Chapter 29, Problem 33GP

(a)

To determine

The electrostatic potential energy between the ions.

Answer to Problem 33GP

  $\text{-5}\text{.3}\text{eV}$

Explanation of Solution

Given info:

The separation distance between ions is about $\text{0}\text{.27}\text{nm}$ .

Formula used:

Binding energy $=-PE=\frac{k{e}^2}{r}$

Where,

  $PE$ is potential energy

  $r$ is a distance

  $k$ is constant, $k=9.00\times {10}^{9}N\cdot m^2/C^2$

  $e$ is electron, $\text{e}\text{=1}\text{.60}\times {\text{10}}^{-19}C$

Calculation:

The potential energy for the point charges is binding energy:

  $\begin{array}{c}PE=-\frac{k{e}^2}{r}\\ =-\frac{\left(2.30\times {10}^{-28}J\cdot m\right)}{\left(0.27\times {10}^{-9}m\right)}\\ =-8.52\times {10}^{-19}J\\ =-5.3eV\end{array}$

(b)

To determine

The binding energy.

Answer to Problem 33GP

  $\text{5}\text{.0}\text{eV}$

Explanation of Solution

Given info:

The separation distance between ions is about $\text{0}\text{.27}\text{nm}$ .

Formula used:

Binding energy $=-PE=\frac{k{e}^2}{r}$

Where,

  $PE$ is potential energy

  $r$ is a distance

  $k$ is constant, $k=9.00\times {10}^{9}N\cdot m^2/C^2$

  $e$ is electron, $\text{e}\text{=1}\text{.60}\times {\text{10}}^{-19}C$

Calculation:

The potential energy for the point charges is binding energy:

  $PE=-\frac{k{e}^2}{r}-=\frac{\left(2.30\times {10}^{-28}J\cdot m\right)}{\left(0.27\times {10}^{-9}m\right)}=-8.52\times {10}^{-19}J=-5.3eV$

Therefore, potential energy of ions is negative, $\text{5}\text{.3}\text{eV}$ is released at the time when ions are brought together.

A release of energy means that energy must be provided to return the ions to the state of free atoms.

So, the sum of binding energy of the KF ions:

Binding energy $=5.3eV+4.07eV-4.34eV=5.0eV$