Chapter 43, Problem 56AP

(a)

To determine

To show that the force exerted on an ionic solid is $F=-\alpha k_e\frac{e^2}{r}\left[1-{\left(\frac{r_0}{r}\right)}^{m-1}\right]$

Answer to Problem 56AP

It is showed that the force exerted on an ionic solid is $F=-\alpha k_e\frac{e^2}{r}\left[1-{\left(\frac{r_0}{r}\right)}^{m-1}\right]$.

Explanation of Solution

Write the expression for the total potential energy of an ionic solid.

  $U_{\text{total}}=-\alpha \frac{k_e{e}^2}{r}+\frac{B}{r^m}$                                                                                                   (I)

Here, $U_{\text{total}}$ is the potential energy of the solid, $\alpha$ is the Madelung constant, $k_e$ is the coulomb constant, $e$ is the electronic charge, $B$ is a constant and $m$ is a constant.

The ionic bond formation will take place at equilibrium spacing where potential energy takes it minimum value.

Write the expression for the interatomic force.

  $F=-\frac{dU}{dr}$                                                                                                                 (II)

Let $r_0$ is the equilibrium spacing.

Write the expression for the equilibrium condition for the force

  $F=-{\frac{dU}{dr}|}_{r=r_0}$                                                                                                             (III)

Here, $F$ is interatomic force between the ions and $r_0$ is he equilibrium forces.

Conclusion:

Take the derivative of the equation (II) to get expression of force.

  $\begin{array}{c}F=-\frac{d\left(-\alpha \frac{k_e{e}^2}{r}+\frac{B}{r^m}\right)}{dr}\\ =-\left[-\left(-\alpha \frac{k_e{e}^2}{r^2}\right)+\left(-\frac{mB}{r^{m+1}}\right)\right]\\ =\left(-\alpha \frac{k_e{e}^2}{r^2}\right)+\left(\frac{mB}{r^{m+1}}\right)\end{array}$                                                                               (IV)

Use equation (II) in equation (III) to get value of $B$.

  $\begin{array}{c}{\left[\left(-\alpha \frac{k_e{e}^2}{r^2}\right)+\left(\frac{mB}{r^{m+1}}\right)\right]}_{r=r_0}=0\\ \left[\left(-\alpha \frac{k_e{e}^2}{r_0^2}\right)+\left(\frac{mB}{r_0^{m+1}}\right)\right]=0\end{array}$

Rearrange above equation to get $B$.

  $\begin{array}{c}\left(\frac{mB}{r_0^{m+1}}\right)=\left(\alpha \frac{k_e{e}^2}{r_0^2}\right)\\ B=\left(\frac{\alpha }{m}\frac{k_e{e}^2}{r_0^2}\left(r_0^{m+1}\right)\right)\\ =\alpha \frac{k_e{e}^2}{m}{r}_0^{m-1}\end{array}$

Substitute $\alpha \frac{k_e{e}^2}{m}{r}_0^{m-1}$ for $B$ in equation (I) to get $F$.

  $\begin{array}{c}F=\left(-\alpha \frac{k_e{e}^2}{r^2}\right)+\left(\frac{m\alpha \frac{k_e{e}^2}{m}{r}_0^{m-1}}{r^{m+1}}\right)\\ =-\alpha \frac{k_e{e}^2}{r^2}\left[1-{\left(\frac{r_0}{r}\right)}^{m-1}\right]\end{array}$                                                                              (V)

Therefore, it is showed that the force exerted on an ionic solid is $F=-\alpha k_e\frac{e^2}{r}\left[1-{\left(\frac{r_0}{r}\right)}^{m-1}\right]$.

(b)

To determine

To show that the restoring force experienced by ion is $F=-Ks$, where $K=\frac{\alpha k_e{e}^2}{r_0^3}\left(m-1\right)$, by assuming that an ion in the solid is displaced a small distance $s$ from $r_0$.

Answer to Problem 56AP

It is showed that the restoring force experienced by ion is $F=-Ks$, where $K=\frac{\alpha k_e{e}^2}{r_0^3}\left(m-1\right)$.

Explanation of Solution

Rewrite expression for the force from equation (V).

  $F=-\alpha k_e\frac{e^2}{r}\left[1-{\left(\frac{r_0}{r}\right)}^{m-1}\right]$

Assume that the ion in the solid is displaced a small distance $s$ from $r_0$.

Conclusion:

Let $r=r_0+s$, so that $r_0=r-s$.

Substitute $r-s$ for $r_0$ in above equation to get expression for the force.

  $\begin{array}{c}F=-\alpha \frac{k_e{e}^2}{r^2}\left[1-{\left(\frac{r-s}{r}\right)}^{m-1}\right]\\ =-\alpha \frac{k_e{e}^2}{r^2}\left[1-{\left(1-\frac{s}{r}\right)}^{m-1}\right]\end{array}$                                                                                    (VI)

Expand ${\left(1-\frac{s}{r}\right)}^{m-1}$ using binomial expansion.

