Chapter 16.13, Problem 65AAP

To determine

The average magnetic moment for cobalt.

Answer to Problem 65AAP

The average magnetic moment for cobalt is $1.7\text{Bohr magnetons}/\text{atom}$.

Explanation of Solution

Write the expression to calculate average magnetic moment for cobalt $\left(N\right)$.

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

Here, saturation magnetization is $M_s$ 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:

Refer "Problem 16.64" in the textbook, obtain the values for lattice constant $\left(a\right)$ and height $\left(c\right)$ for cobalt.

  $\begin{array}{l}a=\text{0}\text{.25071}\text{nm}\\ c=\text{0}\text{.40686}\text{nm}\end{array}$

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 $1.42\times {10}^6\text{A}/\text{m}$ for $M_s$, $9.030\times {10}^{28}\text{atoms}/{\text{m}}^3$ for $\text{Atomic density}$, and $9.27\times {10}^{-24}\text{A}\cdot {\text{m}}^2$ for ${\mu }_B$ in Equation (I).

 $\begin{array}{c}N=\frac{1.42\times {10}^6\text{A}/\text{m}}{\left(9.27\times {10}^{-24}\text{A}\cdot {\text{m}}^2\right)\left(9.030\times {10}^{28}\text{atoms}/{\text{m}}^3\right)}\\ =1.7\text{Bohr magnetons}/\text{atom}\end{array}$

Thus, the average magnetic moment for cobalt is $1.7\text{Bohr magnetons}/\text{atom}$.