Chapter 29, Problem 1P

To determine

The binding energy.

Answer to Problem 1P

  $5.13eV$ .

Explanation of Solution

Given info:

Charge on $K^{+}$ and $C{l}^{-}$ ions are $\text{q = e = 1}{\text{.6 × 10}}^{\text{-19}}\text{ C}$

Distance $\left(r\right)$ between the ions $=0.28nm=0.28\times {10}^{-9}m$

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:

According to the question, with the reference level at infinity binding energy of two ions will be

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

  $=\frac{\left(9.00\times {10}^{9}N\cdot m^2/C^2\right){\left(1.60\times {10}^{-19}C\right)}^2}{0.28\times {10}^{-9}m}$

  $=\frac{\left(2.30\times {10}^{-28}J\cdot m\right)}{\left(0.28\times {10}^{-9}m\right)}$

  $=8.21\times {10}^{-19}J$

  $=5.13eV$

Conclusion: The required value is $5.13eV$ .