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Chapter 1 • Theory & Derivations

Fundamentals of Nuclear Energy, Atom Densities & Fission Fuel Breeding

Foundational principles of nuclear reactor physics: comparison of fission and fusion energetics, microscopic nuclear force characteristics, classification of nuclear materials into fissile (U-235, Pu-239, U-233), fissionable (U-238, Th-232), and fertile isotopes; breeding reaction chains and breeding ratio/doubling time kinetics in fast breeder reactors; relativistic mass-energy equivalence, mass defect, binding energy systematics, and nucleon separation energy criteria; Maxwell-Boltzmann thermal neutron gas velocity distributions; and rigorous mathematical formulation of elemental, isotopic, and molecular atom densities for nuclear fuel and moderator compounds.

§1.1Nuclear Reactor Physics Overview, Fission vs Fusion & Nuclear Force

1. Scope of Nuclear Reactor Physics

Nuclear reactor physics is the branch of applied nuclear physics and engineering that governs the distribution, transport, slowing down, and multiplication of neutrons inside a nuclear reactor core. The central objective is to sustain and control a steady-state or time-dependent chain reaction, converting microscopic nuclear binding energy into macroscopic thermal power: $$\text{Nuclear Energy} \longrightarrow \text{Thermal Energy} \longrightarrow \text{Mechanical Turbine Work} \longrightarrow \text{Electrical Power}$$

At the subatomic level, atomic nuclei are held together by the residual strong nuclear force, a phenomenological interaction mediated primarily by virtual meson exchange between nucleons (protons and neutrons). Key physical characteristics include:

  • Colossal Magnitude: Approximately $100$ to $1000$ times stronger than the electromagnetic Coulomb repulsion at distances $r \approx 1\text{ fm}$.
  • Extremely Short Range: Effective only over nuclear dimensions ($r \sim 1\text{ to }2\text{ fm}$), decaying exponentially as $\sim \frac{e^{-\mu r}}{r}$ for $r > 2\text{ fm}$.
  • Saturation Property: A nucleon interacts only with its immediate nearest neighbors, meaning the binding energy per nucleon ($B/A \approx 8\text{ MeV}$) is approximately constant across intermediate and heavy nuclei.
  • Hard Repulsive Core: Becomes fiercely repulsive at $r < 0.4\text{ fm}$, preventing the collapse of atomic nuclei into point singularities.

2. Energetics of Fission vs Fusion

The curve of binding energy per nucleon ($B/A$ versus mass number $A$) exhibits a prominent maximum near iron-56 (${}^{56}\text{Fe}$, where $B/A \approx 8.79\text{ MeV/nucleon}$). Because nature favors higher binding energy per nucleon (deeper potential wells), energy can be liberated exothermically via two distinct nuclear pathways:

  1. Nuclear Fusion (Light Nuclei, $A < 56$): Fusing two light nuclei (such as deuterium and tritium, ${}^2\text{H} + {}^3\text{H} \to {}^4\text{He} + n + 17.6\text{ MeV}$) moves the products up the steep initial incline toward ${}^4\text{He}$ ($B/A \approx 7.07\text{ MeV}$), yielding $\sim 3.5\text{ MeV/nucleon}$.
  2. Nuclear Fission (Heavy Nuclei, $A > 230$): Splitting a massive actinide nucleus (${}^{235}\text{U}$) into two intermediate fragments ($A_1 \sim 95, A_2 \sim 140$) shifts the system from $B/A \approx 7.56\text{ MeV}$ up to $B/A \approx 8.5\text{ MeV}$: $$\Delta(B/A) \approx 8.5\text{ MeV} - 7.56\text{ MeV} \approx 0.94\text{ MeV/nucleon}$$ Multiplying by $A = 236$ nucleons: $$Q_{\text{fission}} \approx 236 \times 0.94\text{ MeV} \approx \mathbf{200\text{ MeV per fission event}}$$

This energy density is astronomical: the complete fission of $1\text{ kg}$ of ${}^{235}\text{U}$ releases $8.2 \times 10^{13}\text{ Joules}$ of energy—equivalent to burning approximately $2,500\text{ metric tons}$ of high-grade coal or $14{,}000\text{ barrels}$ of crude oil!

§1.2Classification of Nuclear Materials: Fissile, Fissionable & Fertile

1. Fissile Isotopes

Fissile materials are nuclides capable of undergoing nuclear fission upon capturing a neutron of any kinetic energy, including thermal neutrons with zero or near-zero kinetic energy ($E \approx 0.0253\text{ eV}$): $$n_{\text{th}} + {}^{A}_Z\text{X} \longrightarrow [{}^{A+1}_Z\text{X}]^* \longrightarrow \text{Fission Fragments} + \nu \, n + Q$$ The three practical fissile fuels of modern nuclear technology are:

  • Uranium-235 (${}^{235}_{92}\text{U}$): The only naturally occurring fissile isotope on Earth, comprising $0.7204\%$ of natural uranium.
  • Plutonium-239 (${}^{239}_{94}\text{Pu}$): Artificially produced via neutron capture in fertile ${}^{238}\text{U}$.
  • Uranium-233 (${}^{233}_{92}\text{U}$): Artificially produced via neutron capture in fertile ${}^{232}\text{Th}$.

