5.3 Binding energy, fission and fusion
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A2 content. The structure of the nucleus and the strong force are in Unit 4; alpha, beta and gamma radiation and half-life are in 5.4.
What this page covers
6 points
- Nuclear binding energy and the mass deficit (mass defect)
- Using \(\Delta E = c^2\Delta m\) 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
In one minute
6 points
- 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, \(\Delta E = c^2\Delta m\).
- The binding energy of a nucleus is the energy needed to separate it completely into its individual protons and neutrons.
- 1 u = 1.66 × 10⁻²⁷ kg, equivalent to about 931.5 MeV. Use kg with \(c^2\) to get joules; divide by 1.60 × 10⁻¹³ for MeV.
- Binding energy per nucleon measures stability. It peaks at about 8.8 MeV near iron-56.
- Fission of heavy nuclei and fusion of light nuclei both move towards the peak: binding energy per nucleon rises, mass falls, energy is released.
- Fusion needs nuclei close enough for the strong force to act, against electrostatic repulsion: very high temperature (fast nuclei) and high density (frequent collisions).
Contents
13 sections
- What is binding energy, and what is the mass deficit?
- How are mass and energy linked?
- How do you convert between u, kg, J and MeV?
- What does the binding energy per nucleon curve show?
- How does fission release energy?
- How does fusion release energy?
- Why does fusion need very high temperatures and densities?
- Exam tips for 5.3
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is binding energy, and what is the mass deficit?
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:
\[\Delta m = Zm_p + (A - Z)m_n - m_{\text{nucleus}}\]Describing binding energy as energy "stored in" the nucleus that comes out when it breaks up. It is the opposite: binding energy must be supplied to break a nucleus apart. A larger binding energy means a nucleus that is harder to take apart, and a lower energy state.
How are mass and energy linked?
\[\Delta E = c^2\Delta m\]\(\Delta E\) is in J, \(\Delta m\) in kg and \(c = 3.00 \times 10^{8}\) 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.
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.
Separate nucleons: \(2 \times 1.007276 + 2 \times 1.008665 = 4.031882\) u.
Mass deficit: \(\Delta m = 4.031882 - 4.00151 = 0.03037\) u \(= 0.03037 \times 1.66 \times 10^{-27} = 5.04 \times 10^{-29}\) kg.
\(\Delta E = (3.00 \times 10^{8})^2 \times 5.04 \times 10^{-29}\): \(4.54 \times 10^{-12}\) J \(= 4.54 \times 10^{-12} \div 1.60 \times 10^{-13}\) = 28.4 MeV.
Per nucleon (A = 4): \(28.4 \div 4\): 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 mass deficit is a small difference between two nearly equal numbers, so keep every figure of the masses until the subtraction is done. Rounding first (for example 1.67 × 10⁻²⁷ kg for both nucleons) can wipe out the answer completely.
How do you convert between u, kg, J and MeV?
- 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.
| From | To | Do this |
|---|---|---|
| mass in u | mass in kg | × 1.66 × 10⁻²⁷ |
| mass in kg | energy in J | × c² = × 9.00 × 10¹⁶ |
| energy in J | energy in MeV | ÷ 1.60 × 10⁻¹³ |
| mass in u | energy in MeV | × 931.5 (only when the question gives this value, or as a check) |
(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.
(a) \(1.66 \times 10^{-27} \times (3.00 \times 10^{8})^2 = 1.494 \times 10^{-10}\) J; \(\div 1.60 \times 10^{-13}\): 934 MeV ≈ 930 MeV.
(b) Total binding energy \(= 7.98 \times 16 = 127.7\) MeV \(= 127.7 \times 1.60 \times 10^{-13} = 2.04 \times 10^{-11}\) J. \(\Delta m = 2.04 \times 10^{-11} \div 9.00 \times 10^{16}\): \(2.27 \times 10^{-28}\) 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.
Multiplying a mass in u by \(c^2\). \(\Delta E = c^2\Delta m\) gives joules only when \(\Delta m\) is in kilograms. Convert first, or use 931.5 MeV per u, but never both in the same step.
What does the binding energy per nucleon curve show?
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.
