Welcome to Nuclear Fission and Fusion
Welcome to one of the most exciting and powerful topics in A2 Physics! In this chapter, we will explore how tiny changes in atomic nuclei can release mind-boggling amounts of energy. From powering the Sun to generating electricity in nuclear power stations, the processes of nuclear fission and nuclear fusion shape our universe and our everyday technology.
Don't worry if this topic sounds intimidating at first! We will break down every concept step-by-step, use intuitive everyday analogies, and walk through the calculations carefully so you can master the CCEA exam questions with confidence.
1. Mass-Energy Equivalence and Nuclear Binding Energy
Mass and Energy: Two Sides of the Same Coin
Before 1905, scientists believed that mass and energy were completely separate quantities. Albert Einstein changed everything with his famous equation:
\(\Delta E = \Delta m c^2\)
Where:
• \(\Delta E\) is the change in energy measured in joules (\(\text{J}\))
• \(\Delta m\) is the change in mass measured in kilograms (\(\text{kg}\))
• \(c\) is the speed of light in a vacuum (\(3.00 \times 10^8\text{ m s}^{-1}\))
Because \(c^2\) is an enormous number (\(9.00 \times 10^{16}\text{ m}^2\text{ s}^{-2}\)), even a tiny amount of mass represents a colossal amount of energy!
The Unified Atomic Mass Unit (\(\text{u}\))
Dealing with masses of individual protons, neutrons, and nuclei in kilograms leads to tiny numbers like \(10^{-27}\text{ kg}\). To make life easier, nuclear physicists define the unified atomic mass unit (\(\text{u}\)):
Definition: One unified atomic mass unit (\(1\text{ u}\)) is defined as exactly \(\frac{1}{12}\)th of the mass of an unbound neutral carbon-12 (\(^{12}_{6}\text{C}\)) atom.
\(1\text{ u} = 1.661 \times 10^{-27}\text{ kg}\)
Using \(\Delta E = \Delta m c^2\), we can convert \(1\text{ u}\) directly into energy:
\(\Delta E = (1.661 \times 10^{-27}\text{ kg}) \times (3.00 \times 10^8\text{ m s}^{-1})^2 \approx 1.495 \times 10^{-10}\text{ J}\)
In nuclear physics, we often express energy in mega-electronvolts (\(\text{MeV}\)), where \(1\text{ eV} = 1.60 \times 10^{-19}\text{ J}\), so \(1\text{ MeV} = 1.60 \times 10^{-13}\text{ J}\). Converting to \(\text{MeV}\):
\(1\text{ u} \equiv 931.5\text{ MeV}\) (or \(931\text{ MeV}\) on standard data sheets)
Top Tip: This conversion is a massive time-saver in exams! If you calculate a mass change in \(\text{u}\), simply multiply by \(931.5\text{ MeV u}^{-1}\) to get the energy released in \(\text{MeV}\).
Mass Defect (\(\Delta m\))
Did you know? If you weigh the individual constituent protons and neutrons of a nucleus separately, their total combined mass is greater than the mass of the assembled nucleus itself!
Definition: The mass defect (\(\Delta m\)) of a nucleus is the difference between the total mass of the individual, separated nucleons and the actual mass of the bound nucleus.
\(\Delta m = \left( Z m_p + (A - Z) m_n \right) - m_{\text{nucleus}}\)
Where:
• \(Z\) is the proton number (atomic number)
• \(A\) is the nucleon number (mass number)
• \((A - Z)\) is the number of neutrons
• \(m_p\) is the mass of a proton
• \(m_n\) is the mass of a neutron
• \(m_{\text{nucleus}}\) is the measured mass of the nucleus
Binding Energy (\(E_B\))
Where did that "missing" mass go? When nucleons bind together via the strong nuclear force, potential energy is released. Because energy has left the system, the mass of the nucleus decreases by \(\Delta m\).
Definition: The binding energy of a nucleus is defined as the minimum work or energy required to completely separate a nucleus into its individual constituent protons and neutrons (or the energy released when the nucleus is assembled from separated nucleons).
\(E_B = \Delta m \times c^2\)
Everyday Analogy: Imagine building a Lego tower using special magnetized bricks. When they snap together, they click loudly (releasing sound/energy). To pull the tower apart into single bricks again, you must put energy back in. The binding energy is simply the energy needed to pull those nuclear "bricks" completely apart.
Binding Energy per Nucleon
Total binding energy alone does not tell us how stable a nucleus is. A huge nucleus with 200 nucleons might have a large total binding energy simply because it has many particles, but it might still be easy to break apart.
To compare stability, we use the binding energy per nucleon:
\(\text{Binding Energy per Nucleon} = \frac{\text{Total Binding Energy } (E_B)}{\text{Nucleon Number } (A)}\)
Crucial Rule: The higher the binding energy per nucleon, the more tightly bound and more stable the nucleus is!
