Nuclear Energy

Nuclear Energy

Info: Board Coverage AQA Paper 2 | Edexcel CP3 | OCR (A) Paper 2 | CIE P4

Explore the simulation above to develop intuition for this topic.

Intuition

A tiny bit of mass contains enormous energy: Einstein’s $E = mc^2$ means mass and energy are equivalent. When a uranium nucleus splits, the fragments weigh slightly less than the original nucleus — that “missing” mass has been converted to energy. Because $c^2$ is enormous ($9 \times 10^{16}$ m²/s²), even a tiny mass defect releases huge energy.

Why it matters: Nuclear energy provides about 10% of the world’s electricity. Understanding binding energy explains why some nuclei are stable and others decay, why nuclear power plants work, and why nuclear weapons are so destructive.

The key insight: Iron-56 has the highest binding energy per nucleon — it’s the most stable nucleus. Elements lighter than iron can release energy through fusion (combining nuclei), while elements heavier than iron can release energy through fission (splitting nuclei). This is why the Sun fuses hydrogen into helium, not the other way around.

1. Mass Defect and Binding Energy

Mass Defect

Definition. The mass defect $\Delta m$ is the difference between the mass of a nucleus and the Total mass of its separated constituent nucleons. It represents the mass equivalent of the binding Energy that holds the nucleus together.

The mass defect $\Delta m$ is the difference between the mass of a nucleus and the sum of the Masses of its constituent nucleons:

$\Delta m = Zm_p + Nm_n - m_{\mathrm{nucleus}}$

Where $Z$ is the number of protons, $N$ is the number of neutrons, $m_p$ is the proton mass, $m_n$ Is the neutron mass, and $m_{\mathrm{nucleus}}$ is the actual nuclear mass.

The mass defect is always positive for stable nuclei — the nucleus is lighter than its Constituent parts.

Einstein’s Mass-Energy Equation

Definition. Binding energy is the minimum energy required to completely separate a nucleus into Its individual protons and neutrons, or equivalently, the energy released when a nucleus is formed From its constituents.

Derivation of Mass-Energy Equivalence

1. From Einstein’s special relativity, the total energy of a body at rest is $E = mc^2$.
2. A nucleus of mass $m_{\mathrm{nucleus}}$ is lighter than its constituent nucleons by the mass defect $\Delta m$.
3. The “missing mass” has been converted to energy that holds the nucleus together.
4. The energy equivalent of the mass defect is the binding energy:

$\boxed{E = \Delta m \cdot c^2}$

$\square$

$\boxed{E = mc^2}$

The energy equivalent of the mass defect is the binding energy:

$\boxed{E_b = \Delta m \cdot c^2}$

This is the energy that would be required to completely separate the nucleus into its individual Protons and neutrons. Equivalently, it is the energy released when the nucleus is formed from its Constituents.

Calculating mass defect. Use atomic mass units (u), where $1\,\mathrm u = 1.661 \times 10^{-27}$ kg, and $1\,\mathrm u \cdot c^2 = 931.5$ MeV.

2. Binding Energy per Nucleon

Definition. The binding energy per nucleon is the binding energy of a nucleus divided by its Mass number $A$: $E_b/A$. It is a measure of nuclear stability — the higher the value, the more Tightly bound the nucleus.

The binding energy per nucleon is a measure of nuclear stability:

$\frac{E_b}{A} = \frac{\Delta m \cdot c^2}{A}$

The Binding Energy Curve

The plot of $E_b/A$ versus mass number $A$ has the following key features:

• Light nuclei ($A < 20$): binding energy per nucleon rises rapidly with $A$. Nuclei become more stable by fusion (combining light nuclei to reach higher $E_b/A$).
• Iron-56 ($\prescript{56}{26}\mathrm{Fe}$): the peak of the curve at $\sim 8.8$ MeV/nucleon. Iron-56 is the most stable nucleus.
• Heavy nuclei ($A > 60$): binding energy per nucleon gradually decreases. Nuclei become more stable by fission (splitting heavy nuclei to reach higher $E_b/A$).

Intuition. Both fusion and fission release energy because they move nuclei towards the peak of The binding energy curve, where $E_b/A$ is maximum. The released energy equals the increase in total Binding energy.

