Discussion
Calculate the saturation magnetization (Msat) for bcc iron of lattice parameter 2.866 Å.
Answer: C
The key's answer is \(3.15 \times 10^{6}\) A/m.
A bcc cell holds 2 atoms, so the number of atoms per unit volume is
\[ \begin{aligned} N &= \frac{2}{(2.866 \times 10^{-10})^{3}} = 8.50 \times 10^{28}\ \text{m}^{-3} \\ M_{sat} &= N n_{B}\mu_{B} = 8.50 \times 10^{28}(4)(9.274 \times 10^{-24}) \\ &\approx 3.15 \times 10^{6}\ \text{A/m} \end{aligned} \]
The key takes 4 Bohr magnetons per atom, the spin-only moment of the four unpaired 3d electrons of an isolated Fe atom. In metallic iron the measured moment is about 2.2 Bohr magnetons, which gives about \(1.7 \times 10^{6}\) A/m; the paper's options follow the four-electron model. 0.79 and 4.73 correspond to one and six magnetons per atom.
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