Question

Calculate the number of Bohr magnetons per atom of iron, given that the saturation magnetization Ms=1.70×106 A/m, that
iron has a BCC crystal structure, and that that the edge length of the cubic unit cell is 0.287 nm.
a, 3.60
b, 2.16
c, 3.48
d, 2.84

Answer 1

Density of atoms in one cubic metre:

$$N=\frac2{a^3},$$

since there are 2 atoms per a unit cell in BCC crystal.

The number of Bohr magnetons per atom:

$$n_B=\frac{M_S}{\mu_BN}=\frac{M_S}{\mu_B(2/a^3)}=\frac{M_Sa^3}{2\mu_B},\\\space\\ n_B=\frac{1.7·10^6}{2(9.274·10^{-24})}=2.16.$$