Account for the energy released in nuclear fission and fusion in terms of mass defect and binding energy, using E = mc^2 and the binding energy curve

What this dot point is asking

NESA wants you to define mass defect and binding energy, calculate them from atomic mass data using $E = m c^2$ (with $1 \text{ u} \to 931.5$ MeV), describe the shape of the binding-energy-per-nucleon curve, and use it to explain why both nuclear fission of heavy nuclei and nuclear fusion of light nuclei release energy.

The answer

Mass defect and binding energy

The mass of a nucleus is always slightly less than the sum of the masses of its constituent protons and neutrons. The difference is the mass defect:

By $E = m c^2$, this "missing" mass corresponds to the binding energy of the nucleus:

This is the energy required to separate the nucleus into free protons and neutrons. Equivalently, it is the energy released when free nucleons assemble into the bound nucleus.

A useful unit conversion: $1 \text{ u} \cdot c^2 = 931.5$ MeV, so a mass defect in atomic mass units converts directly to a binding energy in MeV.

Binding energy per nucleon

The bound state of a nucleus is more meaningful when normalised by the number of nucleons:

This is the average energy needed to remove one nucleon. Plotted against mass number $A$, it produces the famous binding-energy-per-nucleon curve:

• $A = 1$ (hydrogen-1): 0 (single proton, no binding).
• $A = 2$ (deuterium): 1.11 MeV per nucleon.
• $A = 3$ (helium-3, tritium): 2.6-2.8 MeV per nucleon.
• $A = 4$ (helium-4): 7.07 MeV per nucleon (a local peak, very tightly bound).
• $A = 12$ (carbon-12): 7.7 MeV per nucleon.
• $A = 56$ (iron-56): 8.79 MeV per nucleon. Near the maximum.
• $A = 235$ (uranium-235): 7.6 MeV per nucleon. Declining.

The curve rises steeply for light nuclei, peaks near $A = 56$, and falls slowly for heavy nuclei.

Rule for releasing energy

A nuclear process releases energy if the products have higher binding energy per nucleon than the reactants. Geometrically, this means the reaction moves the nucleons "uphill" on the binding-energy-per-nucleon curve, toward the iron peak.

• Light side of the peak. Combining light nuclei (fusion) moves up toward iron: energy is released.
• Heavy side of the peak. Splitting heavy nuclei (fission) moves up toward iron: energy is released.
• At the peak (iron). Neither fission nor fusion releases energy. Iron is the end point of stellar nucleosynthesis (see the related dot point on stars).

Nuclear fission

A heavy unstable nucleus splits into two medium-mass fragments and a few neutrons. Example, induced fission of uranium-235 by a thermal neutron:

The barium and krypton fragments have $E_B / A \approx 8.5$ MeV per nucleon, while uranium has 7.6 MeV per nucleon. The increase of about 0.9 MeV per nucleon over 235 nucleons gives about 200 MeV released per fission event, carried away as kinetic energy of the fragments and neutrons plus prompt and delayed gamma radiation.

A chain reaction can occur if the released neutrons trigger further fissions. In a controlled reactor, a moderator (water, graphite) slows neutrons to thermal energies for efficient capture by $^{235}$U, and control rods absorb excess neutrons to keep the reaction critical (one fission triggers exactly one more). In a fission weapon, no control is used.

Nuclear fusion

Two light nuclei combine into a heavier one. Example, deuterium-tritium fusion (the easiest practical fusion reaction):

Deuterium has 1.11 MeV per nucleon and tritium 2.83 MeV per nucleon. Helium-4 has 7.07 MeV per nucleon. The energy released is $\Delta m \, c^2 = 17.6$ MeV per reaction. Even though only 5 nucleons rearrange, the change per nucleon is much larger than for fission, so fusion is more energy-dense per kilogram of fuel.

Fusion requires temperatures of $\sim 10^8$ K to overcome the Coulomb repulsion between the positively charged nuclei, plus high densities and confinement times. In stars (Sun and similar), the proton-proton chain fuses four protons into a helium-4 nucleus, releasing 26.7 MeV per cycle, the energy source that sustains stellar luminosity.

On Earth, achieving net-energy-positive fusion is the long-running goal of magnetic-confinement (tokamak) and inertial-confinement (laser-driven) research.

Worked example: deuterium-tritium fusion

Calculate the energy released. Atomic masses: $m(^2$H$) = 2.01410$ u, $m(^3$H$) = 3.01605$ u, $m(^4$He$) = 4.00260$ u, $m(n) = 1.00867$ u.