Say what happens to the binding energy per nucleon (it increases), then the consequence: the products are more tightly bound, their total mass is less than that of the reactants, and the mass difference appears as kinetic energy of the products.
How does fission release energy?
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.
One possible fission of uranium-235 is \({}^{235}\text{U} + \text{n} \rightarrow {}^{140}\text{Xe} + {}^{94}\text{Sr} + 2\text{n}\). 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.
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: \(139.921646 + 93.915356 + 2 \times 1.008665 = 235.854332\) u. \(\Delta m = 0.198263\) u \(= 3.29 \times 10^{-28}\) kg, so \(\Delta E = c^2\Delta m\): \(2.96 \times 10^{-11}\) J (185 MeV).
(b) Mass of one atom: \(235.04393 \times 1.66 \times 10^{-27} = 3.90 \times 10^{-25}\) kg. \(2.96 \times 10^{-11} \div 3.90 \times 10^{-25}\): \(7.59 \times 10^{13}\) J kg⁻¹.
For comparison, burning a kilogram of coal releases about 3 × 10⁷ J, a factor of more than a million less.
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.
Total binding energy = per-nucleon value × A. U: \(235 \times 7.59 = 1784\) MeV; Xe: \(140 \times 8.29 = 1161\) MeV; Sr: \(94 \times 8.59 = 807\) MeV.
Released = after − before: \((1161 + 807) - 1784\): 184 MeV, agreeing with Worked example 3 to within rounding.
Subtracting the binding energies the wrong way round, getting a negative answer, and reporting the energy as "absorbed". Energy is released when the total binding energy increases: released = after − before.
How does fusion release energy?
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.
\({}^{2}\text{H} + {}^{3}\text{H} \rightarrow {}^{4}\text{He} + \text{n}\). 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.
(a) Before 5.030151 u, after 5.011268 u, so \(\Delta m = 0.018883\) u. \(\Delta E = 0.018883 \times 1.66 \times 10^{-27} \times (3.00 \times 10^{8})^2\): \(2.82 \times 10^{-12}\) J (17.6 MeV).
(b) Fuel mass (one deuterium plus one tritium atom): \(5.030151 \times 1.66 \times 10^{-27} = 8.35 \times 10^{-27}\) kg, so \(2.82 \times 10^{-12} \div 8.35 \times 10^{-27} = 3.38 \times 10^{14}\) J kg⁻¹. Then \(3.38 \times 10^{14} \div 7.59 \times 10^{13} \approx 4.5\): about four and a half times as much energy per kilogram as uranium fission.
Why does fusion need very high temperatures and densities?
- 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, \(kQ_1Q_2/r\), 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, \(\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT\). 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.
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. \(k = 8.99 \times 10^{9}\) N m² C⁻², \(k_B = 1.38 \times 10^{-23}\) 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.
(a) \(E = \dfrac{kQ_1Q_2}{r} = \dfrac{8.99 \times 10^{9} \times (1.60 \times 10^{-19})^2}{3.0 \times 10^{-15}}\): \(7.7 \times 10^{-14}\) J (≈ 0.48 MeV).
(b) Each deuteron supplies half, so \(\tfrac{3}{2}k_BT = 3.84 \times 10^{-14}\) J and \(T = \dfrac{2 \times 3.84 \times 10^{-14}}{3 \times 1.38 \times 10^{-23}}\): \(1.9 \times 10^{9}\) K.
(c) Mean kinetic energy in the Sun's core: \(\tfrac{3}{2} \times 1.38 \times 10^{-23} \times 1.5 \times 10^{7} = 3.1 \times 10^{-16}\) 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.)
Nuclei are positive → they repel → they must get within the short range of the strong force → so they need a large kinetic energy → mean kinetic energy ∝ T. For density, say it increases the rate of collisions, and so the rate of fusion.
Exam tips for 5.3
Check that nucleon numbers and proton numbers balance on both sides, and count the free neutrons. A missing neutron changes \(\Delta m\) by about 1 u, hundreds of times the answer.
Copy masses with every figure given and round only the final energy. The answer is the small difference between two large numbers.
Either u → kg → (× \(c^2\)) J → MeV, or u × 931.5 → MeV when that value is given. State the route in your working so each conversion can be followed.
