Question

Assume there exists some hypothetical metal that exhibits ferromagnetic behavior and that has (1) a simple cubic crystal structure, (2) an atomic radius of 0.154 nm, and (3) a saturation flux density of 0.83 tesla. Determine the number of Bohr magnetons per atom for this material. A: What is the volume, in m3, for this unit cell?B: How many atoms are there per m3 in this material?C: What is the number of Bohr magentons per atom for this material?

Answer 1

a. 2.9218x10^(-29) m^3

b. 7.1230x10^(28) atoms/m^3

c. 2.0812 BM/atom

Step-by-step explanation:

Atomic radius r = 0.154 nm

saturation flux density Bs = 0.83 tesla

- the formula for the volume of the simple cubic crystal (Vc) = a^3 = (2r)^3

= (2 x 0.154x10^-9)^3

= 2.9218x10^-29 m^3

- The formula for the Bohr magneton per atom with respect to VC, Bs, permeability of the vacuum Uo and magnetic moment per Bohr magneton Ub is;

Bohr magneton per atom nb = (Bs x VC) / (Ub x Uo)

= (0.83 x 2.9218x10^-29) / (9.27x10^-24) x(1.257x10^-6)

= 2.4251x10^-29 / 1.1652x10^-29

= 2.0812 BM/atom

- Number of atoms per Vc, N = nb / Vc

= 2.0812 / 2.9218x10^-29 = 7.1230x10^28 atoms