Basic Physics of Nuclear Medicine/Print version

Binding Energy

We are now in a position to consider the subject of nuclear stability. From what we have covered so far, we have seen that the nucleus is a tiny region in the centre of an atom and that it is composed of neutrally and positively charged particles. So, in a large nucleus such as that of uranium (Z=92) we have a large number of positively charged protons concentrated into a tiny region in the centre of the atom. An obvious question which arises is that with all these positive charges in close proximity, why doesn't the nucleus fly apart? How can a nucleus remain as an entity with such electrostatic repulsion between the components? Should the orbiting negatively-charged electrons not attract the protons away from the atoms centre?

Let us take the case of the helium-4 nucleus as an example. This nucleus contains two protons and two neutrons so that in terms of amu we can figure out from what we covered earlier that the

and the

Therefore we would expect the total mass of the nucleus to be 4.03298 amu.

The experimentally determined mass of a helium-4 nucleus is a bit less - just 4.00260 amu. In other words there is a difference of 0.03038 amu between what we might expect as the mass of this nucleus and what we actually measure. You might think of this difference as very small at just 0.75%. But remember that since the mass of one electron is 0.00055 amu the difference is actually equivalent to the mass of about 55 electrons. Therefore it is significant enough to wonder about.

It is possible to consider that this missing mass is converted to energy which is used to hold the nucleus together; it is converted to a form of energy called Binding Energy. You could say, as with all relationships, energy must be expended in order to maintain them!

Like the gram in terms of the mass of nuclei, the common unit of energy, the joule is rather cumbersome when we consider the energy needed to bind a nucleus together. The unit used to express energies on the atomic scale is the electron volt, symbol: eV.

One electron volt is defined as the amount of energy gained by an electron as it falls through a potential difference of one volt. This definition on its own is not of great help to us here and it is stated purely for the sake of completeness. So do not worry about it for the time being. Just appreciate that it is a unit representing a tiny amount of energy which is useful on the atomic scale. It is a bit too small in the case of binding energies however and the mega-electron volt (MeV) is often used.

Albert Einstein introduced us to the equivalence of mass, m, and energy, E, at the atomic level using the following equation:

where c is the velocity of light.

It is possible to show that 1 amu is equivalent to 931.48 MeV. Therefore, the mass difference we discussed earlier between the expected and measured mass of the helium-4 nucleus of 0.03038 amu is equivalent to about 28 MeV. This represents about 7 MeV for each of the four nucleons contained in the nucleus.