LECTURE 4: FUSION IN NUCLEAR PHYSICS

1. Fusion Industry Context

• Helion and Project Orion (Fusion Pilot Plant)

  • Site: Mulligan, Washington.

  • Chosen for grid access via Rock Island Dam hydroelectric plant.

  • Enables Helion to connect to Washington power networks and sell to Microsoft data centers.

  • Microsoft Power Purchase Agreement (PPA) to buy electricity with Helion announced in 2023.

  • Orion is Helion’s eighth device, building on the Polaris device.

  • No net energy fusion achieved yet.

  • Orion uses the field-reversed configuration (FRC), a relatively less-studied fusion approach.

  • Ambitious timeline: selling electricity by 2028 represents a very aggressive schedule with notable risk.

2. Germany’s Fusion Strategy

• Germany aims to establish a fusion power plant as a central part of its technology strategy.

• Includes hub networking for magnetic and laser fusion.

• Expansion of research infrastructures and technology demonstrators.

• The sensational framing in the agenda suggests desktop fusion-like progress, but practical feasibility is not imminent; a desktop device powering a toaster is not expected soon.

3. Avalanche Energy Milestone and Business Outlook

• Avalanche announced a milestone of $300 \text{ kilovolts}$ in a steady-state device.

• Advances the feasibility of high-performance plasmas in their orbitron concept.

• Robin Langley (CEO) framed the goal as delivering high-flux neutrons at low cost to a wide range of customers needing reliable neutron sources.

• Avalanche projects profitability as early as 2028 via neutron-based applications (radioisotope production, facility rentals, etc.).

• Desktop electricity-generating fusion plant remains very distant, but progress toward compact fusion devices that produce neutrons is notable.

4. Basic Plasma and Fusion Energy Overview

• Plasma Definition and Composition

  • Plasma is the fourth state of matter.

  • Consists of highly energetic charged particles: cations (positively charged), electrons (negatively charged), and other ions.

• Core Fusion Challenges (Two Main Tasks)

  1. Raise the temperature to enable fusion reactions.

  2. Create a containment mechanism that holds material at extremely high temperatures (order of magnitude of $10^8 \text{ K}$).

• Historical Context: Energy Balance for Fusion

  • It has taken ~100 years to obtain the first net energy output from fusion, largely because more energy had to be put in to reach and sustain the necessary conditions.

• Fusion Energy Ambition vs. Historical Energy Balance

  • Early attempts produced less energy than was required to initiate fusion.

  • Current efforts seek net energy gain and practical electricity production, with ambitious timelines (e.g., 2028 for some projects).

• Laser Fusion (General Idea)

  • In laser-driven fusion, a laser targets a micron-scale pellet containing the fusion fuel (often deuterium-tritium).

  • The laser heats the pellet specifically, not the surrounding infrastructure, avoiding overheating surrounding components.

5. Nuclear Binding Energy and Stability

• Definition and Formula

  • Binding energy (BE) of a nucleus is the energy needed to separate all nucleons.

  • Per-nucleon binding energy is given by:
    $\text{BE per nucleon} = \frac{E<em>b}{A} = \frac{E</em>b}{N+Z}$
    where $A = N+Z$ is the mass number.

• Helium-4 Example ($^4\text{He}$)

  • Total binding energy: $E_b\left( ^4\text{He} \right) \approx 28.3 \text{ MeV}$.

  • Since He-4 has 4 nucleons, the binding energy per nucleon is:
    $\frac{E_b}{A} = \frac{28.3}{4} \text{ MeV} \approx 7.075 \text{ MeV per nucleon}.$

• Energy Scale and the Chart of BE Per Nucleon

  • The BE per nucleon increases with mass number up to a peak near iron (Fe-56 region).

  • It then gradually decreases for heavier nuclei.

  • He-4 sits on the left side of the curve with a BE/A of about ~7 MeV.

  • Fe-56 is near the maximum BE/A (~8.7–8.8 MeV per nucleon).

