Question

How does the finite range of the nuclear force relate to the fact that $\mathrm{B}\mathrm{E}/\mathrm{A}$ is greatest for $\mathrm{A}$near$60$.

Short answer

Because $\mathrm{B}\mathrm{E}/\mathrm{A}$is greatest for nuclei with a number of nucleons close to $60$.

Step-by-step solution

Definition of Nuclear force

The nuclear force is a force that exists between atoms' protons and neutrons. The nuclear force affects neutrons and protons, both of which are nucleons.

Explanation

We know that as the number of protons increases, so do the nuclear charge. This charge allows the nucleus to attract electrons, which is known as nuclear attraction. Binding energy is also defined as the energy that separates particles from a system. Furthermore, it is proportional to the number of nucleons.

Some nucleons in a stable nucleus with more than 40 nucleons would be separated from other nucleons and thus would not feel nuclear attraction. Protons, on the other hand, are repulsed because Coulomb force has a greater range.

Because of coulomb repulsion between protons and nuclear attraction, the relation $\mathrm{B}\mathrm{E}/\mathrm{A}$ is greatest for nuclei with a number of nucleons close to 60.