Chapter 29, Problem 3P

To determine

The binding energy.

Answer to Problem 3P

  $4.6eV$

Explanation of Solution

Given info:

Two electrons spend $\text{33\%}$ of their time midway between them.

Distance $\left(r\right)$ between the ions $=0.074nm=0.074\times {10}^{-9}m$

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, At the time when electrons are midway between the protons then each electron will have potential energy due to two protons. So,

  $P{E}_{ep}=\frac{-\left(2\right)\left(0.33\right)k{e}^2}{\left(\frac{r}{2}\right)}=-\frac{\left(4\right)\left(0.33\right)\left(2.30\times {10}^{-28}J\cdot m\right)}{\left(0.074\times {10}^{-9}m\right)\left(1.60\times {10}^{-19}J/eV\right)}=-25.6eV$

The protons have potential energy:

  $P{E}_{pp}=+\frac{k{e}^2}{r}=+\frac{\left(2.30\times {10}^{-28}J\cdot m\right)}{\left(0.074\times {10}^{-9}m\right)\left(1.60\times {10}^{-19}J/eV\right)}=+19.4eV$

When bond breaks, each hydrogen atom will be in ground state with energy $E_1=-13.6eV$

So, the binding energy will be

  $2E_1-\left(2P{E}_{ep}+P{E}_{pp}\right)=2\left(-13.6eV\right)-\left[2\left(-25.6eV\right)+19.4eV\right]=4.6eV$

Conclusion: The required value is $4.6eV$ .