Section 1 Key Takeaway: Mass and energy are interchangeable. The mass of a nucleus is always less than the mass of its separate nucleons; this difference is the mass defect, which corresponds to the binding energy holding the nucleus together.
2. The Binding Energy per Nucleon Curve
Key Features of the Curve
When we plot the binding energy per nucleon against nucleon number (\(A\)) for all known nuclides, we get one of the most famous graphs in physics:
• Rapid rise for light nuclei: For small values of \(A\) (such as \(^{2}_{1}\text{H}\), \(^{3}_{1}\text{H}\), \(^{4}_{2}\text{He}\)), the curve rises steeply. There are distinct local peaks for exceptionally stable light nuclei, notably Helium-4 (\(^{4}_{2}\text{He}\)), Carbon-12 (\(^{12}_{6}\text{C}\)), and Oxygen-16 (\(^{16}_{8}\text{O}\)).
• The Maximum Peak at Iron-56 (\(^{56}_{26}\text{Fe}\)): The curve reaches a maximum peak at \(A = 56\) with a value of approximately \(8.8\text{ MeV per nucleon}\). Iron-56 is the most stable nucleus in the universe.
• Gradual decline for heavy nuclei: For \(A > 56\), the curve slopes gently downward, falling to around \(7.6\text{ MeV per nucleon}\) for heavy elements like Uranium-238 (\(^{238}_{92}\text{U}\)).
Why Energy is Released in Fission and Fusion
Any nuclear reaction that moves products closer to the peak at Iron-56 results in products with a higher binding energy per nucleon. Higher binding energy per nucleon means the nucleons become more tightly bound, their mass per nucleon decreases, and the difference in mass is released as energy!
• Nuclear Fusion (Left of Peak): Two light nuclei with low \(A\) combine to form a heavier, more stable nucleus. Because they move up the steep left side of the curve towards iron, there is a large increase in binding energy per nucleon, releasing huge amounts of energy.
• Nuclear Fission (Right of Peak): A heavy, unstable nucleus with large \(A\) splits into two lighter, more stable nuclei. Moving from the right side upward towards the peak increases the binding energy per nucleon, releasing energy.
Common Mistake to Avoid: Students often write "energy is released because binding energy is destroyed." This is incorrect! Energy is released because the products are more tightly bound (have a higher total binding energy) than the reactants, meaning total mass has decreased.
Section 2 Key Takeaway: Iron-56 sits at the peak of stability. Light nuclei undergo fusion and heavy nuclei undergo fission to move toward this peak, releasing energy in the process.
3. Nuclear Fission
What is Nuclear Fission?
Definition: Nuclear fission is the splitting of a large, heavy, unstable nucleus into two smaller, more stable daughter nuclei (fission fragments), accompanied by the release of several neutrons and energy.
While some heavy isotopes undergo spontaneous fission, in nuclear reactors we use induced fission.
Induced Fission of Uranium-235
In induced fission, a slow-moving neutron (called a thermal neutron) is captured by a Uranium-235 nucleus. This forms an unstable compound nucleus, Uranium-236, which oscillates and promptly splits:
\(^{1}_{0}\text{n} + ^{235}_{92}\text{U} \to ^{236}_{92}\text{U}^* \to ^{141}_{56}\text{Ba} + ^{92}_{36}\text{Kr} + 3\,^{1}_{0}\text{n} + \text{Energy}\)
Note: The daughter fragments (like Barium-141 and Krypton-92) can vary, but the total nucleon number (\(A\)) and proton number (\(Z\)) are always conserved on both sides of the equation.
Chain Reactions
Each fission event produces \(2\) to \(3\) fast neutrons. If these released neutrons are captured by other \(^{235}\text{U}\) nuclei, they cause further fissions, which produce even more neutrons. This self-sustaining process is called a chain reaction.
• Subcritical: On average, fewer than 1 emitted neutron causes another fission; the reaction dies out.
• Critical: Exactly 1 emitted neutron per fission goes on to cause another fission; the reaction continues at a steady, controlled rate (normal operating condition of a power station).
• Supercritical: More than 1 neutron per fission causes further fission; the reaction rate grows exponentially (used in nuclear weapons).
Components of a Thermal Nuclear Reactor
A commercial thermal reactor is carefully designed to keep the chain reaction steady and safe. You must know the function of each main component for your CCEA exam:
1. Fuel Rods:
Contain enriched uranium (typically \(2\text{–}5\%\) \(^{235}\text{U}\) and \(95\text{–}98\%\) non-fissile \(^{238}\text{U}\)). Naturally occurring uranium contains only about \(0.7\%\) \(^{235}\text{U}\).
2. Moderator:
• Material: Water, heavy water, or graphite.