• Stellar Nucleosynthesis and the Iron Peak

  • In stars, light nuclei fuse to form heavier nuclei, releasing energy as BE per nucleon generally increases toward Fe-56.

  • Once iron-56 is produced, further fusion does not yield net energy, marking a turning point in stellar evolution.

  • In about ~6 billion years, the Sun would begin producing iron-56; at that stage, the Sun’s life as a stable star would be effectively over.

  • Elements heavier than iron (e.g., lead, gold) are formed primarily in supernova explosions, via rapid neutron capture and other processes.

  • Concept of stardust: atoms in our bodies originate from multiple generations of stars and supernovae; some atoms in one finger may have originated in different stars.

• Nuclear Stability and Magic Numbers

  • Certain numbers of protons or neutrons yield extra stability, referred to as magic numbers:
    $2, 8, 20, 28, 50, 82, 126$

  • These stable numbers can refer to either protons or neutrons, and they arise from closed nuclear shells in the shell model.

• Even-Even vs Odd-Even vs Odd-Odd Stability Tendencies

  • Nuclei with both proton and neutron pairs (even Z and even N) tend to be more stable (paired nucleons lower energy).

  • Odd numbers of protons and neutrons, especially odd-odd nuclei, tend to be less stable.

  • Helium-4 as a highly stable nucleus: He-4 has 2 protons and 2 neutrons (an even-even nucleus with pairing), contributing to its high stability relative to many light nuclei.

• Left of the Iron Peak vs Right of the Iron Peak (Fusion vs Fission Tendency)

  • Nuclei lighter than Fe-56 tend to gain energy via fusion (fusing light nuclei increases BE/A).

  • Nuclei heavier than Fe-56 tend to gain energy via fission (fission into lighter nuclei increases BE/A).

6. Nuclear Decay, Reactions, and Energy Considerations

• Spontaneity Criterion for Nuclear Processes (Thermodynamics Perspective)

  • A nuclear reaction is spontaneous (energetically favorable) if the energy released is positive, i.e., the Q-value is positive:
    Q = \left(m{\text{reactants}} - m{\text{products}}\right) c^2 > 0

  • In general, spontaneous processes correspond to a decrease in the system’s Gibbs free energy ($G$) for chemical processes; for nuclear processes, the focus is typically on energy release (Q-value) rather than chemical Gibbs energy.

• Binding Energy and Energy Release in Reactions

  • The total binding energy relates to the mass defect: mass of nucleus is less than the sum of its constituent nucleons; the difference is released as energy when nucleons bind.

• Nuclear Decay Modes

  • Alpha decay: emission of an alpha particle ($^4\text{He}$ nucleus); reduces A by 4 and Z by 2.

  • Beta minus decay ($ \beta^-$): conversion of a neutron to a proton with emission of an electron ($e^-$) and an antineutrino; increases Z by 1, A unchanged.

  • Beta plus decay ($ \beta^+$)/electron capture: conversion of a proton to a neutron; decreases Z by 1, A unchanged.

• Practical Example: Determining the Number of Alpha and Beta Decays in a Decay Chain ($^{238}\text{U} \rightarrow ^{222}\text{Rn}$)

  • Problem setup: Determine how many alpha particles are emitted when $^{238}<em>{92}\text{U} \rightarrow ^{222}</em>{86}\text{Rn}$.

  1. Step 1: Count alpha decays from A change

     • The mass number drops from 238 to 222, so the total number of alpha emissions is:
       $n<em>{\alpha} = \frac{A</em>i - A_f}{4} = \frac{238 - 222}{4} = 4.$

  2. Step 2: Account for charge change due to alpha emissions

     • Each alpha emission reduces Z by 2, so after 4 alphas:
       $\Delta Z_{\alpha} = -2 \times 4 = -8.$

     • Starting from $Z_i = 92$, the Z after four alphas would be $Z' = 92 - 8 = 84.$

  3. Step 3: Determine how many beta decays are needed to reach the final nucleus

     • Final nucleus is $^{222}<em>{86}\text{Rn}$ with $Z</em>f = 86$.