Binding energy per nucleon is the total divided by the nucleon number \(A\), not the proton number \(Z\). To go back, multiply by \(A\).
Binding energy per nucleon increases → total binding energy increases → mass of products is less → \(\Delta E = c^2\Delta m\) released as kinetic energy. Each link is a separate point.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Binding energy is energy stored in a nucleus and released when it breaks up. | Work must be done against the strong force to separate nucleons. | Binding energy is the energy needed to pull a nucleus apart; it is released when the nucleus forms. |
| The nucleus with the largest binding energy is the most stable. | Total binding energy grows with the number of nucleons. | Compare binding energy per nucleon; iron-56 is near the top. |
| In fission, some nucleons are destroyed and turned into energy. | Nucleon number and proton number are conserved. | Each nucleon ends up more tightly bound, and the total mass falls slightly. |
| Mass changes only in nuclear reactions. | \(\Delta E = c^2\Delta m\) applies to every energy change. | Chemical mass changes exist but are far too small to measure. |
| Fusion of any two nuclei releases energy. | Beyond the iron peak, the product would have a lower binding energy per nucleon. | Fusion releases energy only for light nuclei; fission only for heavy ones. |
| Fusion needs a high temperature to "ignite" the fuel like a fire. | Nothing is burning; the issue is electrostatic repulsion. | High temperature gives nuclei enough kinetic energy to get close enough for the strong force to act. |
How this topic connects
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.
5.2 Internal energy and the ideal gas
\(\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT\) comes from kinetic theory; here it turns a required kinetic energy into a temperature.
5.4 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.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(\Delta E = c^2\Delta m\) | energy from a mass difference (mass in kg, energy in J) | yes |
| \(\Delta m = Zm_p + (A - Z)m_n - m_{\text{nucleus}}\) | mass deficit of a nucleus | no — learn it |
| binding energy per nucleon = binding energy ÷ A | comparing stability | no — learn it |
| 1 u = 1.66 × 10⁻²⁷ kg | converting masses | given in the data list |
| 1 u ≡ 931.5 MeV | quick u → MeV conversion | no — derive it from 1 u × c², or use when given |
| \(E = kQ_1Q_2/r\) | Coulomb barrier height | no — derive it from V = kQ/r (in the list) × charge |
| \(\tfrac{1}{2}m\langle c^2\rangle = \tfrac{3}{2}kT\) | mean kinetic energy ↔ temperature | yes |
Check your understanding
1. Define the binding energy of a nucleus.
The energy needed to separate the nucleus completely into its individual protons and neutrons.
2. Explain why the mass of a nucleus is less than the total mass of its separate nucleons.
Energy is released when the nucleons bind together, so the nucleus has less energy than the separated nucleons. By \(\Delta E = c^2\Delta m\), less energy means less mass.
3. A reaction loses 0.0500 u of mass. Calculate the energy released in MeV.
\(0.0500 \times 931.5 = 46.6\) MeV (via kg and \(c^2\) with the data-list values: 46.7 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.
\(39.2 \div 7 = 5.60\) MeV per nucleon, less than helium-4's 7.09 MeV, so lithium-7 is less stable.
5. Explain why iron-56 cannot release energy by either fission or fusion.
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. 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.
Fusion: \(17.6 \div 5 = 3.52\) MeV per nucleon. Fission: \(185 \div 236 = 0.784\) MeV per nucleon. Fusion releases about 4.5 times more per nucleon, matching the per-kilogram comparison.
Specification checklist
Wording is ours; numbers refer to the specification points.
- (133) I can explain nuclear binding energy and the mass deficit, and use \(\Delta E = c^2\Delta m\) to link a mass difference to an energy for nuclei and nuclear reactions.
- (134) I can use the atomic mass unit and convert masses and energies between u, kg, J and MeV.
- (135) I can use the binding energy per nucleon curve, with its peak near iron, to explain why fission of heavy nuclei and fusion of light nuclei release energy.
- (136) I can describe fusion and explain why it needs very high temperatures and densities, using the electrostatic repulsion between nuclei and the link between kinetic energy and temperature.