Chapter 2, Problem 2.26P

To determine

The equilibrium inter-ionic separation of ${\text{K}}^{+}$ and a ${\text{Cl}}^{-}$ and to compare the value to the experimental value of $0.266\text{nm}$.

Answer to Problem 2.26P

The calculated value of $r$ is approximately same as the experimental value of $0.266\text{nm}$.

Explanation of Solution

Given:

The experimental value of equilibrium inter-ionic separation is $0.266\text{nm}$.

Formula used:

The formula to calculate the inter-ionic potential is given by,

$V_{\text{pion}}\left(r\right)=-\frac{{\left(Ze\right)}^2}{4\pi {\epsilon }_{0}r}+\frac{B}{r^m}$      ......... (I)

Here, $V_{\text{pion}}\left(r\right)$ is the inter-ionic potential, $Ze$ is the electron charge of ${\text{K}}^{+}$ ion, ${\epsilon }_0$ is the permittivity of vacuum, $r$ is inter atomic separationand $B$ ,and $m$ are constants.

Differentiate the equation (I) with respect to $r$.

$\begin{array}{c}\frac{dV}{dr}=\frac{d}{dr}\left(-\frac{{\left(Ze\right)}^2}{4\pi {\epsilon }_{0}r}+\frac{B}{r^m}\right)\\ F=\left(-\left(-1\right)\frac{{\left(Ze\right)}^2{r}^{-2}}{4\pi {\epsilon }_0}+\left(-12\right)B{r}^{-13}\right)\end{array}$      ......... (II)

Here, $F$ is the interionic force.

Calculation:

The interatomic separation is calculated as,

Substitute $1.602\times {10}^{-19}\text{C}$ for $Ze$, $8.85\times {10}^{-12}\text{F}/\text{m}$ for ${\epsilon }_0$, $8.3497\times {10}^{-134}\text{J}\cdot {\text{m}}^{12}$ for $B$, and $12$ for $m$ in equation (II).

$F=\left(-\left(-1\right)\frac{{\left(1.602\times {10}^{-19}\text{C}\right)}^2{r}^{-2}}{4\pi \left(8.85\times {10}^{-12}\text{F}/\text{m}\right)}-12\left(8.3497\times {10}^{-134}\text{J}\cdot {\text{m}}^{12}\right)r^{-13}\right)$

At equilibrium the force is $0$.

Substitute $0$ for $F$ in the above equation.

$\begin{array}{c}0=\left(-\left(-1\right)\frac{{\left(1.602\times {10}^{-19}\text{C}\right)}^2{r}^{-2}}{4\pi \left(8.85\times {10}^{-12}\text{F}/\text{m}\right)}-12\left(8.3497\times {10}^{-134}\text{J}\cdot {\text{m}}^{12}\right)r^{-13}\right)\\ \frac{2{\left(1.602\times {10}^{-19}\text{C}\right)}^2}{4\pi \left(8.85\times {10}^{-12}\text{F}/\text{m}\right)\times r}=\frac{12\left(8.3497\times {10}^{-134}\text{J}\cdot {\text{m}}^{12}\right)}{r^{12}}\\ \frac{r^{12}}{r}=4.3419\times {10}^{-106}\\ r=\sqrt[11]{4.3419\times {10}^{-106}}\end{array}$

Solve further as,

$\begin{array}{c}r=0.264\times {10}^{-9}\text{m}\\ =0.264\times {10}^{-9}\text{m}\left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ =0.264\text{nm}\end{array}$

Conclusion:

The calculated value of $r$ is approximately same as the experimental value of $0.266\text{nm}$.