Derivation of Energy Released from the Binding Energy Curve

1. For any nuclear process, the total number of nucleons is conserved: $A_{\mathrm{products}} = A_{\mathrm{reactants}}$.
2. The binding energy per nucleon changes from $(E_b/A)_{\mathrm{initial}}$ to $(E_b/A)_{\mathrm{final}}$.
3. Total binding energy before: $E_{b,\mathrm{initial}} = (E_b/A)_{\mathrm{initial}} \times A$.
4. Total binding energy after: $E_{b,\mathrm{final}} = (E_b/A)_{\mathrm{final}} \times A$.
5. Energy released equals the increase in total binding energy:

$\Delta E = \left[(E_b/A)_{\mathrm{final}} - (E_b/A)_{\mathrm{initial}}\right] \times A$

$\boxed{\Delta E = \Delta(E_b/A) \times A}$

For fission of heavy nuclei: $(E_b/A)_{\mathrm{final}} > (E_b/A)_{\mathrm{initial}}$So $\Delta E > 0$.

For fusion of light nuclei: $(E_b/A)_{\mathrm{final}} > (E_b/A)_{\mathrm{initial}}$So $\Delta E > 0$.

$\square$

3. Nuclear Fission

Definition. Nuclear fission is the process in which a heavy nucleus splits into two (or more) Lighter nuclei, releasing energy and one or more neutrons.

Process

A heavy nucleus (e.g., uranium-235) absorbs a neutron and splits into two lighter nuclei (fission Fragments), releasing energy and more neutrons:

$\prescript{1}{0}\mathrm n + \prescript{235}{92}\mathrm U \to \prescript{236}{92}\mathrm U^* \to \prescript{141}{56}\mathrm{Ba} + \prescript{92}{36}\mathrm{Kr} + 3\prescript{1}{0}\mathrm n + \mathrm{energy}$

Energy Released

The fission fragments have a higher binding energy per nucleon than the parent nucleus. The energy Released is:

$\Delta E = (E_b/A)_{\mathrm{products}} \times A_{\mathrm{products}} - (E_b/A)_{\mathrm{parent}} \times A_{\mathrm{parent}}$

For U-235 fission: $\Delta E \approx 200$ MeV per fission event.

Chain Reaction

Definition. A chain reaction is a self-sustaining series of nuclear fission events in which Neutrons released from each fission induce further fission events.

Definition. The critical mass is the minimum mass of fissile material required to sustain a Nuclear chain reaction under given conditions.

Each fission event releases 2-3 neutrons, which can induce further fission events. If each fission Causes exactly one further fission, the reaction is critical (steady). If more than one, it is supercritical (increasing). If fewer, it is subcritical (dying out).

Controlled chain reaction: nuclear power stations use control rods (e.g., boron or cadmium, Which absorb neutrons) to maintain a critical state.

Uncontrolled chain reaction: nuclear weapons.

4. Induced Fission Cross-Sections

The Concept of Nuclear Cross-Section

Definition. The nuclear cross-section $\sigma$ quantifies the probability that a specific Nuclear reaction occurs when a projectile strikes a target nucleus. It has dimensions of area.

The cross-section is defined operationally as:

$\sigma = \frac{\mathrm{number of reactions per unit time}}{\mathrm{incident flux} \times \mathrm{number of target nuclei}}$

The SI unit would be m$^2$But nuclear cross-sections are so small that the standard unit is the barn:

$1\,\mathrm{barn} = 10^{-28}\,\mathrm m^2$

The name is deliberate: a typical nuclear cross-section is “as big as a barn” compared to the Geometric cross-section of a nucleus ($\sim \pi R^2 \sim 10^{-30}$ m$^2$). Quantum mechanical Effects --- resonances, tunnelling, and the wave nature of particles --- make the effective Interaction area much larger than the physical size.

Fission Cross-Section versus Neutron Energy

The fission cross-section depends critically on neutron energy. This is the fundamental reason why Some isotopes are “fissile” and others are merely “fertile.”

For $\prescript{235}{92}\mathrm U$:

• Thermal neutron cross-section ($E \approx 0.025$ eV): $\sigma_f \approx 585$ barns
• Fast neutron cross-section ($E \approx 1$ MeV): $\sigma_f \approx 1$ barn

This roughly 500-fold decrease from thermal to fast energies is why a moderator is essential in Thermal reactors.

Why U-235 is Fissile but U-238 is Not

Definition. A fissile nucleus can undergo fission after absorbing a neutron of any energy, Including thermal. A fertile nucleus cannot fission with thermal neutrons but can be converted Into a fissile isotope.

