Chapter 16.13, Problem 63AAP

To determine

The theoretical value for the saturation magnetization for cobalt metal.

Answer to Problem 63AAP

The theoretical value for the saturation magnetization for cobalt metal is $2.51\times {10}^6\text{A}/\text{m}$.

Explanation of Solution

Write the expression to calculate saturation magnetization for cobalt metal $\left(M_s\right)$.

  $M_s=\left(\text{Atomic density}\right)N{\mu }_B$                                                                                  (I)

Here, magnetic moment is $N$ and Bohr magneton is ${\mu }_B$.

Write the expression to calculate atomic density of the cobalt.

  $\text{Atomic density}=\frac{n}{V}$                                                                                             (II)

Here, number of atoms per unit cell is $n$ and volume of the unit cell is $V$.

Write the expression to calculate volume of the unit cell for the cobalt $\left(V\right)$.

  $\begin{array}{c}V=\text{Area}\times \text{height}\\ =\left(3a^2\sin 60°\right)\times c\end{array}$                                                                                            (III)

Here, lattice constant of the unit cell is $a$ and height of the unit cell is $c$.

Conclusion:

The magnetic moment of the cobalt atom is considered to be $3\text{Bohr magnetons}$.

Substitute $\text{0}\text{.25071}\text{nm}$ for $a$ and $\text{0}\text{.40686}\text{nm}$ for $c$ in Equation (III).

 $\begin{array}{c}V=\left(3{\left(\text{0}\text{.25071}\text{nm}\right)}^2\sin 60°\right)\times \text{0}\text{.40689}\text{nm}\\ =\left(3{\left(\text{0}\text{.25071}\text{nm}\times \frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)}^2\sin 60°\right)\times \left(\text{0}\text{.40689}\text{nm}\times \frac{1\times {10}^{-9}\text{m}}{1\text{nm}}\right)\\ =6.644\times {10}^{-29}{\text{m}}^3\end{array}$

Substitute $6\text{atoms}/\text{unit cell}$ for n and $6.644\times {10}^{-29}{\text{m}}^3$ for $V$ in Equation (II).

 $\begin{array}{c}\text{Atomic density}=\frac{6\text{atoms}/\text{unit cell}}{6.644\times {10}^{-29}{\text{m}}^3}\\ =9.030\times {10}^{28}\text{atoms}/{\text{m}}^3\end{array}$

The value of Bohr magneton $\left({\mu }_B\right)$ is taken to be $9.27\times {10}^{-24}\text{A}\cdot {\text{m}}^2$.

Substitute $9.030\times {10}^{28}\text{atoms}/{\text{m}}^3$ for $\text{Atomic density}$, $3\text{Bohr magnetons}/\text{atom}$ for $N$ and $9.27\times {10}^{-24}\text{A}\cdot {\text{m}}^2$ for ${\mu }_B$ in Equation (I).

 $\begin{array}{c}{M}_s=\left(9.030\times {10}^{28}\text{atoms}/{\text{m}}^3\right)\left(3\text{Bohr magnetons}/\text{atom}\right)\left(9.27\times {10}^{-24}\text{A}\cdot {\text{m}}^2\right)\\ =2.51\times {10}^6\text{A}/\text{m}\end{array}$

Thus, the theoretical value for the saturation magnetization for cobalt metal is $2.51\times {10}^6\text{A}/\text{m}$.