For a Nucleus This Difference in Mass Is Called the Mass Defect of the Nucleus

Binding energy

One reason for the stability of a nucleus can be seen if we look more closely at a particular nucleus such as helium.

Imagine that you were asked to make such a nucleus. You are given the four pieces (two protons and two neutrons), asked to measure their masses, make them stick them together somehow and then measure the mass of the finished helium nucleus. You will find that the mass of the completed nucleus is different from that of the total mass of the four particles from which it was made! The mass of the nucleus is less than the total mass of the four particles.

This is really quite surprising - it is like taking a cake, weighing it, then cutting up into slices, weighing them and then finding that the cake had a different mass from the sum of the masses of the slices.

For a nucleus this difference in mass is called the mass defect of the nucleus.

The mass defect for a number of nuclei is shown below.

This can best be explained by looking at how easy it would be to split the alpha particle apart again. If we think of this mass difference as a difference of energy (using E = mc2) then the alpha particle has 28.3 MeV less energy than the four particles. It means that this energy would be needed to split up the helium nucleus. This is called the binding energy of a nucleus.

Sample binding energy problem – working in MeV

Using the data below and that the mass of deuterium (2H) is 2.014102 u calculate the binding energies of 42He and 31 H from the following reactions:

21H + 21H 32He + 10n + 3.34 MeV

21H + 21H 31H + 11H + 4.0 MeV

21H + 31H 42He + 10n + 17.6 MeV

Data:

1 u = 1.66x10-27 kg c = 3x108 ms-1 Masses: Proton 1.007 273 u Neutron 1.008 665 u

Taking 1u = 931 MeV we will do this version in MeV throughout.

2x1875.129 = 3750.258 gives 32He + 939.067 + 3.34 MeV (1)

2x1875.129 = 3750.258 gives 31H + 937.771 + 4.0 MeV (2)

Therefore from equation (1):

3750.258 – 939.067 – 3.34 = 32He

Mass equivalent of 32He = 2807.851 MeV

Mass equivalent of 2p + 1n = 2814.609 MeV

Binding energy of 32He = 6.758 MeV

Therefore from equation (2):

3750.258 – 937.771 – 4.0 = 31H

Mass equivalent of 31H = 2808.487 MeV

Mass equivalent of 1p + 2n = 2815.905 MeV

Binding energy of 31H = 7.418 MeV

Also: 1875.129 + 31H gives 42He + 939.067 + 17.6 MeV(3)

Therefore: 1875.129 + 2808.487 = 42He + 939.067 + 17.6

Mass equivalent of 42He = 3726.949 MeV

Mass equivalent of 2p + 2n = 3753.682 MeV

Binding energy of 42He = 0.013398 = 26.73 MeV