Nuclear fission and fusion (QCE Physics Unit 1)

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 $E = mc^2$ 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:

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

Binding energy per nucleon curve

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

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

$^{235}_{92}\text{U} + ^1_0n \to ^{141}_{56}\text{Ba} + ^{92}_{36}\text{Kr} + 3 ^1_0n + \text{energy}$

Roughly $200$ 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:

$4 ^1_1\text{H} \to ^4_2\text{He} + 2 e^+ + 2 \nu_e + \text{energy}$

About $26$ MeV per net fusion. Earthbound fusion research (ITER, JET) uses the deuterium-tritium reaction:

$^2_1\text{H} + ^3_1\text{H} \to ^4_2\text{He} + ^1_0n + 17.6$ MeV

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

Mass-energy equivalence

The conversion factor is:

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

Examples in context

Example 1. A fission of one $^{235}\text{U}$ nucleus releases about $200 \text{ MeV}$ ($3.2 \times 10^{-11} \text{ J}$), where the mass defect is $\Delta m \approx 0.215 \text{ u}$ and $E = \Delta m c^2$. A Gladstone-scale $700 \text{ MW}$ electrical plant (typical for the proposed coal-to-nuclear transition modelling) would require about $700 \text{ MW} \times (1/0.33) / (3.2 \times 10^{-11} \text{ J})$ which is $6.6 \times 10^{19}$ fissions per second, or roughly $2.6 \text{ kg per day}$ of $^{235}\text{U}$. Coal equivalent is around $7000 \text{ tonnes per day}$, 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 $\Delta m = 0.0287 \text{ u}$, releasing about $26.7 \text{ MeV}$ per cycle. A Sunshine Coast tidal-research station drawing $4 \text{ kW}$ from solar panels at $20$ per cent efficiency intercepts $20 \text{ kW}$ of sunlight, the surface manifestation of roughly $4.7 \times 10^{15}$ fusions per second occurring in the solar core to support that local power flux. The binding-energy-per-nucleon curve peaking near iron-$56$ 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-$56$ (about $8.8 \text{ MeV}$).

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

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

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

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