Nuclear binding energy and the mass deficit (mass defect)

Using ΔE = c²Δm for nuclear masses and energies

The atomic mass unit u, and converting between u, kg, J and MeV

The binding energy per nucleon curve and the iron peak

Energy release in fission and fusion, from masses and from the curve

Why fusion needs very high temperatures and densities

1. A nucleus has less mass than the protons and neutrons it is made from, that missing mass is the binding energy, and every fission or fusion calculation is the same three steps: mass before, mass after, ΔE = c²Δm.

2. The binding energy of a nucleus is the energy needed to separate it completely into its individual protons and neutrons.

3. 1 u = 1.66 × 10⁻²⁷ kg, equivalent to about 931.5 MeV. Use kg with c² to get joules; divide by 1.60 × 10⁻¹³ for MeV.

4. Binding energy per nucleon measures stability. It peaks at about 8.8 MeV near iron-56.

5. Fission of heavy nuclei and fusion of light nuclei both move towards the peak: binding energy per nucleon rises, mass falls, energy is released.

6. Fusion needs nuclei close enough for the strong force to act, against electrostatic repulsion: very high temperature (fast nuclei) and high density (frequent collisions).

The strong nuclear force holds the nucleons together, so work must be done to pull them apart; that work is the binding energy. Run the process backwards and the same energy is released when free nucleons come together to form a nucleus. Energy has mass, so the bound nucleus, which has less energy than the separated nucleons, also has less mass. For a nucleus with Z protons and N = A - Z neutrons:

ΔE is in J, Δm in kg and c = 3.00 ×10⁸ m s⁻¹. This equation is in the formulae list. Any change in energy comes with a change in mass. In chemical reactions the change is far too small to measure; in nuclear reactions it is about a million times larger per particle and easily measured.

WORKED EXAMPLE · QUESTION

Mass of a proton = 1.007276 u, mass of a neutron = 1.008665 u, mass of a helium-4 nucleus = 4.00151 u; 1 u = 1.66 × 10⁻²⁷ kg. Calculate the binding energy of a helium-4 nucleus in J and in MeV, and its binding energy per nucleon.

SOLUTION

Separate nucleons: 2 ×1.007276 + 2 ×1.008665 = 4.031882 u.

Mass deficit: Δm = 4.031882 - 4.00151 = 0.03037 u = 0.03037 ×1.66 ×10⁻²⁷ = 5.04 ×10⁻²⁹ kg.

ΔE = (3.00 ×10⁸)² ×5.04 ×10⁻²⁹: 4.54 ×10⁻¹² J = 4.54 ×10⁻¹² ÷1.60 ×10⁻¹³ = 28.4 MeV.

Per nucleon (A = 4): 28.4 ÷4: 7.09 MeV.

Using 1 u = 931.5 MeV directly gives 28.3 MeV; the small difference comes from rounding the data-list constants to 3 s.f., and either route is acceptable if you show it.

The atomic mass unit u is one twelfth of the mass of a carbon-12 atom: 1 u = 1.66 × 10⁻²⁷ kg.

Nuclear masses are quoted in u because they are then close to whole numbers (a proton or neutron ≈ 1 u).

Nuclear energies are quoted in MeV: 1 MeV = 10⁶ × 1.60 × 10⁻¹⁹ J = 1.60 × 10⁻¹³ J.

WORKED EXAMPLE · QUESTION

(a) Show that 1 u is equivalent to about 930 MeV. (b) The binding energy per nucleon of oxygen-16 is 7.98 MeV. Calculate the mass deficit of an oxygen-16 nucleus in kg and in u.

SOLUTION

(a) 1.66 ×10⁻²⁷ ×(3.00 ×10⁸)² = 1.494 ×10⁻¹⁰ J; ÷1.60 ×10⁻¹³: 934 MeV ≈ 930 MeV.

(b) Total binding energy = 7.98 ×16 = 127.7 MeV = 127.7 ×1.60 ×10⁻¹³ = 2.04 ×10⁻¹¹ J. Δm = 2.04 ×10⁻¹¹ ÷9.00 ×10¹⁶: 2.27 ×10⁻²⁸ kg = 0.137 u.

The accepted conversion is 931.5 MeV per u; the 934 MeV in (a) differs only because the data-list constants are rounded.

Binding energy per nucleon = total binding energy ÷ nucleon number A. It is the average energy needed to remove one nucleon, and it is the right measure of stability. Total binding energy simply grows with the number of nucleons, so it cannot compare nuclei of different sizes.

