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.
N
P
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.
P
N
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.
Helium 4
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.
Element
12 Carbon 6
16 Oxygen 8
40 Calcium 20
56 Iron 26
208 Lead 82
235 Uranium 92
Mass defect (u)
0.099 00
0.137 08
0.367 41
0.528 75
1.757 84
1.935 38
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 = mc 2) 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.
Example
These are the results you would get (all masses in unified atomic mass units (u).)
protons (2) 2 x 1.007 276
neutrons (2) 2 x 1.008 665
Mass of two protons + two neutrons
Mass of "completed" helium nucleus
=
=
=
=
2.014 552
2.017 330
4.031 882
4.001 508
There is a difference of 0.030374 u between them, the helium nucleus is lighter than the four
particles that made it!
Example
The mass of the isotope 73Li is 7.016 u. Find its binding energy.
Protons
3x1.007 273 = 3.021 819
Neutrons
4x1.008 665 = 4.034 66
Total
= 7.056 479
Nucleus
= 7.016
Mass defect
= 0.040 479 u
Binding energy
= 0.040 479x931 = 37.69 MeV
1
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:
2
1H
2
1H
2
1H
+ 21H
+ 21H
+ 31H
Data:
1 u = 1.66x10-27 kg
3
1
2He + 0n + 3.34 MeV
3
1
1H + 1H + 4.0 MeV
4
1
2He + 0n + 17.6 MeV
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
2x1875.129 = 3750.258 gives
3
2He
3
1H
+ 939.067 + 3.34 MeV
+ 937.771 + 4.0 MeV
(1)
(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
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
(3)
2