• Purpose: Slows down fast fission neutrons to thermal speeds (around \(2\text{ km s}^{-1}\), kinetic energy \(\approx 0.025\text{ eV}\)).
• How it works: Fast neutrons collide elastically with light nuclei (like \(\text{H}\) or \(\text{C}\)) in the moderator, transferring kinetic energy and slowing down. Fast neutrons are simply captured by \(^{238}\text{U}\) without causing fission; only slow (thermal) neutrons are readily absorbed by \(^{235}\text{U}\) to cause fission.
3. Control Rods:
• Material: Boron or Cadmium (materials that are excellent neutron absorbers).
• Purpose: Control the rate of the chain reaction by absorbing excess neutrons.
• Operation: Lowering the rods into the core absorbs more neutrons and slows down the reaction. Raising the rods increases the reaction rate. In an emergency, rods are fully inserted to shut down the reactor instantly (a "SCRAM").
4. Coolant:
• Material: Pressurized water, liquid sodium, or carbon dioxide gas.
• Purpose: Carries thermal energy away from the reactor core to a heat exchanger/boiler to produce steam, which drives turbines to generate electricity.
5. Radiation Shielding:
• Material: Thick steel pressure vessel surrounded by high-density reinforced concrete (several metres thick).
• Purpose: Absorbs dangerous escaping neutrons and high-energy gamma (\(\gamma\)) radiation to protect workers and the environment.
Section 3 Key Takeaway: Fission splits heavy nuclei using thermal neutrons. Reactors control this chain reaction using moderators (to slow neutrons), control rods (to absorb neutrons), coolant (to remove heat), and thick shielding (for safety).
4. Nuclear Fusion
What is Nuclear Fusion?
Definition: Nuclear fusion is the process where two small, light nuclei join together (fuse) to form a single, heavier, more stable nucleus, releasing vast amounts of energy.
Fusion powers the Sun and all main-sequence stars. A typical fusion reaction being developed for Earth-based power stations is the fusion of two hydrogen isotopes, deuterium (\(^{2}_{1}\text{H}\)) and tritium (\(^{3}_{1}\text{H}\)):
\(^{2}_{1}\text{H} + ^{3}_{1}\text{H} \to ^{4}_{2}\text{He} + ^{1}_{0}\text{n} + 17.6\text{ MeV}\)
Why is Fusion so Difficult to Achieve? (The Coulomb Barrier)
Nuclei have positive electrical charges because they contain protons. According to Coulomb's Law, two positively charged particles repel each other with an electrostatic force that increases sharply as they get closer:
\(F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}\)
For fusion to occur, the two nuclei must get close enough (within \(\approx 10^{-15}\text{ m}\), or \(1\text{ fm}\)) for the short-range strong nuclear force to overcome this electrostatic repulsion and pull them together.
Conditions Required for Terrestrial Fusion
To overcome this massive Coulomb repulsion barrier, fusion reactors require extreme conditions:
1. Extremely High Temperatures (\(T \approx 10^8\text{ K}\)):
High temperatures give the nuclei tremendous kinetic energy (\(E_k \approx \frac{3}{2}kT\)) so they travel fast enough to overcome electrostatic repulsion during head-on collisions.
2. High Particle Density and Confinement:
At these temperatures, matter exists as a plasma (a gas of stripped ions and free electrons). The plasma must be dense enough and held together long enough for frequent collisions to occur.
Technical Challenge: No physical material can withstand contact with plasma at \(100\text{ million Kelvin}\). Scientists use strong magnetic fields in devices called Tokamaks (magnetic confinement) or high-powered lasers (inertial confinement) to keep the plasma suspended away from reactor walls.
Fusion vs Fission: A Quick Comparison
• Fuel Abundance: Fusion uses isotopes of hydrogen (deuterium can be extracted easily from seawater, tritium can be bred from lithium), whereas fission relies on finite supplies of uranium ore.
• Waste Products: Fusion produces non-toxic, non-radioactive Helium-4 (though reactor walls become activated by neutrons). Fission produces highly radioactive fission fragments with half-lives of thousands of years.
• Safety: Fusion cannot run away into a meltdown; if containment fails, the plasma cools instantly and the reaction stops. Fission requires continuous cooling to prevent core damage from decay heat.
• Current Feasibility: Fission is mature commercial technology; fusion is still experimental because sustaining net energy gain is technically difficult.
Section 4 Key Takeaway: Fusion joins light nuclei, releasing more energy per unit mass than fission. It requires temperatures of millions of Kelvin to overcome electrostatic Coulomb repulsion between positive nuclei.
5. Step-by-Step Energy Calculations
Let's look at standard exam calculation styles you will encounter.