Helium-4 sits well above its neighbours: an unusually tightly bound nucleus, which is why alpha particles are emitted as a unit.

Why the curve has this shape (briefly)

The strong force is short range: a nucleon is attracted only by its nearest neighbours. In a small nucleus a large fraction of nucleons are at the surface with fewer neighbours, so binding per nucleon is low. The electrostatic repulsion between protons acts across the whole nucleus; in large nuclei it grows faster than the binding, so the binding energy per nucleon falls again.

In fission a heavy nucleus splits into two middle-mass nuclei, usually with two or three free neutrons. In uranium-235 it is triggered by absorbing a slow neutron.

The products are nearer the peak, so their binding energy per nucleon is higher (about 8.3–8.6 MeV against 7.6 MeV). Nucleon number is conserved, so a higher binding energy per nucleon means a larger total binding energy after the split. The increase is released, mostly as kinetic energy of the fragments. Free neutrons have zero binding energy, since they are not bound to anything.

WORKED EXAMPLE · QUESTION

One possible fission of uranium-235 is ²³⁵U + n →¹⁴⁰Xe + ⁹⁴Sr + 2n. Atomic masses: ²³⁵U 235.04393 u, ¹⁴⁰Xe 139.921646 u, ⁹⁴Sr 93.915356 u; neutron 1.008665 u. Calculate (a) the energy released in one fission and (b) the energy released per kilogram of uranium-235 that undergoes fission.

SOLUTION

Check the balance: nucleons 235 + 1 = 140 + 94 + 2; protons 92 = 54 + 38. The electrons balance too, so atomic masses can be used.

(a) Before: 235.04393 + 1.008665 = 236.052595 u. After: 139.921646 + 93.915356 + 2 ×1.008665 = 235.854332 u. Δm = 0.198263 u = 3.29 ×10⁻²⁸ kg, so ΔE = c²Δm: 2.96 ×10⁻¹¹ J (185 MeV).

(b) Mass of one atom: 235.04393 ×1.66 ×10⁻²⁷ = 3.90 ×10⁻²⁵ kg. 2.96 ×10⁻¹¹ ÷3.90 ×10⁻²⁵: 7.59 ×10¹³ J kg⁻¹.

For comparison, burning a kilogram of coal releases about 3 × 10⁷ J, a factor of more than a million less.

WORKED EXAMPLE · QUESTION

Binding energies per nucleon: ²³⁵U 7.59 MeV, ¹⁴⁰Xe 8.29 MeV, ⁹⁴Sr 8.59 MeV. Estimate the energy released in the fission of Worked example 3.

SOLUTION

Total binding energy = per-nucleon value × A. U: 235 ×7.59 = 1784 MeV; Xe: 140 ×8.29 = 1161 MeV; Sr: 94 ×8.59 = 807 MeV.

Released = after − before: (1161 + 807) - 1784: 184 MeV, agreeing with Worked example 3 to within rounding.

In fusion two light nuclei join to form a heavier nucleus. On the steep left-hand side of the curve, the product has a much higher binding energy per nucleon than the reactants, so the total binding energy increases and energy is released.

Per reaction, fusion releases less energy than fission (about 18 MeV against about 200 MeV), but per kilogram of fuel it releases more, because the fuel nuclei are so light.

Fusion of nuclei beyond the iron peak, or fission of nuclei lighter than it, would move away from the peak and would need an energy input.

WORKED EXAMPLE · QUESTION

²H + ³H →⁴He + n. Atomic masses: ²H 2.014102 u, ³H 3.016049 u, ⁴He 4.002603 u; neutron 1.008665 u. (a) Calculate the energy released in one reaction. (b) Compare the energy released per kilogram of fuel with the fission of Worked example 3.

SOLUTION

(a) Before 5.030151 u, after 5.011268 u, so Δm = 0.018883 u. ΔE = 0.018883 ×1.66 ×10⁻²⁷ ×(3.00 ×10⁸)²: 2.82 ×10⁻¹² J (17.6 MeV).

(b) Fuel mass (one deuterium plus one tritium atom): 5.030151 ×1.66 ×10⁻²⁷ = 8.35 ×10⁻²⁷ kg, so 2.82 ×10⁻¹² ÷8.35 ×10⁻²⁷ = 3.38 ×10¹⁴ J kg⁻¹. Then 3.38 ×10¹⁴ ÷7.59 ×10¹³ ≈4.5: about four and a half times as much energy per kilogram as uranium fission.

