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)
Raise the temperature to enable fusion reactions.
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{Eb}{A} = \frac{Eb}{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, 126These 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 > 0In 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}{92}\text{U} \rightarrow ^{222}{86}\text{Rn} .
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{\alpha} = \frac{Ai - A_f}{4} = \frac{238 - 222}{4} = 4.
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.
Step 3: Determine how many beta decays are needed to reach the final nucleus
Final nucleus is ^{222}{86}\text{Rn} with Zf = 86.
The remaining difference to reach Zf is n{\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 (Ai, Zi) and end nucleus (Af, Zf), compute:
Number of alpha decays: n{\alpha} = \frac{Ai - A_f}{4}
Z after alpha decays: Z' = Zi - 2 n{\alpha}
Number of beta decays needed: n{\beta} = Zf - 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{Eb}{A} = \frac{Eb}{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):
Eb = (\text{mass defect}) \times c^2 = (m{\text{initial}} - \text{sum of nucleon masses}) c^2Alpha 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{\alpha} = \frac{Ai - Af}{4}, Z' = Zi - 2 n{\alpha}, n{\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}