1) The saturation magnetic induction of Nickel is 0.65 Wb/m^2. If the density of Nickel is 8906 kg/m^3 and atomic weight is 58.7, calculate the magnetic moment of Nickel atom in Bohr magneton.
I solved it in the following way:
Let M(s) & B(s) be the saturation magnetization and magnetic flux density respectively, p be the magnetic moment of the atom, m(o) be the permeability of free space,p( is the Bohr magneton and N be the number of atoms per unit volume.
B(s) = m(o)xM(s)
M(s) = 5.17 x 10^5 A/m
Now,
Nickel is a face centered cubic(fcc) lattice and its lattice spacing is 0.3524 nm
N = 4/(0.3524)^3
= 9 x 10^28 atoms/m^3
M(s) = p x N
p = M(s)/N
In terms of Bohr magneton,
p = M(s)/[N x p(]
= 0.61 Bohr magneton
Though my answer is in agreement with the book answer, I haven’t used the density and atomic weight values given in the question. Is there any mistake in my solution?

No, you haven't made any mistakes, you've just used alternate data as a starting point. The problem didn't give you the crystal lattice spacing nor the unit cell type, so you must have got them from some outside reference. Of course, this is one way to compute N.

Another way, using just the data given, would be to simply divide the density by the mass of 1 Ni atom. Since this mass is 58.7grams / Avogadro's number, you have everything you need to find N without going outside the given data of the problem (assuming you know Avogadro's number, which is a safe bet).

Hope this helps!

--Stuart Anderson

That was of great help to me.Thanx.