The saturation magnetization of nickel measured at room temperature is 5.1×105 A/m.

Question

(a) Calculate the average number of Bohr magnetons on a nickel atom, knowing that nickel has an FCC structure with an atom at 0,0,0. At room temperature the lattice parameter is 0.352 nm.
(b) Compare your answer to part (a) with the theoretical number of Bohr magnetons based upon the atomic electron structure of nickel, and explain any possible difference.

Answer 1

The saturation magnetization is calculated from Equation 17.12 assuming that at

saturation all of the spins are oriented parallel to the applied magnetic field. The value of Ms is given, the value of nB is the unknown, and Nv is calculated from the information given about the Ni crystal structure. At room temperature, the lattice parameter of Ni is 0.352 nm and the lattice is face centered cubic with atoms at the position 0,0,0. There are 4 atoms per FCC unit cell. The number of atoms per unit volume (Nv) is









In metallic Ni atoms there are on the average 0.60 aligned spins. In atomic Ni which is atomic number 28 there are eight 3d electrons. According to Hund’s rule the first 5 will be all parallel and the next 3 will be antiparallel to the first 5. There are two remaining parallel 3d electrons in a free atom of Ni. A free atom of Ni should have two Bohr magnetons. In metallic Ni at room temperature, however, two is not the average number of aligned spins, because the 3d electrons are involved in the bonding in the transition metals and thermal vibrations at room temperature also disorder the spins.