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Chapter 20: Problem 14

The formula for yttrium iron garnet \(\left(\mathrm{Y}_{3} \mathrm{Fe}_{5} \mathrm{O}_{12}\right)\) may be written in the form \(\mathrm{Y}_{3}^{c} \mathrm{Fe}_{2}^{a} \mathrm{Fe}_{3}^{d} \mathrm{O}_{12}\), where the superscripts \(a, c\), and \(d\) represent different sites on which the \(\mathrm{Y}^{3+}\) and \(\mathrm{Fe}^{3+}\) ions are located. The spin magnetic moments for the \(\mathrm{Y}^{3+}\) 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 \(\mathrm{Y}^{3+}\) ion, given the following information: (1) each unit cell consists of eight formula \(\left(\mathrm{Y}_{3} \mathrm{Fe}_{5} \mathrm{O}_{12}\right)\) units; (2) the unit cell is cubic with an edge length of \(1.2376 \mathrm{~nm} ;\) (3) the saturation magnetization for this material is. \(1.0 \times 10^{4} \mathrm{~A} / \mathrm{m}\); and (4) there are five Bohr magnetons associated with each \(\mathrm{Fe}^{3+}\) ion

Short Answer

Expert verified
Question: Calculate the number of Bohr magnetons associated with each Yttrium ion (Y³⁺) in a Yttrium Iron Garnet unit cell, with the given saturation magnetization of 1.0 x 10^4 A/m, edge length of 1.2376 nm, and considering that there are five Bohr magnetons associated with each Fe³⁺ ion.

Step by step solution

01

Calculating the volume of the unit cell

To calculate the volume of the unit cell, we will use the formula for the volume of a cube: volume = a³ where a is the edge length. a = 1.2376 nm = 1.2376 x 10^(-9) m volume = (1.2376 x 10^(-9))³ m³
02

Calculate the total magnetic moment in the unit cell

Using the saturation magnetization and the cell volume, we can find the total magnetic moment in the unit cell. M_s = 1.0 x 10^4 A/m (saturation magnetization) Total magnetic moment = M_s x Volume Total magnetic moment = (1.0 x 10^4 A/m) x (1.2376 x 10^(-9))³ m³
03

Calculate the magnetic moment of one formula unit

According to the problem statement, each unit cell consists of eight Y3Fe5O12 formula units. Hence, we need to divide the total magnetic moment by 8 to find the magnetic moment of one formula unit. Magnetic moment of one formula unit = Total magnetic moment / 8
04

Calculating the magnetic moments of Fe³⁺ ions

We know that there are five Bohr magnetons associated with each Fe³⁺ ion. As there are 5 Fe³⁺ ions in each Y3Fe5O12 formula unit, the total number of Bohr magnetons in all Fe³⁺ ions in a formula unit is 5 times 5, which is 25.
05

Calculating the number of Bohr magnetons associated with each Yttrium ion

Since the total number of Bohr magnetons in a Y3Fe5O12 formula unit can be divided between its 3 Yttrium ions, we can find the number of Bohr magnetons per Yttrium ion by dividing the difference in magnetic moments of the formula unit and the Fe³⁺ ions by 3. Bohr magnetons in Y3Fe5O12 formula unit = Magnetic moment of one formula unit - (5 x Bohr magnetons in Fe³⁺ ions) Bohr magnetons per Y³⁺ ion = (Bohr magnetons in Y3Fe5O12 formula unit) / 3 By calculating the values in the steps above, we will get the number of Bohr magnetons associated with each Yttrium ion.

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Most popular questions from this chapter

Estimate saturation values of \(H\) for singlecrystal iron in [100], [110], and [111] directions.

The magnetization within a bar of some metal alloy is \(3.2 \times 10^{5} \mathrm{~A} / \mathrm{m}\) at an \(H\) field of \(50 \mathrm{~A} / \mathrm{m}\). Compute the following: (a) the magnetic susceptibility, (b) the permeability, and (c) the magnetic flux density within this material. (d) What type(s) of magnetism would you suggest as being displayed by this material? Why?

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.

(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

It is possible to express the magnetic susceptibility \(\chi_{m}\) in several different units. For the discussion of this chapter, \(\chi_{m}\) was used to designate the volume susceptibility in SI units, that is, the quantity that gives the magnetization per unit volume \(\left(\mathrm{m}^{3}\right)\) of material when multiplied by \(H\). The mass susceptibility \(\chi_{m}(\mathrm{~kg})\) yields the magnetic moment (or magnetization) per kilogram of material when multiplied by \(H ;\) similarly, the atomic susceptibility \(\chi_{m}\) (a) gives the magnetization per kilogram-mole. The latter two quantities are related to \(\chi_{m}\) through the relationships $$ \begin{aligned} &\chi_{m}=\chi_{m}(\mathrm{~kg}) \times \text { mass density (in } \mathrm{kg} / \mathrm{m}^{3} \text { ) } \\ &\left.\chi_{m}(\mathrm{a})=\chi_{m}(\mathrm{~kg}) \times \text { atomic weight (in } \mathrm{kg}\right) \end{aligned} $$ When using the cgs-emu system, comparable parameters exist, which may be designated by \(\chi_{m}^{\prime}, \chi_{m}^{\prime}(\mathrm{g})\), and \(\chi_{m}^{\prime}(\mathrm{a})\); the \(\chi_{m}\) and \(\chi_{m}^{\prime}\) are related in accordance with Table 20.1. From Table \(20.2, \chi_{m}\) for silver is \(-2.38 \times 10^{-5}\); convert this value into the other five susceptibilities
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