The distinction follows from the odd-even binding energy effect. When a nucleus absorbs a neutron, The compound nucleus has excitation energy equal to the binding energy of the added neutron $E_b^{\mathrm{added}}$. Fission occurs only if this exceeds the fission barrier $E_f$.

For $\prescript{235}{92}\mathrm U + \mathrm n \to \prescript{236}{92}\mathrm U^*$:

• $\prescript{235}{92}\mathrm U$ has 143 neutrons (odd). Adding a neutron pairs the last neutron, gaining extra pairing energy.
• $E_b^{\mathrm{added}} \approx 6.5$ MeV while $E_f \approx 5.3$ MeV.
• Since $E_b^{\mathrm{added}} \gt E_f$Fission proceeds even with thermal neutrons.

For $\prescript{238}{92}\mathrm U + \mathrm n \to \prescript{239}{92}\mathrm U^*$:

• $\prescript{238}{92}\mathrm U$ has 146 neutrons (even). Adding a neutron creates an unpaired neutron with less pairing energy gain.
• $E_b^{\mathrm{added}} \approx 4.9$ MeV while $E_f \approx 5.5$ MeV.
• Since $E_b^{\mathrm{added}} \lt E_f$Thermal neutrons cannot induce fission.

Fast Fission of U-238

U-238 can fission, but only with neutrons above approximately 1 MeV. The condition is:

$E_b^{\mathrm{added}} + E_n^{\mathrm{kinetic}} \gt E_f \implies 4.9 + E_n \gt 5.5 \implies E_n \gt 0.6\,\mathrm{MeV}$

The practical threshold is quoted as $\sim 1$ MeV to account for the distribution of fission barrier Heights and the very small cross-section just above threshold.

The Conversion Chain: U-238 to Pu-239

Although U-238 is not fissile, it is fertile. It captures a neutron and is transmuted into Pu-239 through beta decays:

$\prescript{238}{92}\mathrm U + \mathrm n \to \prescript{239}{92}\mathrm U \xrightarrow{\beta^-,\; 23.5\,\mathrm{min}} \prescript{239}{93}\mathrm{Np} \xrightarrow{\beta^-,\; 2.36\,\mathrm{days}} \prescript{239}{94}\mathrm{Pu}$

This conversion chain is the basis of breeder reactors, discussed in Section 5.

5. Nuclear Fission Mechanics

Fission Fragment Mass Distribution

When a heavy nucleus fissions, the two fragments are not of equal mass. The distribution is Asymmetric, producing one “heavy” fragment ($A \approx 135$—$145$) and one “light” fragment ($A \approx 90$—$100$).

The nuclear shell model explains this qualitatively. Magic numbers (2, 8, 20, 28, 50, 82, 126) Correspond to filled nuclear shells, and nuclei near magic numbers are significantly more stable. Asymmetric splitting allows one fragment to approach a magic neutron number ($N = 82$), which is Energetically favourable. Symmetric splitting ($A \approx 118$ each) is roughly 100 times less Probable.

Prompt Neutrons versus Delayed Neutrons

Prompt neutrons are emitted within $\sim 10^{-14}$ s of fission as the highly deformed fragments De-excite. 2—3 are emitted per fission.

Delayed neutrons are emitted on timescales of seconds to minutes by certain fission products That undergo beta decay to excited states above the neutron separation energy. Key precursors for U-235 thermal fission:

The delayed neutron fraction $\beta$ is the fraction of all fission neutrons that are delayed. For U-235, $\beta \approx 0.0065$ (0.65%). Although tiny, these neutrons are essential for reactor Control (see Section 6).

Fission Product Poisoning

Some fission products have enormous neutron absorption cross-sections and can shut down the chain Reaction.

Xenon-135 is the worst offender:

• Thermal absorption cross-section: $\sigma_a \approx 2.65 \times 10^6$ barns (the largest known for any stable nuclide)
• Produced mainly by decay: $\prescript\{135\}\{52\}\mathrm\{Te\} \to \prescript\{135\}\{53\}\mathrm I \xrightarrow\{\beta^-\} \prescript\{135\}\{54\}\mathrm\{Xe\} \xrightarrow\{\beta^-\} \prescript\{135\}\{55\}\mathrm\{Cs\}$
• $\prescript{135}{54}\mathrm{Xe}$ has $t_{1/2} = 9.2$ hours

After a reactor shutdown, xenon-135 builds up from iodine-135 decay faster than it decays away. This iodine pit can prevent restart for 24—48 hours.

