Mass Defect and Nuclear Binding Energy
The mass defect is the difference between the mass of the separate nucleons and the mass of the nucleus, and the nuclear binding energy, E = Δm·c2, is the energy that would be released in forming the nucleus from its nucleons; it is greatest per nucleon for iron-56. In brief: a helium-4 nucleus is lighter than two protons and two neutrons by 0.030377 u, which corresponds to 28.30 MeV, or 7.07 MeV per nucleon.
The mass of a nucleus is smaller than the sum of the masses of its nucleons. The missing mass has been converted to the energy that holds the nucleus together, according to Einstein’s relation. This is the source of the energy of nuclear fission and fusion, and it explains why heavy nuclei release energy when they split and light ones when they fuse. This article explains how to calculate the mass defect and binding energy, gives a table for important nuclides, describes the binding energy curve and works through examples.
What is the mass defect?
The mass of a nucleus is always smaller than the sum of the masses of the free protons and neutrons in it. The difference is the mass defect:
Δm = [Z·mp + N·mn] − mnucleus
When the nucleons come together to form the nucleus, energy is released, and the mass of the nucleus is smaller by the corresponding mass (E = mc2). To take the nucleus apart again, the same energy must be supplied. This is the nuclear binding energy: Eb = Δm · c2. A large binding energy means that the nucleus is very stable. A convenient measure for comparison is the binding energy per nucleon, Eb/A.
Which units are used?
- 1 u = 1.6605 × 10−27 kg = 931.494 MeV/c2. A mass defect of 1 u thus corresponds to 931.494 MeV.
- 1 MeV = 1.602 × 10−13 J.
- For one mole of nuclei: 1 MeV per nucleus = 9.649 × 107 kJ/mol.
Tables normally give atomic masses (including electrons), and not nuclear masses. The calculation is then done with the mass of the hydrogen atom (1.007825 u) instead of the proton, so that the electron masses cancel: Δm = Z·m(1H) + N·mn − m(atom). The small electron binding energies are neglected.
How do you calculate the binding energy?
- Find Z and N for the nuclide.
- Calculate the total mass of Z hydrogen atoms and N neutrons.
- Subtract the atomic mass of the nuclide to get Δm in u.
- Multiply by 931.494 MeV/u to get the binding energy in MeV.
- Divide by A for the binding energy per nucleon.
Binding energies of important nuclides
| Nuclide | A | Binding energy (MeV) | Binding energy per nucleon (MeV) |
|---|---|---|---|
| 2H | 2 | 2.22 | 1.11 |
| 4He | 4 | 28.30 | 7.07 |
| 12C | 12 | 92.16 | 7.68 |
| 16O | 16 | 127.62 | 7.98 |
| 56Fe | 56 | 492.26 | 8.79 |
| 235U | 235 | 1783.9 | 7.59 |
What does the binding energy curve show?
A plot of binding energy per nucleon against mass number rises steeply for light nuclei, passes through a maximum near iron-56 (and nickel-62, which has the highest value of about 8.79 MeV per nucleon) and falls slowly for heavier nuclei, to about 7.6 MeV for uranium. The curve explains the sources of nuclear energy:
- Fusion: when light nuclei combine (for example hydrogen to helium), the binding energy per nucleon increases, and the difference is released.
- Fission: when a very heavy nucleus splits into two medium-sized fragments, the binding energy per nucleon increases from about 7.6 to about 8.5 MeV, and the difference is released.
- Iron: no energy can be obtained from the fusion or fission of iron-group nuclei, which is why fusion in the core of massive stars stops at iron.
- Helium-4 stands out: its high binding energy per nucleon (7.07 MeV) for such a light nucleus explains why alpha particles are emitted in decay, and why helium is so common as the product of fusion.
Worked examples
Example 1: helium-4
Z = 2, N = 2. Mass of 2 hydrogen atoms: 2 × 1.007825 = 2.015650 u. Mass of 2 neutrons: 2 × 1.008665 = 2.017330 u. Total: 4.032980 u. The atomic mass of 4He is 4.002603 u. Δm = 4.032980 − 4.002603 = 0.030377 u. Eb = 0.030377 × 931.494 = 28.30 MeV. Per nucleon: 28.30 / 4 = 7.07 MeV.
Example 2: iron-56
Z = 26, N = 30. 26 × 1.007825 = 26.20345 u; 30 × 1.008665 = 30.25995 u; the total is 56.46340 u. The atomic mass of 56Fe is 55.934936 u. Δm = 0.528464 u. Eb = 0.528464 × 931.494 = 492.26 MeV, and per nucleon 492.26 / 56 = 8.790 MeV.
Example 3: binding energy per mole
One mole of helium-4 nuclei has a binding energy of 28.30 MeV × 9.649 × 107 kJ/mol per MeV = 2.73 × 109 kJ/mol. For comparison, the energy to break a C–H bond is about 413 kJ/mol, so nuclear energies are about a million times larger than chemical energies.
Example 4: the energy from a mass change
What energy corresponds to a mass of 1.00 g? E = mc2 = 1.00 × 10−3 kg × (3.00 × 108 m/s)2 = 9.0 × 1013 J, or 25 GWh, which is enough to power about 2000 homes for a year.
Example 5: energy released by combining nuclei
When two deuterons form a helium-4 nucleus (2H + 2H → 4He), the binding energy rises from 2 × 2.22 = 4.44 MeV to 28.30 MeV, and the energy released is 28.30 − 4.44 = 23.9 MeV.
Where does the idea come from?
Albert Einstein derived the equivalence of mass and energy in 1905. Francis Aston’s precise measurements of atomic masses in the 1920s showed that the masses of nuclei are smaller than the sums of the masses of their parts, with “packing fractions” that varied in a regular way, and Arthur Eddington used the data in 1920 to suggest that stars shine by turning hydrogen into helium. Carl von Weizsäcker gave a semi-empirical mass formula in 1935, which describes the binding energy in terms of volume, surface, Coulomb, asymmetry and pairing terms, and was refined by Hans Bethe.
Which mistakes are common?
- Using the proton mass with atomic masses: use the hydrogen atom mass to account for the electrons.
- Giving the answer in u instead of MeV: multiply by 931.494.
- Mixing up total and per-nucleon binding energy: stability is compared per nucleon.
- Thinking that mass is lost: it is converted into the energy released and carried away.
- Forgetting the sign: the binding energy is the energy needed to separate the nucleons, a positive quantity.