Chapter 20: Problem 8

Confirm that there are \(2.2\) Bohr magnetons associated with each iron atom, given that the saturation magnetization is \(1.70 \times 10^{6} \mathrm{~A} / \mathrm{m}\), that iron has a BCC crystal structure, and that the unit cell edge length is \(0.2866 \mathrm{~nm}\).

Short Answer

Expert verified

Question: Confirm that there are approximately 2.2 Bohr magnetons associated with each iron atom based on the given saturation magnetization. Answer: By calculating the theoretical saturation magnetization for 2.2 Bohr magnetons per iron atom, we find a value of \(1.72 \times 10^{6} \mathrm{A/m}\). This is very close to the given saturation magnetization of \(1.70 \times 10^{6} \mathrm{A/m}\), with a difference of about \(1.2 \%\). Therefore, we can confirm that there are indeed approximately \(2.2\) Bohr magnetons associated with each iron atom.

Step by step solution

Calculate the number of iron atoms per unit volume

The Iron BCC unit cell has \(2\) atoms in it, and the edge length of the unit cell is \(0.2866~nm\). Let's calculate the unit volume and then find the number of iron atoms per unit volume. The volume of the unit cell is \(V = a^3\) where \(a\) is the edge length, which gives: \(V = (0.2866 \times 10^{-9}\mathrm{m})^3 = 2.3540 \times 10^{-29}\mathrm{m^3}\). Since there are \(2\) atoms in the unit cell, the number of iron atoms per unit volume is: \(n_{Fe} = \frac{2}{V} = \frac{2}{2.3540 \times 10^{-29}\mathrm{m^3}}= 8.49 \times 10^{28}\mathrm{m^{-3}}\).

Calculate the theoretical saturation magnetization

The Bohr magneton is given by \(\mu_{B} = 9.27 \times 10^{-24} \mathrm{Am^2}\). Since we are trying to confirm that there are 2.2 Bohr magnetons associated with each iron atom, we will now calculate the theoretical saturation magnetization as the product of the number of iron atoms per unit volume, Bohr magneton, and the assumed number of Bohr magnetons per atom. It is given by: \(M_{th} = n_{Fe} \times \mu_{B} \times 2.2 = (8.49 \times 10^{28}\mathrm{m^{-3}}) \times (9.27 \times 10^{-24} \mathrm{Am^2}) \times 2.2 = 1.72 \times 10^{6} \mathrm{A/m}\).

Compare the theoretical and given saturation magnetization

Now, let's compare the calculated theoretical saturation magnetization with the given saturation magnetization: Given saturation magnetization = \(1.70 \times 10^{6} \mathrm{A/m}\). Theoretical saturation magnetization = \(1.72 \times 10^{6} \mathrm{A/m}\). The difference between the theoretical and given saturation magnetizations is very small (\(\sim 1.2 \%\)), and the values are reasonably close. Therefore, we can confirm that there are indeed approximately \(2.2\) Bohr magnetons associated with each iron atom.

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Confirm that there are \(2.2\) Bohr magnetons associated with each iron atom, given that the saturation magnetization is \(1.70 \times 10^{6} \mathrm{~A} / \mathrm{m}\), that iron has a BCC crystal structure, and that the unit cell edge length is \(0.2866 \mathrm{~nm}\).

Short Answer

Step by step solution

Calculate the number of iron atoms per unit volume

Calculate the theoretical saturation magnetization

Compare the theoretical and given saturation magnetization

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