Magnetic Moment of a Ferrite Molecule

MAGNETIC MOMENT OF A FERRITE MOLECULE

Bohr Magneton

The orbital magnetic moment and the spin magnetic moment of an electron in an atom can be expressed in terms of smallest atomic unit of magnetic moment called Bohr magneton.

 1 Bohr Magneton = e / 2m

 ⇒ μB⇒ 9.27 × 10^‒24 Am²

Saturation magnetization of a ferrite molecule can be calculated from the number of unpaired spins of Fe²⁺ and Fe³⁺

Let us consider an example of Ferrite say Ferrous ferrite having the formula Fe²⁺ Fe₂³⁺O₄ for calculating the magnetic moment.

In Ferrous ferrite we have two types of ions viz. Fe²⁺ and Fe³⁺

(i) Here Fe²⁺ ions has six electrons in 3d shell. Out of 6 electrons two electrons are paired with each other and hence left with 4 unpaired electrons as shown in Fig. 1.18.

The Fe²⁺ gives rise to 4 Bohr magneton.

(ii) Fe³⁺ ions has five electrons in 3d shell and hence all these 5 are unpaired electrons as shown in Fig. 1.19.

∴ The Fe³⁺ gives rise to 5 Bohr magneton.

Since we have two Fe³⁺, totally, the Fe₂³⁺ gives rise to 2×5=10 Bohr magnetons.

 ∴ Total magnetisation of Fe²⁺ + Fe₂³⁺ = 4+ 10 = 14 Bohr magneton. (i.e.,) 14 μB.

Theoretically we get 14 μB but experimetally the total magnetic moment got is only 4.08 μB. The reason for this discrepancy is as follows:

If all the spins are aligned parallel then we will get the total magnetisation as 14μB. But in ferrites half of the magnetic spins of Fe₂³⁺ ions are parallel to one direction and the remaining half of Fe₂³⁺ ions are parallel in opposite direction as shown in Fig. 1.20 and hence they cancel each other. Therefore, the net magnetic moment is only due to Fe²⁺ ions alone (i.e.,) hence we get the total magnetisation as 4μB, which has a good agreement with the experiment value i.e., 4.08 μB.

Similarly we can calculate the total magnetic moment of any ferrite molecule with respect to the number of unpaired electrons in the divalent metal ions.