Q51P - Fundamentals Of Physics

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 $\mathbf{58.71}\mathbf{ }\mathbf{g}\mathbf{/}\mathbf{mol}$ .)

Step-by-Step Solution

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 1: 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  $58.71\mathrm{g}/\mathrm{mole}$

Step 2: Understanding the concepts of magnetic dipole moment

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

Formulae:

 

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

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

Step 3: 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,  ${\mathrm{\mu }}_{\mathrm{B}}=5.15\times {10}^{-24} \mathrm{A} {\mathrm{m}}^2$.