Question

The saturation magnetization ${\mathbf{M}}_{\mathbf{max}}$ of the ferromagnetic metal nickel is$\mathbf{4}\mathbf{.}\mathbf{70}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{A}\mathbf{/}\mathbf{m}$ . Calculate the magnetic dipole moment of a single nickel atom. (The density of nickel is$\mathbf{8}\mathbf{.}\mathbf{90}\mathbf{}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$, and its molar mass is$\mathbf{58.71}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol}$ .)

Short answer

Magnetic dipole moment of a single nickel atom is, ${\mathrm{\mu }}_{\mathrm{B}}=5.15\times {10}^{-24}\mathrm{A}{\mathrm{m}}^2$.

Step-by-step solution

Listing the given quantities

Saturation magnetization ${\mathrm{M}}_{\max }=4.70\times {10}^5\mathrm{A}/\mathrm{m}$

Density of nickel is$\mathrm{\rho }=8.90\mathrm{g}/{\mathrm{cm}}^3$

Molar mass of nickel is $58.71\mathrm{g}/\mathrm{mole}$

Understanding the concepts of magnetic dipole moment

We use the concept of magnetic dipole moment. Using the equations, we findthenumber of atoms per unit volume, and then we can findthemagnetic dipole moment.

Formulae:

${\mathrm{\mu }}_{\mathrm{B}}=\frac{{\mathrm{M}}_{\max }}{\mathrm{n}}$

$\mathrm{n}=\frac{{\mathrm{\rho N}}_{\mathrm{A}}}{\mathrm{M}}$

Calculations of the magnetic dipole moment of a single nickel atom

We find the number of atoms per unit volume.

Using equation

$\begin{array}{c}\mathrm{n}=\frac{{\mathrm{\rho N}}_{\mathrm{A}}}{\mathrm{M}}\\ =\frac{\left(8.90\right)(6.023\times {10}^{23})}{58.71}\\ =9.126\times {10}^{22}\frac{\mathrm{atoms}}{{\mathrm{cm}}^3}\end{array}$

We can convert it in $\mathrm{atoms}/{\mathrm{m}}^3$ .We can write

$\begin{array}{c}\mathrm{n}=9.126\times {10}^{22}\frac{\mathrm{atoms}}{{\mathrm{cm}}^3}\times {\left(\frac{1\mathrm{cm}}{{10}^{-2}\mathrm{m}}\right)}^3\\ =9.126\times {10}^{22}\frac{\mathrm{atoms}}{{\mathrm{cm}}^3}\times 1\frac{{\mathrm{cm}}^3}{{10}^{-6}{\mathrm{m}}^3}\\ =9.126\times {10}^{28}\frac{\mathrm{atoms}}{{\mathrm{m}}^3}\end{array}$

Substituting this value in magnetic dipole moment,

$\begin{array}{c}{\mathrm{\mu }}_{\mathrm{B}}=\frac{{\mathrm{M}}_{\max }}{\mathrm{n}}\\ =\frac{4.70\times {10}^5}{9.126\times {10}^{28}}\\ =5.15\times {10}^{-24}\mathrm{A}{\mathrm{m}}^2\end{array}$

Magnetic dipole moment of a single nickel atom is, ${\mathrm{\mu }}_{\mathrm{B}}=5.15\times {10}^{-24}\mathrm{A}{\mathrm{m}}^2$.