Assertion : Nuclear binding energy per nucleon is in the order -...

Binding energy per nucleon of ${ }_3 \mathrm{Li}^7(5.38$ $\mathrm{MeV})$ is lesser than ${ }_2 \mathrm{He}^4(7.08 \mathrm{MeV})$ as helium is found to be more stable than Li. As the atomic mass number increases, the binding energy per nucleon decreases. As the atomic number and the atomic mass number increase, the repulsive electrostatic forces within the nucleus increase due to the greater number of protons in the heavy elements. To overcome this increased repulsion, the proportion of neutrons in the nucleus must increase to maintain stability. This increase in the neutron-to-proton ratio only partially compensates for the growing proton-proton repulsive force in the heavier, naturally occurring elements. Because the repulsive forces are increasing, less energy must be supplied, on the average, to remove a nucleon from the nucleus. The binding energy per nucleon has decreased. The binding energy per nucleon of a nucleus is an indication of its degree of stability. Generally, the more stable nuclides have higher binding energy per nucleon than the less stable ones. The increase in binding energy per nucleon as the atomic mass number decreases from 260 to 60 is the primary reason for the energy liberation in the fission process. The increase in the binding energy per nucleon as the atomic mass number increases from 1 to 60 is the reason for the energy liberation in the fusion process, which is the opposite reaction of fission.