     • The remaining difference to reach $Z<em>f$ is $n</em>{\beta} = Z_f - Z' = 86 - 84 = 2.$

  • Therefore, the decay sequence is: 4 alpha decays and 2 beta minus decays ($4\alpha + 2\beta^-$).

• Summary Method (for similar problems)

  • Given a start nucleus $(A<em>i, Z</em>i)$ and end nucleus $(A<em>f, Z</em>f)$, compute:

    • Number of alpha decays: $n<em>{\alpha} = \frac{A</em>i - A_f}{4}$

    • Z after alpha decays: $Z' = Z<em>i - 2 n</em>{\alpha}$

    • Number of beta decays needed: $n<em>{\beta} = Z</em>f - Z'$ (assuming $\beta^-$ decays; adjust if the end state requires $\beta^+$ decays)

7. Important Numerical and Symbolic References (LaTeX-formatted)

• Binding energy per nucleon definition
  $\text{BE per nucleon} = \frac{E<em>b}{A} = \frac{E</em>b}{N+Z}$

• Helium-4 BE and BE per nucleon

  • $E_b\left( ^4\text{He} \right) \approx 28.3 \text{ MeV}$

  • $\frac{E_b}{A} = \frac{28.3}{4} \text{ MeV} \approx 7.075 \text{ MeV per nucleon}$

• Peak BE per nucleon near iron

  • $\text{BE per nucleon}_{\text{max}} \text{ around } ^{56}\text{Fe} \text{ (}\approx 8.7\text{--}8.8 \text{ MeV)}$

• Stellar production and iron-56

  • Fusion in stars forms heavier nuclei up to iron-56; iron is a turning point for energy release in stellar cores.

• Magnetic and laser fusion strategy language

  • Fusion power plant planning often mentions creating research infrastructures and technology demonstrators; direct desktop-scale fusion applications are not imminent.

• Conceptual notes on fusion vs fission

  • Fusion fuels (e.g., deuterium-tritium) can minimize long-lived radioactive waste relative to fission, but some activation of reactor materials by neutrons remains a concern.

• Quick reference recap (key formulas to memorize for exams)

  • BE per nucleon: $\text{BE per nucleon} = \frac{E_b}{A}$

  • Mass-energy relation (binding energy via mass defect):
    $E<em>b = (\text{mass defect}) \times c^2 = (m</em>{\text{initial}} - \text{sum of nucleon masses}) c^2$

  • Alpha decay changes: $A \rightarrow A-4, Z \rightarrow Z-2$

  • Beta minus decay changes: $A \rightarrow A, Z \rightarrow Z+1$

  • Decay-chain analysis (example): $n<em>{\alpha} = \frac{A</em>i - A<em>f}{4}, Z' = Z</em>i - 2 n<em>{\alpha}, n</em>{\beta} = Z_f - Z'$

  • Q-value for spontaneity (energy release): Q = \left(m{\text{reactants}} - m{\text{products}}\right) c^2 > 0 \text{ for exothermic, spontaneous reactions}

• Note on units and conversions

  • $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}; 1 \text{ MeV} = 10^6 \text{ eV}$

  • Speed of light: $c \text{ approximately } 2.998 \times 10^8 \text{ m/s}$

  • Nuclear energy scales are typically given in MeV per nucleus or per nucleon; convert to joules if needed for outside calculations.

8. Final Synthesis for Exam-Style Understanding

• Know why BE per nucleon peaks near Fe-56 and how this explains why fusion of light nuclei releases energy while fission of heavy nuclei also releases energy.

• Be able to identify magic numbers and reason about stability trends (even-even vs odd-even vs odd-odd nuclei).

• Be comfortable with the difference between spontaneous (uncontrolled) fission/fusion and controlled reactor operation, including the concept of Q-values and energy release.

• Be able to perform a simple decay-chain accounting problem to determine numbers of alpha and beta decays given initial and final nuclides, as shown in the $^{238}\text{U} \rightarrow ^{222}\text{Rn}$