2. Fissionable Isotopes

Fissionable materials are nuclides capable of undergoing fission when struck by neutrons, but whose compound nucleus requires a threshold kinetic energy (typically $E_n > 1.0\text{ to }1.5\text{ MeV}$) to overcome the fission activation barrier: $$n_{\text{fast}} (E_n > E_{\text{th}}) + {}^{238}_{92}\text{U} \longrightarrow [{}^{239}_{92}\text{U}]^* \longrightarrow \text{Fission}$$ Examples include even-$N$ actinides such as ${}^{238}_{92}\text{U}$, ${}^{232}_{90}\text{Th}$, and ${}^{240}_{94}\text{Pu}$. All fissile materials are fissionable, but not all fissionable materials are fissile.

3. Fertile Isotopes and Nuclear Transmutation

Fertile materials are nuclides that are not themselves fissile by thermal neutrons, but can be converted into fissile isotopes through neutron radiative capture $(n, \gamma)$ followed by subsequent beta-minus ($\beta^-$) decays:

  1. The Uranium-Plutonium Cycle: $${}^{238}_{92}\text{U} + n \longrightarrow {}^{239}_{92}\text{U} \xrightarrow[\beta^-, \, 23.5\text{ min}]{} {}^{239}_{93}\text{Np} \xrightarrow[\beta^-, \, 2.356\text{ days}]{} {}^{239}_{94}\text{Pu} \quad (\text{Fissile!})$$
  2. The Thorium-Uranium Cycle: $${}^{232}_{90}\text{Th} + n \longrightarrow {}^{233}_{90}\text{Th} \xrightarrow[\beta^-, \, 22.3\text{ min}]{} {}^{233}_{91}\text{Pa} \xrightarrow[\beta^-, \, 26.97\text{ days}]{} {}^{233}_{92}\text{U} \quad (\text{Fissile!})$$
Fertile isotopes constitute $>99.27\%$ of natural uranium (${}^{238}\text{U}$) and $100\%$ of natural thorium (${}^{232}\text{Th}$), representing over $99\%$ of the planet's mineable nuclear fuel reserves.

§1.3Breeding Physics: Conversion Ratio, Breeding Gain & Doubling Time

1. Conversion Ratio and Breeding Ratio

In any nuclear reactor containing fertile material (${}^{238}\text{U}$ or ${}^{232}\text{Th}$), fissile fuel is simultaneously consumed by fission and produced by transmutation. We quantify this balance through the Conversion Ratio ($CR$), defined as: $$CR \equiv \frac{\text{Average rate of production of new fissile nuclei}}{\text{Average rate of consumption (fission + capture) of fissile nuclei}} = \frac{\dot{N}_{\text{fissile, produced}}}{\dot{N}_{\text{fissile, consumed}}}$$ Classification based on $CR$:

  • Burner / Converter Reactor ($CR < 1$): Produces fewer fissile nuclei than it consumes. Commercial Light Water Reactors (LWRs) operate with $CR \approx 0.55\text{ to }0.65$.
  • Break-Even Reactor ($CR = 1$): Exactly replaces every fissile atom consumed.
  • Breeder Reactor ($CR > 1$): Produces more fissile fuel than it consumes. When $CR > 1$, it is designated the Breeding Ratio ($BR \equiv CR$).
The net excess fissile production rate is governed by the Breeding Gain ($G$): $$G \equiv BR - 1$$

2. The Doubling Time ($T_d$)

The doubling time $T_d$ is the operating time required for a breeder reactor to generate enough excess fissile material to completely fuel an identical second reactor (including out-of-pile reprocessing losses): $$T_d \approx \frac{M_{\text{core}}}{G \cdot \dot{M}_{\text{fissile, consumed}}}$$ In terms of reactor electric power $P_e$, thermal efficiency $\eta_{\text{th}}$, and capacity factor $CF$: $$T_d = \frac{M_{\text{fissile, inventory}}}{(BR - 1) \cdot (1 + \alpha) \cdot \dot{N}_{\text{fiss}} \cdot m_{\text{fiss}}}$$ Fast Breeder Reactors (FBRs) utilizing liquid metal cooling (Sodium or Lead) achieve breeding ratios of $BR \approx 1.20\text{ to }1.35$, yielding practical doubling times of $10\text{ to }20\text{ years}$.

