On the Ratio of Nuclear Binding Energy & Protons Kinetic Energy

Abstract

As proton number increases, coulombic repulsion between protons increases and proportionately protons’ kinetic energy increases which seems to be equal to the total nuclear binding energy. Based on the Fermi gas model of nucleons, it is possible to show that, within the nucleus, at the stable mass number, starting from Z=30, protons kinetic energy is proportional to (Zm_pc²)^1/3(Za_c)^2/3 and neutrons kinetic energy is proportional to (Nm_pc²)^1/3(Za_c)^2/3. Thus nucleons total kinetic energy is proportional to (Am_pc²)^1/3(Aa_c)^2/3. Starting from Z = 2 to 100 proportionality coefficient seems to be connected with proton number and can be expressed as  k_Z = (2/30)^1/6ln(4p) if (Z<30) and k_Z = ln(4p) if (Z<=30).