   $\begin{array}{c}{\left(1-\frac{s}{r}\right)}^{m-1}=1-\left(m-1\right)\frac{s}{r}\\ =1-\left(m-1\right)\frac{s}{r}+\frac{\left(m-1\right)\left(m\right)}{2!}{\left(\frac{s}{r}\right)}^2\mathrm{......}\end{array}$

Take the first two terms of the binomial expansion.

Substitute $1-\left(m-1\right)\frac{s}{r}$ for ${\left(1-\frac{s}{r}\right)}^{m-1}$ in equation (VI) to get $F$.

  $\begin{array}{c}F\approx -\alpha \frac{k_e{e}^2}{r^2}\left[1-\left(1-\left(m-1\right)\frac{s}{r}\right)\right]\\ =-\alpha \frac{k_e{e}^2}{r^2}\left[\left(m-1\right)\frac{s}{r}\right]\end{array}$

Substitute $r_0$ for $r$ in above equation to get restoring force at equilibrium spacing.

  $\begin{array}{c}F\approx -\alpha \frac{k_e{e}^2}{r_0^2}\left[\left(m-1\right)\frac{s}{r_0}\right]\\ \approx -\alpha \frac{k_e{e}^2}{r_0^3}\left(m-1\right)s\end{array}$

The above equation indicates that the force is proportional to displacement from equilibrium position and force acts in the opposite direction of displacement. Therefore above equation is equivalent to restoring force.

Write the above expression in generalized form.

  $F=-Ks$

Here, $K$ is stiffness constant.

Compare above two expressions to get $K$.

  $K=\alpha \frac{k_e{e}^2}{r_0^3}\left(m-1\right)$                                                                                                     (V)

Therefore, it is showed that the restoring force experienced by ion is $F=-Ks$, where $K=\frac{\alpha k_e{e}^2}{r_0^3}\left(m-1\right)$.

(c)

To determine

The frequency of vibration of a ${\text{Na}}^{+}$ ion in $\text{NaCl}$ using the result of part(b).

Answer to Problem 56AP

The frequency of vibration of a ${\text{Na}}^{+}$ ion in $\text{NaCl}$ using the result of part(b) is $9.18\text{THz}$.

Explanation of Solution

In figure $\text{38}\text{.22}$, the distance from sodium ion to sodium ion is $0.562737\text{nm}$.

Then interatomic spacing in NaCl is half of the distance between sodium ion to sodium ion.

Therefore, calculate the spacing in $\text{NaCl}$.

  $\begin{array}{c}{r}_0=\frac{0.562737\text{nm}}{2}\\ =0.281369\times {10}^{-9}\text{m}\end{array}$

Write the expression for the ionic cohesive energy.

  $U_0=-\alpha \frac{k_e{e}^2}{r_0}\left(1-\frac{1}{m}\right)$                                                                                                (VI)

Here, $U_0$ is the ionic cohesive energy.

Use expression (V) to get stiffness constant.

Write the expression for frequency of vibration of sodium ion.

  $f=\frac{1}{2\pi }\sqrt{\frac{K}{m}}$                                                                                                             (VII)

Here, $m$ is the mass of eh sodium atom and $f$ is the frequency of vibration.

Conclusion:

The ionic cohesive energy for the crystal is $-7.84\text{eV}$.

Substitute $1.7476$ for $\alpha$, $-7.84\text{eV}$ for $U_0$, $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_e$, $1.60\times {10}^{-19}\text{C}$ for $e$ and $8$ for $m$ in equation (VI).

  $\begin{array}{c}-7.84\text{eV}=-\left(1.7476\right)\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right){\left(1.60\times {10}^{-19}\text{C}\right)}^2}{r_0}\left(1-\frac{1}{8}\right)\\ r_0=-\left(1.7476\right)\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right){\left(1.60\times {10}^{-19}\text{C}\right)}^2}{-7.84\text{eV}}\left(1-\frac{1}{8}\right)\\ =2.81\times {10}^{-10}\text{m}\end{array}$

Substitute $1.7476$ for $\alpha$, $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_e$, $1.60\times {10}^{-19}\text{C}$ for $e$, $8$ for $m$ and $2.81\times {10}^{-10}\text{m}$ for $r_0$ in equation (V) to get $K$.

  $\begin{array}{c}K=\left(1.7476\right)\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right){\left(1.60\times {10}^{-19}\text{C}\right)}^2}{{\left(2.81\times {10}^{-10}\text{m}\right)}^3}\left(8-1\right)\\ =127\text{N}/\text{m}\end{array}$

Substitute $127\text{N}/\text{m}$ for $K$ and $23.0\text{u}$ for $m$ in equation (VII) to get $f$.

  $\begin{array}{c}f=\frac{1}{2\pi }\sqrt{\frac{127\text{N}/\text{m}}{23.0\text{u}\times \frac{1.66\times {10}^{-27}\text{kg}}{1\text{u}}}}\\ =9.18\times {10}^{12}\text{Hz}\times \frac{1\text{THz}}{{10}^{12}\text{Hz}}\\ =9.18\text{THz}\end{array}$

Therefore, the frequency of vibration of a ${\text{Na}}^{+}$ ion in $\text{NaCl}$ using the result of part(b) is $9.18\text{THz}$.