Samarium-149: $\sigma_a \approx 4.1 \times 10^4$ barns, stable, and accumulates permanently.

Energy Distribution in Fission

The $\sim 200$ MeV released per U-235 fission is distributed as follows:

Breeder Reactors

A breeder reactor produces more fissile material than it consumes by placing fertile material Around the core where excess neutrons convert it to fissile isotopes.

Uranium-plutonium cycle:

$\prescript{238}{92}\mathrm U + \mathrm n \to \prescript{239}{92}\mathrm U \to \prescript{239}{93}\mathrm{Np} \to \prescript{239}{94}\mathrm{Pu}$

Thorium-uranium cycle:

$\prescript{232}{90}\mathrm{Th} + \mathrm n \to \prescript{233}{90}\mathrm{Th} \to \prescript{233}{91}\mathrm{Pa} \to \prescript{233}{92}\mathrm U$

The breeding ratio $\mathrm{BR} = \mathrm{fissile atoms produced} / \mathrm{fissile atoms consumed}$. For BR $\gt 1$ The reactor is a net producer. Fast breeder reactors using liquid sodium achieve BR $\approx 1.2$—$1.3$. The thorium cycle is attractive because thorium is roughly $3\times$ more Abundant than uranium and produces less long-lived transuranic waste.

Nuclear Waste Classification

6. Nuclear Reactor Design

Why Neutrons Must Be Moderated

Fission neutrons are born with energies peaking around 1—2 MeV. As established in Section 4, the U-235 fission cross-section at thermal energies ($\sigma_f \approx 585$ barns) is roughly 500 times Larger than at fast energies ($\sigma_f \approx 1$ barn). Without moderation, the probability of Inducing further fission is too low to sustain a chain reaction with natural or low-enriched Uranium.

Moderation by Elastic Collisions

Neutrons are slowed by elastic collisions with light moderator nuclei. Consider a neutron of Mass $m_n$ and kinetic energy $E$ colliding elastically with a stationary nucleus of mass $M$.

Theorem (maximum energy loss). The maximum fractional energy loss per head-on collision is:

$\frac{\Delta E_{\max}}{E} = \frac{4\,m_n\,M}{(m_n + M)^2}$

Proof. In the lab frame, the neutron has initial velocity $v$ and the target is at rest. After a Head-on elastic collision, conservation of momentum and kinetic energy give the neutron’s final Velocity:

$v' = \frac{m_n - M}{m_n + M}\,v$

The neutron’s final kinetic energy is $E' = \frac{1}{2}m_n(v')^2 = \left(\frac{m_n - M}{m_n + M}\right)^2 E$.

Therefore:

$\frac{\Delta E}{E} = \frac{E - E'}{E} = 1 - \left(\frac{m_n - M}{m_n + M}\right)^2 = \frac{4m_n M}{(m_n + M)^2}$

This is maximised when $m_n = M$ (equal masses), giving $\Delta E/E = 1$: the neutron transfers all Its energy in a single collision. $\square$

Consequences for moderator choice:

The average logarithmic energy decrement per collision is $\xi$And the number of collisions to Thermalise from $E_0 = 2$ MeV to $E_{\mathrm{th}} = 0.025$ eV is $n = \ln(E_0/E_{\mathrm{th}})/\xi$. Hydrogen is the best moderator by energy loss per collision, but it has a non-negligible absorption Cross-section. Deuterium (in heavy water) is the best practical compromise: high energy loss with Negligible absorption.

Control Rods and the Multiplication Factor

Control rods are made of materials with very high neutron absorption cross-sections: boron-10 ($\sigma_a \approx 3840$ barns), cadmium-113 ($\sigma_a \approx 20\,800$ barns), or hafnium.

The multiplication factor is:

$k = \frac{\mathrm{neutrons in generation } n + 1}{\mathrm{neutrons in generation } n}$

The effective multiplication factor $k_{\mathrm{eff}}$ accounts for neutron leakage and non-fuel Absorption: $k_{\mathrm{eff}} = k_{\infty} \cdot P_{\mathrm{non-leak}}$Where $k_{\infty}$ is the Infinite-medium factor and $P_{\mathrm{non-leak}}$ is the non-leakage probability.

