimate the number of Bohr magnetons per \(\mathrm{Cu}^{2+}\) ion

To find the estimated number of Bohr magnetons per \(\mathrm{Cu}^{2+}\) ion, we need to carry out the division from Step 4: Number of Bohr magnetons per \(\mathrm{Cu}^{2+}\) ion \(\approx \frac{1.6875 \times 10^{4} \mathrm{~A} / \mathrm{m}}{9.27 \times 10^{-24} \mathrm{~A} \cdot \mathrm{m}^{2}} = \frac{1.6875 \times 10^{4}}{9.27 \times 10^{-24}}\) Number of Bohr magnetons per \(\mathrm{Cu}^{2+}\) ion \(\approx 1.819 \times 10^{27}\) The estimated number of Bohr magnetons associated with each \(\mathrm{Cu}^{2+}\) ion is approximately \(1.819 \times 10^{27}\).

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Most popular questions from this chapter

Confirm that there are \(1.72\) Bohr magnetons associated with each cobalt atom, given that the saturation magnetization is \(1.45 \times 10^{6} \mathrm{~A} / \mathrm{m}\), that cobalt has an HCP crystal structure with an atomic radius of \(0.1253 \mathrm{~nm}\) and a \(c / a\) ratio of \(1.623\).

The formula for samarium iron garnet \(\left(\mathrm{Sm}_{3} \mathrm{Fe}_{5} \mathrm{O}_{12}\right)\) may be written in the form \(\mathrm{Sm}_{3}^{c} \mathrm{Fe}_{2}^{a} \mathrm{Fe}_{3}^{d} \mathrm{O}_{12}\), where the superscripts \(a, c\), and \(d\) represent different sites on which the \(\mathrm{Sm}^{3+}\) and \(\mathrm{Fe}^{3+}\) ions are located. The spin magnetic moments for the \(\mathrm{Sm}^{3+}\) and \(\mathrm{Fe}^{3+}\) ions positioned in the \(a\) and \(c\) sites are oriented parallel to one another and antiparallel to the \(\mathrm{Fe}^{3+}\) ions in \(d\) sites. Compute the number of Bohr magnetons associated with each \(\mathrm{Sm}^{3+}\) ion, given the following information: (1) each unit cell consists of eight formula \(\left(\mathrm{Sm}_{3} \mathrm{Fe}_{5} \mathrm{O}_{12}\right)\) units; (2) the unit cell is cubic with an edge length of \(1.2529 \mathrm{~nm} ;\) (3) the saturation magnetization for this material is \(1.35 \times 10^{5} \mathrm{~A} / \mathrm{m} ;\) and (4) there are 5 Bohr magnetons associated with each \(\mathrm{Fe}^{3+}\) ion.

Briefly describe the phenomenon of magnetic hysteresis and why it occurs for ferro- and ferrimagnetic materials.

(a) Explain the two sources of magnetic moments for electrons. (b) Do all electrons have a net magnetic moment? Why or why not? (c) Do all atoms have a net magnetic moment? Why or why not?

Cite the differences between type I and type II superconductors.
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