Reactants: $2.01410 + 3.01605 = 5.03015$ u.

Products: $4.00260 + 1.00867 = 5.01127$ u.

Mass defect: $\Delta m = 5.03015 - 5.01127 = 0.01888$ u.

Energy: $E = \Delta m \times 931.5 = 0.01888 \times 931.5 = 17.6$ MeV.

Most of this energy is carried by the neutron (about 14.1 MeV) because the helium-4 nucleus is heavier and recoils more slowly. In a power reactor the neutron would deposit its energy in a blanket of lithium to breed more tritium and heat a working fluid.

Try it: Mass-energy calculator to convert between mass defects (in u or kg) and energy releases (in MeV or J) for any nuclear reaction.

Worked example: uranium-235 fission

Approximate energy per fission. Take the change in $E_B / A$ as 0.9 MeV per nucleon and $A = 235$ nucleons.

$E \approx 235 \times 0.9 = 210$ MeV. (Typical experimental value 200 MeV.)

For comparison, burning a single carbon atom in a chemical reaction releases only a few eV. Fission releases about $10^8$ times more energy per atom, which is why nuclear processes can power cities and weapons with kilograms of fuel.

Examples in context

Example 1. Binding energy per nucleon for $^{56}$Fe via Lucas Heights mass-spectrometer data. ANSTO Lucas Heights mass spectrometers measure $^{56}$Fe at $m = 55.9349 \text{ u}$. Total constituent mass: $26 m_p + 30 m_n = 26 \times 1.00728 + 30 \times 1.00867 = 56.4615 \text{ u}$. Mass defect $\Delta m = 56.4615 - 55.9349 = 0.5266 \text{ u}$. Binding energy $E_b = \Delta m \times 931.5 \text{ MeV} = 491 \text{ MeV}$. Per nucleon: $E_b / A = 491 / 56 = 8.77 \text{ MeV/nucleon}$, the peak of the binding-energy curve. This is why $^{56}$Fe is the cosmic "ash" of stellar fusion - no further energy can be released by fusing or splitting iron nuclei.

Example 2. Energy budget of a $^{235}$U fission at OPAL. A typical OPAL fission $^{235}_{92}\text{U} + n \to {}^{141}_{56}\text{Ba} + {}^{92}_{36}\text{Kr} + 3n$ has reactant mass $235.0439 + 1.0087 = 236.053 \text{ u}$. Products: $140.914 + 91.926 + 3 \times 1.0087 = 235.866 \text{ u}$. Mass defect $\Delta m = 236.053 - 235.866 = 0.187 \text{ u} = 174 \text{ MeV}$. Including delayed energy from fission-product decay, total per event is $\sim 200 \text{ MeV}$. Per kg of $^{235}$U: $E = 200 \times 1.6 \times 10^{-13} \times 6.02 \times 10^{23} / 0.235 = 8.2 \times 10^{13} \text{ J/kg}$, equal to $\sim 2 \times 10^6$ kg of coal.

Try this

Q1. Define mass defect and binding energy and state their relationship via $E = mc^2$. [2 marks]

• Cue. $\Delta m$: difference between sum of free-nucleon masses and bound-nucleus mass. $E_b = \Delta m c^2$ (or $\Delta m \times 931.5 \text{ MeV}$).

Q2. Calculate the binding energy per nucleon for $^4$He given mass $= 4.0026 \text{ u}$, with $m_p = 1.00728 \text{ u}$ and $m_n = 1.00867 \text{ u}$. [3 marks]

• Cue. Free nucleons: $2 \times 1.00728 + 2 \times 1.00867 = 4.0319 \text{ u}$. $\Delta m = 0.0293 \text{ u} = 27.3 \text{ MeV}$. Per nucleon: $6.83 \text{ MeV}$.

Q3. A D-T fusion reaction $^2_1 \text{H} + ^3_1 \text{H} \to {}^4_2 \text{He} + n$ releases $17.6 \text{ MeV}$. (a) Identify why fusion releases energy despite both reactants being light. (b) Calculate the mass defect in kg. (c) Compare fusion vs fission energy yield per unit reactant mass. [2+2+3 marks]

• Cue. (a) $^4$He has higher binding energy per nucleon than D and T. (b) $\Delta m = 17.6 / 931.5 = 0.0189 \text{ u} = 3.14 \times 10^{-29} \text{ kg}$. (c) Fusion: $\sim 3.4 \times 10^{14} \text{ J/kg}$; fission $\sim 8 \times 10^{13} \text{ J/kg}$; fusion $\sim 4 \times$ more energy per kg of reactants.