Nuclei are positively charged and repel each other electrostatically.

The strong force that pulls them together has a range of only about 10⁻¹⁵ m, so the nuclei must get that close before they can fuse.

The electrostatic potential energy at that separation, kQ₁Q₂/r, is the Coulomb barrier. The nuclei need enough kinetic energy to climb it.

The mean kinetic energy of particles in a gas is proportional to the absolute temperature, ½m⟨c²⟩= (3/2)kT. Fast enough nuclei mean a very high temperature, and the material is a plasma of bare nuclei and electrons.

A high density (many nuclei per unit volume) makes collisions frequent, so enough reactions happen each second to release a useful power and to keep the plasma hot.

WORKED EXAMPLE · QUESTION

Two deuterons (charge +e each) must approach to a separation of 3.0 × 10⁻¹⁵ m to fuse. They approach head-on with equal speeds. k = 8.99 ×10⁹ N m² C⁻², k_B = 1.38 ×10⁻²³ J K⁻¹. (a) Calculate the electrostatic potential energy at this separation. (b) Estimate the temperature at which the mean kinetic energy of a deuteron would be enough. (c) The core of the Sun is at about 1.5 × 10⁷ K. Comment.

SOLUTION

(a) E = kQ₁Q₂/r = (8.99 ×10⁹ ×(1.60 ×10⁻¹⁹)²)/(3.0 ×10⁻¹⁵): 7.7 ×10⁻¹⁴ J (≈ 0.48 MeV).

(b) Each deuteron supplies half, so (3/2)k_BT = 3.84 ×10⁻¹⁴ J and T = (2 ×3.84 ×10⁻¹⁴)/(3 ×1.38 ×10⁻²³): 1.9 ×10⁹ K.

(c) Mean kinetic energy in the Sun's core: (3/2) ×1.38 ×10⁻²³ ×1.5 ×10⁷ = 3.1 ×10⁻¹⁶ J, only about 0.8% of what is needed.

Fusion still happens because the particles have a spread of speeds: a tiny fraction move much faster than average, and with an enormous number of nuclei in a very dense core, that fraction is enough. (Quantum tunnelling also helps; it is beyond this specification.)

Atomic structure and electric fields (Unit 4)

The nucleus, nucleon and proton numbers, isotopes and the strong force are set up in Unit 4, along with the electronvolt and units such as MeV/c². The Coulomb barrier is the electric potential V = kQ/r of one nucleus multiplied by the charge of the other.

2 Internal energy and the ideal gas

½m⟨c²⟩= (3/2)kT comes from kinetic theory; here it turns a required kinetic energy into a temperature.

4 Radioactive decay, and stars

Every decay releases energy because the products have less mass, and hydrogen fusion powers main-sequence stars on the Hertzsprung–Russell diagram.

1. Define the binding energy of a nucleus.

2. Explain why the mass of a nucleus is less than the total mass of its separate nucleons.

3. A reaction loses 0.0500 u of mass. Calculate the energy released in MeV.

4. The binding energy of lithium-7 is 39.2 MeV. Calculate its binding energy per nucleon and state whether it is more or less stable than helium-4.

5. Explain why iron-56 cannot release energy by either fission or fusion.

6. Use the D–T result (17.6 MeV per reaction) to find the energy released per nucleon of fuel, and compare it with the 185 MeV fission of Worked example 3.

1. The energy needed to separate the nucleus completely into its individual protons and neutrons.

2. Energy is released when the nucleons bind together, so the nucleus has less energy than the separated nucleons. By ΔE = c²Δm, less energy means less mass.

3. 0.0500 ×931.5 = 46.6 MeV (via kg and c² with the data-list values: 46.7 MeV).

4. 39.2 ÷7 = 5.60 MeV per nucleon, less than helium-4's 7.09 MeV, so lithium-7 is less stable.

5. It is at (very nearly) the peak of the binding energy per nucleon curve. Splitting it or joining it with another nucleus would give products with a lower binding energy per nucleon, so energy would have to be supplied.

6. Fusion: 17.6 ÷5 = 3.52 MeV per nucleon. Fission: 185 ÷236 = 0.784 MeV per nucleon. Fusion releases about 4.5 times more per nucleon, matching the per-kilogram comparison.