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Answer:
[tex]n_b=0.947 Bohr\ magnetons/atom[/tex]
Explanation:
From the question we are told that:
Atomic radius [tex]r=0.126 nm[/tex]
Saturation flux density [tex]B= 0.69 T[/tex]
Generally the equation for number of Bohr magnetons per atom is mathematically given by
[tex]n_b=\frac{B_s*Vc}{\mu_0*\mu_b}[/tex]
Where
[tex]\mu_0=Magnetic\ permeability[/tex]
[tex]\mu_0=1.257 * 10^{-6}[/tex]
[tex]\mu_b=Magnetic\ Moment\ of\ an\ Atom[/tex]
[tex]\mu_b=9.27* 10^{-24}[/tex]
[tex]Vc=d^3[/tex]
[tex]Vc=(2(0.126*10^-9))^3[/tex]
[tex]Vc=1.6*10^{-29}[/tex]
Therefore the number of Bohr magnetons per atom for this material
[tex]n_b=\frac{0.69*1.6*10^{-29}}{1.257 * 10^{-6}*9.27* 10^{-24}}[/tex]
[tex]n_b=0.947 Bohr\ magnetons/atom[/tex]