Chapter 29, Problem 31GP

To determine

The binding energy of the $H_2$ molecule.

Answer to Problem 31GP

  $13eV$ .

Explanation of Solution

Given:

Radius of electrons in the atom, $\Delta x=0.053\times {10}^{-9}m$

Radius of electrons in the molecule, $\Delta x=0.074\times {10}^{-9}m$

Formula used:

It is known that: $\text{KE}=\frac{{\text{2h}}^{\text{2}}}{\text{2m}{\left(\text{Δx}\right)}^{\text{2}}}$

Where,

  $\text{KE}$ is the kinetic energy.

  $\text{Δx}$ is the radius.

  $\text{h}$ is Planck’s constant.

  $\text{m}$ is mass.

Calculation:

As the electrons are in separate atoms: $\text{KE}=\frac{{\text{2h}}^{\text{2}}}{\text{2m}{\left(\text{Δx}\right)}^{\text{2}}}$

  $\therefore {\text{KE}}_1=\frac{\text{2}{\left(1.055\times {10}^{-34}\text{J•s}\right)}^{\text{2}}}{\text{2}\left(9.11\times {10}^{-31}\text{kg}\right){\left(0.053\times {10}^{-9}\text{m}\right)}^{\text{2}}}=4.35\times {10}^{-18}\text{J}=\text{27}\text{.2}\text{eV}$

When the electrons are in separate atoms:

  $\therefore {\text{KE}}_2=\frac{\text{2}{\left(1.055\times {10}^{-34}\text{J•s}\right)}^{\text{2}}}{\text{2}\left(9.11\times {10}^{-31}\text{kg}\right){\left(0.074\times {10}^{-9}\text{m}\right)}^{\text{2}}}=2.23\times {10}^{-18}\text{J}=\text{14}\text{.0}\text{eV}$

Therefore, the binding energy : $K{E}_2-K{E}_1$

  $K{E}_2-K{E}_1=27.2eV-14.0eV=13.0eV$

Conclusion: The binding energy of the $H_2$ molecule is $13.0eV$ .