Worked Example 1: Binding Energy of a Helium-4 Nucleus
Question:
Calculate the binding energy per nucleon of a Helium-4 (\(^{4}_{2}\text{He}\)) nucleus in \(\text{MeV}\).
Data given:
• Mass of proton (\(m_p\)) = \(1.00728\text{ u}\)
• Mass of neutron (\(m_n\)) = \(1.00866\text{ u}\)
• Mass of \(^{4}_{2}\text{He}\) nucleus = \(4.00150\text{ u}\)
• \(1\text{ u} = 931.5\text{ MeV}\)
Step 1: Count the nucleons.
Helium-4 has \(Z = 2\) protons and \(A - Z = 4 - 2 = 2\) neutrons.
Step 2: Calculate the mass of separate nucleons.
\(\text{Mass of constituents} = 2(1.00728\text{ u}) + 2(1.00866\text{ u}) = 2.01456\text{ u} + 2.01732\text{ u} = 4.03188\text{ u}\)
Step 3: Find the mass defect (\(\Delta m\)).
\(\Delta m = 4.03188\text{ u} - 4.00150\text{ u} = 0.03038\text{ u}\)
Step 4: Convert mass defect to binding energy (\(E_B\)).
\(E_B = 0.03038\text{ u} \times 931.5\text{ MeV u}^{-1} = 28.30\text{ MeV}\)
Step 5: Find the binding energy per nucleon.
\(\text{Binding Energy per nucleon} = \frac{28.30\text{ MeV}}{4} = 7.08\text{ MeV per nucleon}\)
Worked Example 2: Energy Released in a Fission Reaction
Question:
Calculate the energy released (in Joules) by the following fission reaction:
\(^{1}_{0}\text{n} + ^{235}_{92}\text{U} \to ^{144}_{56}\text{Ba} + ^{89}_{36}\text{Kr} + 3\,^{1}_{0}\text{n}\)
Data:
• Mass of \(^{235}\text{U}\) = \(235.0439\text{ u}\)
• Mass of \(^{144}\text{Ba}\) = \(143.9229\text{ u}\)
• Mass of \(^{89}\text{Kr}\) = \(88.9176\text{ u}\)
• Mass of \(^{1}_{0}\text{n}\) = \(1.0087\text{ u}\)
• \(1\text{ u} = 1.661 \times 10^{-27}\text{ kg}\), \(c = 3.00 \times 10^8\text{ m s}^{-1}\)
Step 1: Total initial mass before fission:
\(m_{\text{initial}} = m(\text{U}) + m(\text{n}) = 235.0439\text{ u} + 1.0087\text{ u} = 236.0526\text{ u}\)
Step 2: Total final mass after fission:
\(m_{\text{final}} = m(\text{Ba}) + m(\text{Kr}) + 3 \times m(\text{n}) = 143.9229\text{ u} + 88.9176\text{ u} + 3(1.0087\text{ u}) = 235.8666\text{ u}\)
Step 3: Calculate the mass change (\(\Delta m\)):
\(\Delta m = m_{\text{initial}} - m_{\text{final}} = 236.0526\text{ u} - 235.8666\text{ u} = 0.1860\text{ u}\)
Step 4: Convert mass defect to kilograms and use \(\Delta E = \Delta m c^2\):
\(\Delta m = 0.1860 \times (1.661 \times 10^{-27}\text{ kg}) = 3.089 \times 10^{-28}\text{ kg}\)
\(\Delta E = (3.089 \times 10^{-28}\text{ kg}) \times (3.00 \times 10^8\text{ m s}^{-1})^2 = 2.78 \times 10^{-11}\text{ J}\)
Alternative Quick Method: \(\Delta E = 0.1860\text{ u} \times 931.5\text{ MeV} = 173.3\text{ MeV}\). Then convert to Joules: \(173.3 \times 10^6 \times 1.60 \times 10^{-19}\text{ J} = 2.77 \times 10^{-11}\text{ J}\).
6. Summary and Quick Revision Checklist
Before sitting your exam, make sure you can confidently check off each of these points:
• Einstein's Equation: State and apply \(\Delta E = \Delta m c^2\) and know that \(1\text{ u} \equiv 931.5\text{ MeV}\).
• Mass Defect & Binding Energy: Define both terms accurately and calculate them given atomic/nuclear masses.
• Binding Energy Curve: Sketch the curve against \(A\), locate the peak at Iron-56 (\(^{56}_{26}\text{Fe}\)), and explain why fission occurs for large \(A\) and fusion for small \(A\).
• Nuclear Fission: Write balanced fission equations and explain chain reactions.
• Reactor Core Parts: State the role and materials for fuel rods, moderator (slows neutrons down by elastic collision), control rods (absorbs neutrons), coolant, and shielding.
• Nuclear Fusion: Explain why extreme temperature and density are needed to overcome the electrostatic Coulomb repulsion barrier.