Unit 1: Thermal, nuclear and electrical physics

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 ($E = mc^2$) 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 $E = mc^2$ to mass defect, and works the QCAA-style energy-per-reaction problem from EA Paper 2 with worked U-235 numbers.

Generated by Claude OpusReviewed by Better Tuition Academy6 min answer

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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.

Worked example

A U-235 fission releases on average 200200 MeV per nucleus. Find the energy released per kilogram of U-235.

Number of nuclei in 11 kg: N=(1000 g)/(235 g mol1)×6.02×1023=2.56×1024N = (1000 \text{ g}) / (235 \text{ g mol}^{-1}) \times 6.02 \times 10^{23} = 2.56 \times 10^{24}.

Energy: E=N×200 MeV=5.12×1026E = N \times 200 \text{ MeV} = 5.12 \times 10^{26} MeV =8.2×1013= 8.2 \times 10^{13} J.

For comparison, burning 11 kg of coal releases roughly 3×1073 \times 10^7 J, so fission is more than a million times more energy-dense.

Common traps

Saying fusion is the opposite of fission. Both release energy by moving toward iron on the binding-energy curve. Fusion combines light nuclei, fission splits heavy ones. They are different routes to the same peak.

Forgetting that mass is "lost". Mass defect converts to kinetic energy of the products. The total relativistic mass-energy is conserved.

Treating E=mc2E = mc^2 as requiring high speed. E=mc2E = mc^2 applies to rest mass. Speed enters through the relativistic mass-energy formula, which is the Unit 4 extension.

Confusing the chain reaction with the energy release. The chain reaction is the multiplication of fissions; the energy per fission is set by the nuclear physics and is not affected by the chain.

In one sentence

Fission splits a heavy nucleus (such as U-235) into medium-mass fragments and fusion combines light nuclei (such as deuterium and tritium) into a heavier one; both release energy because the binding energy per nucleon is higher for the products than the reactants, and the energy released equals Δmc2\Delta m c^2 where Δm\Delta m is the mass defect.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

Year 11 SAC4 marksIn a fusion reaction, two deuterium nuclei combine: $^2_1\text{H} + ^2_1\text{H} \to ^3_2\text{He} + ^1_0n$. The mass defect is $3.27 \times 10^{-3}$ u. Calculate the energy released, in MeV. ($1$ u $= 931.5$ MeV/c$^2$.)
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.

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