91Ó°ÊÓ

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,

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

For, N-H-N,

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

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

Thus, the total potential energy is,

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

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

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

Therefore, the potential energy is,

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

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.