Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The repulsive force between two protons is given by
$$F=\dfrac{ke^2}{r^2}$$
Plug the known;
$$F=\dfrac{(9\times 10^9)(1.6\times 10^{-19})^2}{(2\times 10^{-15})^2}$$
$$F=\color{red}{\bf 57.6}\;\rm N$$
$$\color{blue}{\bf [b]}$$
The gravitational force between the two protons is given by
$$F=\dfrac{Gm_p^2}{r^2}$$
Plug the known;
$$F=\dfrac{(6.67\times 10^{-11})(1.67\times 10^{-27})^2}{(2\times 10^{-15})^2}$$
$$F=\color{red}{\bf 4.65\times 10^{-35}}\;\rm N$$
It is obvious that the repulsive force between protons, as calculated, greatly exceeds the gravitational force. Therefore, it is clear that gravitational force is not the mechanism responsible for maintaining the two protons, or the nucleus, as a cohesive unit.
$$\color{blue}{\bf [c]}$$
The calculated repulsive electric force between two protons is significantly greater than the attractive gravitational force between them. This implies that gravity is not the force responsible for holding the nucleus together. Instead, there must be another force at work that overcomes the electrostatic repulsion between protons and binds the nucleus together. This force is the strong nuclear force.
Characteristics of the strong nuclear force:
- It is a short-range force, acting over distances of only a few femtometers.
- It is much stronger than the electromagnetic force at nuclear distances.
- It is attractive at nuclear distances, overcoming the electrostatic repulsion between protons and binding nucleons (protons and neutrons) together to form nuclei.
- It does not depend on the charge of the particles involved, unlike the electromagnetic force.
- It saturates with distance, meaning it remains constant within the nucleus but drops to negligible levels outside the nucleus.
