Chapter 11.13, Problem 75AAP

To determine

The linear densities of cerium oxide $\left({\text{CeO}}_2\right)$ in $\left[111\right]$ and $\left[110\right]$ directions in ions per nanometer.

Answer to Problem 75AAP

The linear densities of cerium oxide $\left({\text{CeO}}_2\right)$ in $\left[111\right]$ and $\left[110\right]$ directions in ions per nanometer are $\left(\text{1}{\text{.07Ce}}^{4+}+2.14{\text{O}}^{2-}\right)/\text{nm}$ and $\text{2}{\text{.62Ce}}^{4+}/\text{nm}$ respectively.

Explanation of Solution

Write the expression to calculate lattice constant of cerium oxide structure $\left(a\right)$.

 $a=\frac{4}{\sqrt{3}}\left(r_{{\text{Ce}}^{4+}}+R_{{\text{O}}^{2-}}\right)$                                                                                                ...... (I)

Here, ionic radius of ${\text{Ce}}^{4+}$ ion is $r_{{\text{Ce}}^{4+}}$ and ionic radius of ${\text{O}}^{2-}$ ion is $R_{{\text{O}}^{2-}}$.

Write the expression to calculate linear density of cerium oxide $\left({\text{CeO}}_2\right)$ in $\left[111\right]$ direction in ions per nanometer $\left({\rho }_L\right)$.

 ${\rho }_L=\frac{n_{{\text{Ce}}^{4+}}{\text{Ce}}^{4+}+n_{{\text{O}}^{2-}}{\text{O}}^{2-}}{\sqrt{3}a}$                                                                                       ...... (II)

Here, number of atoms of ${\text{Ce}}^{4+}$ ion is $n_{{\text{Ce}}^{4+}}$ and number of atoms of ${\text{O}}^{2-}$ ion is $n_{{\text{O}}^{2-}}$.

Write the expression to calculate linear density of cerium oxide $\left({\text{CeO}}_2\right)$ in $\left[110\right]$ direction in ions per nanometer $\left({\rho }_L\right)$.

 ${\rho }_L=\frac{n_{{\text{Ce}}^{4+}}{\text{Ce}}^{4+}}{\sqrt{2}a}$                                                                                       ...... (II)

Conclusion:

Substitute $0.102\text{nm}$ for $r_{{\text{Ce}}^{4+}}$ and $0.132\text{nm}$ for $R_{{\text{O}}^{2-}}$ in Equation (I).

 $\begin{array}{c}a=\frac{4}{\sqrt{3}}\left(0.102\text{nm}+0.132\text{nm}\right)\\ =0.540\text{nm}\end{array}$

For cerium oxide, there will be 1 cerium ion and 2 oxygen ions present in the $\left[111\right]$ direction of the unit cell of cerium oxide.

Substitute $1$ for $n_{{\text{Ce}}^{4+}}$, $2$ for $n_{{\text{O}}^{2-}}$ and $0.540\text{nm}$ for $a$ in Equation (II).

 $\begin{array}{c}{\rho }_L=\frac{\left(1\right){\text{Ce}}^{4+}+\left(2\right){\text{O}}^{2-}}{\sqrt{3}\left(0.540\text{nm}\right)}\\ =\left(\text{1}{\text{.07Ce}}^{4+}+2.14{\text{O}}^{2-}\right)/\text{nm}\end{array}$

For cerium oxide, there will be 2 cerium ions and 0 oxygen ions present in the $\left[110\right]$ direction of the unit cell of cerium oxide.

Substitute $2$ for $n_{{\text{Ce}}^{4+}}$ and $0.540\text{nm}$ for $a$ in Equation (III).

 $\begin{array}{c}{\rho }_L=\frac{\left(2\right){\text{Ce}}^{4+}}{\sqrt{2}\left(0.540\text{nm}\right)}\\ =\text{2}{\text{.62Ce}}^{4+}/\text{nm}\end{array}$

Thus, the linear densities of cerium oxide $\left({\text{CeO}}_2\right)$ in $\left[111\right]$ and $\left[110\right]$ directions in ions per nanometer are $\left(\text{1}{\text{.07Ce}}^{4+}+2.14{\text{O}}^{2-}\right)/\text{nm}$ and $\text{2}{\text{.62Ce}}^{4+}/\text{nm}$ respectively.