a) Calculate the electric potential energy of the adenine–thymine bond, using the same combinations of molecules (O9H9N and N9H9N) as in Exercise 21.21.

(b) Compare this energy with the potential energy of the proton-electron pair in the hydrogen atom.

Charges at a distance apply force on each other, which is equal and opposite in direction. If the charges are of the same nature, this force will be repulsive; if not, it will be an attractive force. This force is known as coulomb force.

The expression of the potential energy for a pair of charges is,

$\mathbf{U}\mathbf{=}\mathbf{k}\frac{\mathbf{qq}\mathbf{\text{'}}}{\mathbf{r}}$

Where q and q^’ are the charges at a distance r.

The expression of the potential energy for a pair of charges is,

$\mathrm{U}=\mathrm{k}\frac{\mathrm{qq}\text{'}}{\mathrm{r}}$

Where the symbols have the usual meaning.

So, for each pair of charges in the molecules, the distance is different, and so are the potential energies.

For O-H-N,

$$\begin{gathered} {\mathrm{O}}^{-}-{\mathrm{H}}^{+} \\ \mathrm{U}=\left(8.99\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{{\left(1.60\times {10}^{-19}\mathrm{C}\right)}^2}{0.170\times {10}^{-9}\mathrm{m}} \\ =1.35\times {10}^{-18}\mathrm{J} \\ {\mathrm{O}}^{-}-{\mathrm{N}}^{-}: \\ \mathrm{U}=\left(8.99\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{{\left(1.60\times {10}^{-19}\mathrm{C}\right)}^2}{0.280\times {10}^{-9}\mathrm{m}} \\ =+8.22\times {10}^{-19}\mathrm{J} \end{gathered}$$

For, N-H-N,

$$\begin{gathered} {\mathrm{N}}^{-}-{\mathrm{H}}^{+}: \\ \mathrm{U}=-\left(8.99\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\times \frac{{\left(1.60\times {10}^{-19}\mathrm{C}\right)}^2}{0.190\times {10}^{-9}\mathrm{m}} \\ =-1.21\times {10}^{-18}\mathrm{J} \end{gathered}$$

$$\begin{gathered} {\mathrm{N}}^{-}-{\mathrm{N}}^{-}: \\ \mathrm{U}=\left(8.99\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\times \frac{{\left(1.60\times {10}^{-19}\mathrm{C}\right)}^2}{0.300\times {10}^{-9}\mathrm{m}} \\ =+7.67\times {10}^{-19}\mathrm{J} \end{gathered}$$

Thus, the total potential energy is,

$$\begin{gathered} {\mathrm{U}}_{\mathrm{tot}}=-1.35\times {10}^{-18}\mathrm{J}+8.22\times {10}^{-19}\mathrm{J}-1.21\times {10}^{-18}\mathrm{J}+7.67\times {10}^{-19}\mathrm{J} \\ =-9.71\times {10}^{-19}\mathrm{J} \end{gathered}$$

Thus, the electrical potential energy of the adenine-thymine bond is$-9.71\times {10}^{-19}\mathrm{J}$

For the Hydrogen atom, the distance between the electron and the proton is 0.0529 nm.

Therefore, the potential energy is,

$$\begin{gathered} \mathrm{U}=\left(8.99\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\times \frac{{\left(1.60\times {10}^{-19}\mathrm{C}\right)}^2}{0.0529\times {10}^{-9}\mathrm{m}} \\ =-4.35\times {10}^{-18}\mathrm{J} \end{gathered}$$

Thus compared with the calculated value of potential energy in part (a), it is found that the U for electron-proton pair in H-atom is about four times larger than for the adenine-thymine bond.