# Nuclear binding energy curve

1. Apr 10, 2012

### huey910

If the total nuclear binding energy of a nucleus increases whenever the number of nucleons increase, why does the nuclear binding energy curve rise and then fall? Don't protons and neutrons 'bring in' the same amount of binding energy every time one of them is added? Please explain. In addition, if fission can release energy, does that not mean that the nucleus itself has a higher mass than the separated, smaller and stabler nuclei produced? Thank you.

Last edited: Apr 10, 2012
2. Apr 10, 2012

### daveb

The curve you are thinking of is the binding energy per nucleon, not the total binding energy.

3. Apr 10, 2012

### sambristol

If you bring a collection of neutrons and protons together to form a stable nucleus then it will have LESS mass than the constituent particles (this is called the mass defect). If the mass defect is m then the binding energy is E=mc2.

If the binding energy of a system was negative ie the nucleus had more mass than the sum of its parts, it would spontaneously decay back into its parts.

The curve by the way is the binding energy divided by the number of nucleons.

The question of why the curve rises and then falls – and on close inspection not without many smaller 'wiggles' is because of the very complex, and not fully understood, interaction between nucleons.

The factors which come into play include the electric repulsion between the protons a number of quantum terms to do with the strong nuclear force (which is not a true central force like electric force) including spin effects and shell effects (similar to but more complex than the shells of electron orbitals in atoms).

The whole lot is summed up in the semi empirical mas formula – look it up. It is called semi empirical because it is the sum of the terms we expect to see each multiplied by a constant. Experimental (ie empirical) information is used to find the best fit for the constants.

The formula works fairly well except for a handful of nuclei (look up nuclear magic numbers)

Hope this helps

Regards

Sam