Question about nuclei?

Question

I think I am correct in my understanding that the binding energy of a nucleon in an atom is measured in much the same way as the potential energy of a mass on the earth's surface. So as a mass on Earth needs a certain amount of energy in order to reach infinity, where the gravitation exerted upon it by Earth is zero, its potential energy is therefore negative. Then the same is true for the binding energy for a nucleon, and therefore the larger (negatively) the binding energy, the more energy required for the nucleon to escape the nucleus. Is this understanding correct?

My problem is this, I have a graph depicting the various binding energy per nucleon for the nuclei of the periodic table, on this graph the radioactive nuclei have larger (negative) BEPN than helium, so how come helium is more stable. Helium nuclei are emitted from some of these elements as the alpha particle, but my graph shows it to have a lower BEPN. Where am I going wrong? Or is my whole understanding flawed?

Answer 1

Nuclei are complex structures and have many energy levels in them. (Look at Iso-spin articles).
You are assuming that all the parts are bound by the same average force, and this is not true. Within this average are well and poorly bound nucleons or aggregates of nucleons.
You can set up oscillations in a nucleus which will lead to its break-up(See water drop theory in nuclei)

To go to your analogy. The moon is bound by the earth's gravitational field but you would only need a fraction of the energy to remove the moon to infinity as you would to remove to infinity an equivalent mass at the earth's surface.
CopyLeft:RufusCat

Answer 2

The larger the binding energy means that there was agreater rmass defect to produce the atom. that mean the nucleides have a smaller mass then their parents which formed them.