Chapter 37, Problem 21P

(a)

To determine

The potential energy of attraction of the ions.

Answer to Problem 21P

Thepotential energy of attraction of the ions is $-5.39\text{eV}$ .

Explanation of Solution

Given:

The equilibrium separation of the ${\text{K}}^{+}$ and ${\text{F}}^{-}$ ions in $\text{KF}$ is $r=0.217\text{nm}$ .

Formula used:

The expression for potential energy of attraction is given by,

  $U_e=-\frac{k{e}^2}{r}$

Calculation:

The potential energy of attraction is calculated as,

  $\begin{array}{c}{U}_e=-\frac{k{e}^2}{r}\\ =-\frac{\left(8.9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right){\left(1.6\times {10}^{-19}\text{C}\right)}^2}{\left(0.217\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\left(1.6\times {10}^{-19}\frac{\text{J}}{\text{eV}}\right)}\\ =-\frac{\left(8.9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}\right){\left(1.6\times {10}^{-19}\right)}^2\text{eV}}{\left(0.217\times {10}^{-9}\text{m}\right)\left(1.6\times {10}^{-19}\text{J}\right)}\\ =-6.64\text{eV}\end{array}$

Conclusion:

Therefore, the potential energy of attraction of the ions is $-6.64\text{eV}$ .

(b)

To determine

The value of dissociation energy.

Answer to Problem 21P

The value of dissociation energy is $5.70\text{eV}$ .

Explanation of Solution

Given:

The ionization energy of potassium is $4.34\text{eV}$

The electron affinity of Fluorine is $-3.40\text{eV}$

Formula used:

The expression for total potential energy of the molecule neglecting energy of repulsion is given by,

  $U_{\text{tot}}=U_e+\Delta E$

The expression for dissociation energy is given by,

  $\begin{array}{c}{E}_d=-U_{\text{tot}}\\ =-\left(U_e+\Delta E\right)\end{array}$

Here, $\Delta E$ is the difference between the ionization energy of potassium and the electron affinity of fluorine.

Calculation:

Difference between the ionization energy of potassium and the electron affinity of fluorine $\Delta E$ is calculated as,

  $\begin{array}{c}\Delta E=4.34\text{eV}-3.40\text{eV}\\ =\text{0}\text{.94}\text{eV}\end{array}$

The dissociation energy is calculated as,

  $\begin{array}{c}{E}_d=-\left(U_e+\Delta E\right)\\ =-\left(-6.64\text{eV}+0.94\text{eV}\right)\\ =5.70\text{eV}\end{array}$

Conclusion:

Therefore, the value of dissociation energy is $5.70\text{eV}$ .

(c)

To determine

The energy due to repulsion of the ions at the equilibrium separation.

Answer to Problem 21P

The energy due to repulsion of the ions at the equilibrium separation is $\text{0}\text{.63}\text{eV}$ .

Explanation of Solution

Given:

The measured dissociation energy $E_{\text{d,measured}}=5.07\text{eV}$

Formula used:

The expression for energy due to repulsion of the ions at equilibrium separation is given by,

  $U_{\text{rep}}=E_{\text{d,calculated}}-E_{\text{d,measured}}$

Calculation:

The energy due to repulsion of the ions at equilibrium separation is calculated as,

  $\begin{array}{c}{U}_{\text{rep}}=E_{\text{d,calculated}}-E_{\text{d,measured}}\\ =5.70\text{eV}-5.07\text{eV}\\ =\text{0}\text{.63}\text{eV}\end{array}$

Conclusion:

Therefore, the energy due to repulsion of the ions at the equilibrium separation is $\text{0}\text{.63}\text{eV}$