Delayed Neutrons and Reactor Control

This is one of the most important engineering facts about nuclear reactors. Without delayed Neutrons, controlling a reactor would be essentially impossible on human timescales.

Theorem. The reactor response time is governed by delayed neutrons, not the prompt neutron Lifetime, provided $k_{\mathrm{eff}} \lt 1 + \beta$.

Proof. The prompt neutron lifetime is $\ell \approx 10^{-4}$ s. If $k_{\mathrm{eff}} = 1.001$ (0.1% supercritical) with only prompt neutrons, the power grows as:

$P(t) = P_0 \, e^{(k-1)t/\ell} = P_0 \, e^{0.001 \times t / 10^{-4}} = P_0 \, e^{10\,t}$

Power doubles every $\ln 2 / 10 \approx 0.069$ s. No mechanical system can respond this fast.

With delayed neutrons (fraction $\beta \approx 0.0065$ for U-235), the reactor is “prompt Subcritical” when $1 \lt k_{\mathrm{eff}} \lt 1 + \beta$. In this regime, the neutron population Grows on the timescale of the longest-lived delayed precursor ($\sim 55$ s for Br-87), not the Prompt lifetime. The effective time constant becomes:

$\tau_{\mathrm{eff}} \approx \frac{\beta / \bar{\lambda}}{k_{\mathrm{eff}} - 1}$

Where $\bar{\lambda} \approx 0.08\,\mathrm s^{-1}$. For $k_{\mathrm{eff}} = 1.001$: $\tau_{\mathrm{eff}} \approx 0.0065 / (0.08 \times 0.001) \approx 81$ s --- manageable by Mechanical control systems. $\square$

Coolant

The coolant transfers heat from the fuel to the steam generators or turbine. Requirements: high Thermal conductivity and specific heat capacity, low neutron absorption, chemical stability under Radiation, and high boiling point.

Fuel and Enrichment

Natural uranium is $\sim 0.72\%$ U-235 and $\sim 99.28\%$ U-238. Thermal reactors require enriched Uranium: light water reactors use 3—5% U-235, research reactors up to 20%. Fuel is fabricated as Ceramic UO$_2$ pellets sealed in zirconium alloy (zircaloy) cladding tubes, assembled into fuel rods And bundles.

Shielding

Radiation shielding protects personnel and the environment. Concrete (cheap, dense, contains Hydrogen for neutron moderation) provides bulk biological shielding. Lead (very dense) is used For compact gamma shielding. Water provides both shielding and moderation in spent fuel pools.

PWR versus AGR Comparison

7. Nuclear Fusion

Definition. Nuclear fusion is the process in which two light nuclei combine to form a heavier Nucleus, releasing energy due to the increase in binding energy per nucleon.

Process

Two light nuclei combine to form a heavier nucleus, releasing energy:

$\prescript{2}{1}\mathrm H + \prescript{3}{1}\mathrm H \to \prescript{4}{2}\mathrm{He} + \prescript{1}{0}\mathrm n + 17.6\,\mathrm{MeV}$

Conditions for Fusion

For fusion to occur, the nuclei must overcome the Coulomb barrier — the electrostatic repulsion Between positively charged nuclei. This requires:

1. Extremely high temperatures ($\sim 10^8$ K) to give nuclei sufficient kinetic energy
2. High particle density to ensure frequent collisions
3. Confinement for long enough for fusion to occur

Magnetic confinement (tokamak) and inertial confinement (laser fusion) are two approaches.

Why Fusion is Harder than Fission

Fission is initiated by a neutral particle (neutron), so there is no Coulomb barrier to overcome. Fusion requires positively charged nuclei to approach within $\sim 10^{-15}$ m, requiring enormous Kinetic energy to overcome the Coulomb repulsion.

8. Nuclear Fusion in Detail

The Proton-Proton Chain

The pp chain is the dominant fusion process in main-sequence stars with mass $\lt 1.5\,M_\odot$ (solar masses). It proceeds in three steps:

Step 1 (rate-limiting, mediated by the weak interaction):

$\prescript{1}{1}\mathrm H + \prescript{1}{1}\mathrm H \to \prescript{2}{1}\mathrm H + \mathrm e^+ + \nu_e + 0.42\,\mathrm{MeV}$

One proton must undergo inverse beta decay ($p \to n + e^+ + \nu_e$), which requires the weak force And is extraordinarily slow --- mean time $\sim 10^{10}$ years in the solar core. This slowness is Why the Sun has a long lifetime.

