91Ó°ÊÓ

Given data:

$$\begin{gathered} R_{N−H}=0.11×{10}^{−9}\mathrm{m} \\ {\mathrm{R}}_{\mathrm{N}−\mathrm{O}}=0.28×{10}^{−9}\mathrm{m} \\ {\mathrm{R}}_{\mathrm{N}−\mathrm{N}}=0.3×{10}^{−9}\mathrm{m} \\ {\mathrm{q}}_{\mathrm{N}}={\mathrm{q}}_{\mathrm{H}}={\mathrm{q}}_{\mathrm{O}}=\mathrm{e} \end{gathered}$$

Force exerting by O on H-N, let the left direction be positive and the right direction be negative.

Therefore, the force between oxygen and hydrogen.

$$\begin{gathered} F_{O−H}=\frac{K{e}^2}{R_{O−H}^2} \\ =\frac{K{e}^2}{{\left(R_{O−N}−R_{N−H}\right)}^2} \\ =\frac{\left(9×{10}^9\right){\left(1.6×{10}^{−19}\right)}^2}{{\left[(0.28−0.11)×{10}^{−9}\right]}^2} \\ =7.97×{10}^{−9}\mathrm{N} \end{gathered}$$

Forces Between oxygen and nitrogen

$$\begin{gathered} F_{O−N}=\frac{K{e}^2}{R_{O−N}^2} \\ =\frac{\left(9×{10}^9\right){\left(1.6×{10}^{−19}\right)}^2}{{\left[0.28×{10}^{−9}\right]}^2} \\ =−2.9×{10}^{−9}\mathrm{N} \end{gathered}$$

Force exerting by H-N on N, let the left direction be positive and the right direction be negative.

Therefore, force between hydrogen and nitrogen is

$$\begin{gathered} F_{N−H}=\frac{K{e}^2}{R_{N−H}^2} \\ =\frac{K{e}^2}{{\left(R_{N−N}−R_{N−H}\right)}^2} \\ =\frac{\left(9×{10}^9\right){\left(1.6×{10}^{−19}\right)}^2}{{\left[(0.3−0.11)×{10}^{−9}\right]}^2} \\ =6.38×{10}^{−9}\mathrm{N} \end{gathered}$$

Force between nitrogen and nitrogen is

$$\begin{gathered} F_{NN}=\frac{K{e}^2}{R_{N−N}^2} \\ =\frac{\left(9×{10}^9\right){\left(1.6×{10}^{−19}\right)}^2}{{\left[0.3×{10}^{−9}\right]}^2} \\ =−2.56×{10}^{−9}N \end{gathered}$$

So, the total net force is

$$\begin{gathered} F_{net}=(7.97−2.9+6.38−2.56)×{10}^{−9} \\ =8.89×{10}^{−9}N \end{gathered}$$

And force is directed towards the left which is attractive.

Hence, the Net force by thymine exerting on adenine is role="math" localid="1668165900263" $8.89×{10}^{-9}N$, and it is an attractive force.

The force between electron and proton in a hydrogen atom

$$\begin{gathered} F_H=\frac{K{e}^2}{r^2} \\ =\frac{9×{10}^9×{\left(1.6×{10}^{−19}\right)}^2}{{\left(0.529×{10}^{−19}\right)}^2} \\ =8.23×{10}^{−8}\mathrm{N} \end{gathered}$$

The ratio is

$$\begin{gathered} R=\frac{8.23×{10}^{−8}}{8.89×{10}^{−9}} \\ ≈9.3 \end{gathered}$$

Therefore, The force between electron and proton in the hydrogen atom is $8.23×{10}^{-8}N$, and its ratio is 9.3.