Skip to main content
QLDPhysicsSyllabus dot point

Topic 2: Ionising radiation and nuclear reactions

Describe nuclear fission and nuclear fusion, including the role of binding energy per nucleon, and apply mass-energy equivalence (E=mc2E = mc^2) to estimate the energy released

A focused answer to the QCE Physics Unit 1 dot point on fission and fusion. Reads the binding-energy curve to show why both reactions release energy, applies E=mc2E = mc^2 to mass defect, and works the QCAA-style energy-per-reaction problem from EA Paper 2 with worked U-235 numbers.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. Binding energy
  3. Binding energy per nucleon curve
  4. Nuclear fission
  5. Nuclear fusion
  6. Mass-energy equivalence
  7. Examples in context
  8. Try this

What this dot point is asking

QCAA wants you to explain fission (splitting of heavy nuclei) and fusion (combining of light nuclei), justify why both release energy by reference to the binding energy per nucleon curve, and apply E=mc2E = mc^2 to convert a mass defect to an energy release.

Binding energy

Binding energy is the energy needed to disassemble a nucleus into its individual nucleons. Equivalently, it is the energy released when the nucleons assemble into the nucleus.

The mass of a nucleus is always less than the sum of the masses of its individual nucleons. The difference is the mass defect:

Δm=Zmp+Nmnmnucleus\Delta m = Z m_p + N m_n - m_{\text{nucleus}}

By E=mc2E = mc^2, this missing mass corresponds to the binding energy:

EB=Δmc2E_B = \Delta m c^2

Binding energy per nucleon curve

If you plot binding energy per nucleon (EB/AE_B / A) against mass number AA, the curve rises sharply for light nuclei, peaks around iron-56 (A56A \approx 56, EB/A8.8E_B / A \approx 8.8 MeV), and slopes gently downward for heavier nuclei.

  • Below iron, combining light nuclei into heavier ones moves toward higher EB/AE_B / A, so energy is released (fusion).
  • Above iron, splitting heavy nuclei into medium-mass fragments also moves toward higher EB/AE_B / A, so energy is released (fission).

Iron is the most tightly bound; no reaction starting from iron and producing iron-and-something releases energy.

Nuclear fission

A heavy nucleus (typically uranium-235 or plutonium-239) absorbs a neutron and splits into two medium-mass fragments plus a few neutrons:

92235U+01n56141Ba+3692Kr+301n+energy^{235}_{92}\text{U} + ^1_0n \to ^{141}_{56}\text{Ba} + ^{92}_{36}\text{Kr} + 3 ^1_0n + \text{energy}

Roughly 200200 MeV per fission, most of it as kinetic energy of the fragments. The released neutrons can trigger further fissions (a chain reaction). Nuclear reactors moderate the neutrons and control the chain to produce steady heat; fission weapons let the chain run away.

Nuclear fusion

Light nuclei combine into a heavier nucleus, releasing energy. The Sun's main fusion path is the proton-proton chain, with net reaction:

411H24He+2e++2νe+energy4 ^1_1\text{H} \to ^4_2\text{He} + 2 e^+ + 2 \nu_e + \text{energy}

About 2626 MeV per net fusion. Earthbound fusion research (ITER, JET) uses the deuterium-tritium reaction:

12H+13H24He+01n+17.6^2_1\text{H} + ^3_1\text{H} \to ^4_2\text{He} + ^1_0n + 17.6 MeV

Fusion releases more energy per kilogram than fission, but requires extreme temperatures (108\sim 10^8 K) to overcome the electrostatic repulsion of nuclei.

Mass-energy equivalence

The conversion factor is:

E=mc2E = m c^2

With c=3.00×108c = 3.00 \times 10^8 m s1^{-1}, 11 kg of mass corresponds to 9×10169 \times 10^{16} J. The mass-MeV conversion is 11 u =931.5= 931.5 MeV/c2^2. Mass defects of millielectronvolts per atom translate to enormous energy releases per kilogram of fuel.