Step 2:

$\prescript{2}{1}\mathrm H + \prescript{1}{1}\mathrm H \to \prescript{3}{2}\mathrm{He} + \gamma + 5.49\,\mathrm{MeV}$

Step 3 (dominant branch, $\sim 85\%$ probability):

$\prescript{3}{2}\mathrm{He} + \prescript{3}{2}\mathrm{He} \to \prescript{4}{2}\mathrm{He} + 2\prescript{1}{1}\mathrm H + 12.86\,\mathrm{MeV}$

Net reaction: $4\prescript{1}{1}\mathrm H \to \prescript{4}{2}\mathrm{He} + 2\mathrm e^+ + 2\nu_e + 2\gamma + 26.7\,\mathrm{MeV}$

The CNO Cycle

In stars more massive than $\sim 1.5\,M_\odot$The CNO cycle dominates. Carbon, nitrogen, and Oxygen act as catalysts:

$\prescript{12}{6}\mathrm C \xrightarrow{+\mathrm p} \prescript{13}{7}\mathrm N \xrightarrow{\beta^+} \prescript{13}{6}\mathrm C \xrightarrow{+\mathrm p} \prescript{14}{7}\mathrm N \xrightarrow{+\mathrm p} \prescript{15}{8}\mathrm O \xrightarrow{\beta^+} \prescript{15}{7}\mathrm N \xrightarrow{+\mathrm p} \prescript{12}{6}\mathrm C + \prescript{4}{2}\mathrm{He}$

Net reaction: $4\mathrm p \to \prescript{4}{2}\mathrm{He} + 26.7$ MeV (identical to the pp Chain).

The CNO cycle rate scales as $\propto T^{16\mathrm{--}20}$ (for the slowest step) versus $\propto T^4$ for the pp chain. At $T \gt 1.5 \times 10^7$ K the CNO cycle dominates. It also Produces a steeper temperature gradient in the stellar core, driving convection in massive stars.

The Lawson Criterion

For net energy output, the fusion power must exceed power losses. John Lawson (1957) derived the Minimum condition for D-T fusion:

$n\tau \gt 10^{20}\,\mathrm s\,\mathrm m^{-3}$

Where $n$ is the ion density (m$^{-3}$) and $\tau$ is the energy confinement time (s).

Derivation sketch. Fusion power density: $P_\{\mathrm\{fus\}\} = \frac\{1\}\{4\}n^2\langle\sigma v\rangle E_\{\mathrm\{fus\}\}$, where $\langle\sigma v\rangle$ is the reactivity (Maxwell-Boltzmann averaged) and the factor of $1/4$ Accounts for equal D and T densities. Power lost by thermal conduction is $P_{\mathrm{loss}} = 3nkT/\tau$. Setting $P_{\mathrm{fus}} \ge P_{\mathrm{loss}}$ and substituting The temperature-dependent $\langle\sigma v\rangle$ yields the Lawson criterion. The exact numerical Value depends on fuel choice and loss model.

The triple product $nT\tau \gt 3 \times 10^{21}\,\mathrm{keV}\,\mathrm s\,\mathrm m^{-3}$ is Sometimes quoted as an equivalent form.

Tokamak Design

A tokamak confines hot plasma in a toroidal (doughnut-shaped) chamber using magnetic fields:

• Toroidal field $B_t$: produced by coils wound around the torus, prevents outward radial drift.
• Poloidal field $B_p$: produced by the plasma current itself (induced by a central solenoid), prevents vertical drift.
• The combined helical field lines confine particles as they spiral along them.

The safety factor $q = rB_t / RB_p$ (minor radius $r$Major radius $R$) must satisfy $q \gt 1$ Everywhere for stability.

Inertial Confinement Fusion

In inertial confinement fusion (ICF), a small D-T pellet is compressed by laser or particle beams. The outer layer ablates outward, driving an inward implosion that compresses the fuel to $\sim 10^3\times$ solid density. The central “hot spot” reaches $\sim 10$ keV ($\sim 10^8$ K), Initiating fusion before the pellet disassembles. Confinement relies on the inertia of the fuel Itself --- no magnetic fields are needed.