Examples in context

Example 1. A fission of one 235U^{235}\text{U} nucleus releases about 200 MeV200 \text{ MeV} (3.2×1011 J3.2 \times 10^{-11} \text{ J}), where the mass defect is Δm0.215 u\Delta m \approx 0.215 \text{ u} and E=Δmc2E = \Delta m c^2. A Gladstone-scale 700 MW700 \text{ MW} electrical plant (typical for the proposed coal-to-nuclear transition modelling) would require about 700 MW×(1/0.33)/(3.2×1011 J)700 \text{ MW} \times (1/0.33) / (3.2 \times 10^{-11} \text{ J}) which is 6.6×10196.6 \times 10^{19} fissions per second, or roughly 2.6 kg per day2.6 \text{ kg per day} of 235U^{235}\text{U}. Coal equivalent is around 7000 tonnes per day7000 \text{ tonnes per day}, a comparison QCAA EA Unit 1 questions exploit to illustrate the mass-energy advantage.

Example 2. Solar fusion in the proton-proton chain converts four protons into one helium nucleus with Δm=0.0287 u\Delta m = 0.0287 \text{ u}, releasing about 26.7 MeV26.7 \text{ MeV} per cycle. A Sunshine Coast tidal-research station drawing 4 kW4 \text{ kW} from solar panels at 2020 per cent efficiency intercepts 20 kW20 \text{ kW} of sunlight, the surface manifestation of roughly 4.7×10154.7 \times 10^{15} fusions per second occurring in the solar core to support that local power flux. The binding-energy-per-nucleon curve peaking near iron-5656 explains why both light-element fusion and heavy-element fission are energetically downhill.

Try this

Q1. Define binding energy per nucleon and identify the element with the maximum value. [2 marks]

  • Cue. Energy required per nucleon to separate into free nucleons; iron-5656 (about 8.8 MeV8.8 \text{ MeV}).

Q2. A fission event releases Δm=0.215 u\Delta m = 0.215 \text{ u}. Calculate the energy released in MeV and in joules. (1 u=931.5 MeV1 \text{ u} = 931.5 \text{ MeV}). [3 marks]

  • Cue. E=0.215×931.5=200 MeV=3.21×1011 JE = 0.215 \times 931.5 = 200 \text{ MeV} = 3.21 \times 10^{-11} \text{ J}.

Q3. Compare the energy release per kilogram of fuel in fission of 235U^{235}\text{U} and fusion of deuterium-tritium. (a) Calculate energy per kilogram for 235U^{235}\text{U} (200 MeV per nucleus200 \text{ MeV per nucleus}). (b) Calculate energy per kilogram for D-T (17.6 MeV per nucleus, average nucleus mass 2.5 u17.6 \text{ MeV per nucleus, average nucleus mass } 2.5 \text{ u}). (c) Comment on which underpins long-term energy-policy modelling for Queensland. [3+3+2 marks; ISMG: Analysis and interpretation, Evaluation]

  • Cue. (a) About 8.2×1013 J kg18.2 \times 10^{13} \text{ J kg}^{-1}; (b) about 6.8×1014 J kg16.8 \times 10^{14} \text{ J kg}^{-1}; (c) fusion higher density but not yet net-positive at grid scale.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Year 11 SAC4 marksIn a fusion reaction, two deuterium nuclei combine: 12H+12H23He+01n^2_1\text{H} + ^2_1\text{H} \to ^3_2\text{He} + ^1_0n. The mass defect is 3.27×1033.27 \times 10^{-3} u. Calculate the energy released, in MeV. (11 u =931.5= 931.5 MeV/c2^2.)
Show worked answer →

E=(Δmc2)E = (\Delta m c^2) in MeV is most easily computed from the unit conversion.

E=(3.27×103 u)×(931.5 MeV/u)=3.05E = (3.27 \times 10^{-3} \text{ u}) \times (931.5 \text{ MeV/u}) = 3.05 MeV.

The mass defect (mass of reactants minus mass of products) has been converted into kinetic energy of the products.

Markers reward use of 11 u =931.5= 931.5 MeV/c2^2, identification of the mass defect as the source of energy, and units in MeV.

Related dot points