Why D-T is the Easiest Fusion Reaction

$\prescript{2}{1}\mathrm H + \prescript{3}{1}\mathrm H \to \prescript{4}{2}\mathrm{He} + \mathrm n + 17.6\,\mathrm{MeV}$

D-T has the lowest Coulomb barrier of any practical fuel combination: tritium has the largest Nuclear radius relative to its charge, the $Q$-value of 17.6 MeV is the highest per reaction of any D-based fuel, and the reactivity $\langle\sigma v\rangle$ peaks at the lowest temperature ($\sim 70$ KeV versus $\sim 500$ keV for D-D). However, tritium ($t_{1/2} = 12.3$ years) must be bred from Lithium:

$\prescript{6}{3}\mathrm{Li} + \mathrm n \to \prescript{4}{2}\mathrm{He} + \prescript{3}{1}\mathrm H + 4.8\,\mathrm{MeV}$

The ITER Project

ITER (International Thermonuclear Experimental Reactor), under construction in Cadarache, France, Aims to demonstrate $Q \ge 10$ (fusion gain: 10 times more power out than heating power in). Key Parameters: major radius 6.2 m, magnetic field 5.3 T, plasma current 15 MA, fusion power 500 MW from 50 MW input heating, pulse duration $\sim 400$ s. ITER is a proof-of-concept; electricity generation Is the goal of the subsequent DEMO reactor.

Stellar Nucleosynthesis: The Iron Peak

The binding energy per nucleon curve peaks near iron-56. This has a profound consequence:

• Fusion of nuclei lighter than iron: exothermic (releases energy)
• Fusion of nuclei heavier than iron: endothermic (absorbs energy)
• Fission of nuclei heavier than iron: exothermic

In massive stars, successive fusion stages build heavier elements: H $\to$ He ($T \sim 10^7$ K), He $\to$ C/O ($T \sim 10^8$ K), C $\to$ Ne/Mg, Ne $\to$ O/Mg, O $\to$ Si/S, Si $\to$ Fe ($T \sim 3 \times 10^9$ K). The process stops at iron because further fusion is endothermic. When The iron core exceeds the Chandrasekhar limit ($\sim 1.4\,M_\odot$), it collapses and triggers a Supernova, whose energy drives the creation of elements heavier than iron via the s-process and R-process.

Problem Set

Common Pitfalls

1. Confusing atomic number (protons) with mass number (protons + neutrons).

2. Forgetting that radioactive decay is random and spontaneous. It cannot be predicted for individual nuclei.

3. Misunderstanding that half-life is constant regardless of the initial amount of substance.

4. Using the wrong equation from the data sheet. Take time to read the full equation, including conditions and variable definitions.

5. Forgetting to include units in final answers, especially when working with derived units like $\text{N}\,\text{kg}^{-1}\,\text{m}^2$.

6. Incorrectly applying $\vec{F} = m\vec{a}$when forces are not collinear. Resolve into components first.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Mistakes

Confusing fission and fusion: Fission is splitting heavy nuclei into lighter ones (uranium-235). Fusion is combining light nuclei into heavier ones (hydrogen into helium). Fission is used in nuclear power plants; fusion powers the Sun. Students often mix up which is which.

Forgetting that mass is converted to energy: In nuclear reactions, the mass of products is less than the mass of reactants. This “missing” mass has been converted to energy via $E = mc^2$. Students often forget to calculate the mass defect.

Misusing $E = mc^2$: This equation relates energy and mass. The mass must be in kilograms and the result is in joules. Students often forget to convert atomic mass units to kilograms or use the wrong units.

Confusing binding energy with binding energy per nucleon: Total binding energy is the energy needed to separate all nucleons. Binding energy per nucleon is total binding energy divided by mass number. The latter determines nuclear stability — higher means more stable. Iron-56 has the highest binding energy per nucleon.

Forgetting that fusion requires extreme temperatures: Fusion requires temperatures of millions of degrees to overcome electrostatic repulsion between positively charged nuclei. This is why fusion reactors are so difficult to build. Students often underestimate the energy requirements.

Cross-References

• Radioactivity: Provides the foundation of nuclear decay processes that underpin both fission and fusion.
• Quantum Physics: Explains the quantum mechanical principles that govern nuclear stability and decay rates.
• Binding Energy: Explores the mass-energy relationship in more detail, including binding energy per nucleon curves.
• Nuclear Reactors: Examines the practical engineering of fission reactors, including moderators and control rods.