§1.4Mass-Energy Equivalence, Nuclear Mass Defect & Binding Energy

1. Mass Defect $\Delta m$

According to Einstein's mass-energy equivalence relation $E = m c^2$, the ground-state mass of any bound nucleus $M(A, Z)$ is strictly less than the combined rest mass of its constituent free protons and neutrons: $$\Delta m \equiv \left[ Z m_p + (A - Z) m_n \right] - M(A, Z)$$ where $m_p = 1.007276466\text{ u}$ is the proton mass, $m_n = 1.008664916\text{ u}$ is the neutron mass, and $1\text{ u} = 931.494\text{ MeV}/c^2$. In terms of neutral atomic masses $M_{\text{atomic}}(A, Z)$ and hydrogen atom mass $m({}^1\text{H}) = 1.007825\text{ u}$: $$\Delta m = \left[ Z m({}^1\text{H}) + (A - Z) m_n \right] - M_{\text{atomic}}(A, Z)$$

2. Total Binding Energy and Binding Energy per Nucleon

The total nuclear binding energy $B(A, Z)$ is the energy required to disassemble the bound nucleus into isolated, non-interacting nucleons: $$B(A, Z) = \Delta m \cdot c^2 = \left( \left[ Z m({}^1\text{H}) + (A - Z) m_n \right] - M_{\text{atomic}}(A, Z) \right) \times 931.494\text{ MeV}$$ The binding energy per nucleon: $$\frac{B}{A} = \frac{B(A, Z)}{A}$$

Isotope Atomic Mass $M$ (u) Total BE $B$ (MeV) BE per Nucleon $B/A$ (MeV)
${}^2_1\text{H}$ (Deuteron) $2.014102$ $2.2245$ $1.112$
${}^4_2\text{He}$ (Alpha) $4.002603$ $28.296$ $7.074$
${}^{56}_{26}\text{Fe}$ (Peak Stability) $55.934937$ $492.26$ $\mathbf{8.790}$
${}^{235}_{92}\text{U}$ $235.043930$ $1783.87$ $7.590$
${}^{238}_{92}\text{U}$ $238.050788$ $1801.69$ $7.570$

§1.5Neutron & Proton Separation Energies and the Fission Barrier

1. Neutron Separation Energy ($S_n$)

The neutron separation energy $S_n$ is the minimum energy required to remove a single neutron from a nucleus $(A, Z)$, ejecting it to infinity at rest. It is the nuclear equivalent of the atomic ionization potential: $${}^{A}_{Z}\text{X} + S_n \longrightarrow {}^{A-1}_{Z}\text{X} + n$$ Using binding energies or atomic masses: $$S_n = B(A, Z) - B(A-1, Z) = \left[ M(A-1, Z) + m_n - M(A, Z) \right] c^2$$ Similarly, the proton separation energy is: $$S_p = B(A, Z) - B(A-1, Z-1) = \left[ M(A-1, Z-1) + m({}^1\text{H}) - M(A, Z) \right] c^2$$

2. The Fission Barrier and Why ${}^{235}\text{U}$ is Fissile but ${}^{238}\text{U}$ is Not

When an incident neutron with kinetic energy $E_n$ is absorbed by target nucleus ${}^{A}_Z\text{X}$, the resulting compound nucleus $[{}^{A+1}_Z\text{X}]^*$ is formed in an excited state with excitation energy $E^*$: $$E^* = S_n({}^{A+1}_Z\text{X}) + E_{\text{cm}} \approx S_n({}^{A+1}_Z\text{X}) + E_n \left( \frac{A}{A + 1} \right)$$ For thermal neutrons ($E_n \approx 0$), the entire excitation energy is simply the separation energy: $E^* \approx S_n$. To induce fission, $E^*$ must exceed the critical fission barrier energy $E_{\text{crit}}$ required to deform the nucleus past the saddle point of the liquid drop potential:

Target Nucleus Compound Nucleus Separation Energy $S_n$ Fission Barrier $E_{\text{crit}}$ $S_n - E_{\text{crit}}$ Fissile by Thermal Neutrons?
${}^{235}_{92}\text{U}$ (Odd $N=143$) ${}^{236}_{92}\text{U}$ (Even-Even) $6.55\text{ MeV}$ $5.70\text{ MeV}$ $+0.85\text{ MeV}$ YES (Fissile!)
${}^{238}_{92}\text{U}$ (Even $N=146$) ${}^{239}_{92}\text{U}$ (Even-Odd) $4.81\text{ MeV}$ $5.85\text{ MeV}$ $-1.04\text{ MeV}$ NO (Threshold $\sim 1.1\text{ MeV}$)
${}^{239}_{94}\text{Pu}$ (Odd $N=145$) ${}^{240}_{94}\text{Pu}$ (Even-Even) $6.53\text{ MeV}$ $5.50\text{ MeV}$ $+1.03\text{ MeV}$ YES (Fissile!)
${}^{232}_{90}\text{Th}$ (Even $N=142$) ${}^{233}_{90}\text{Th}$ (Even-Odd) $4.79\text{ MeV}$ $5.95\text{ MeV}$ $-1.16\text{ MeV}$ NO (Threshold $\sim 1.2\text{ MeV}$)

The profound physical origin is the pairing term in the semi-empirical mass formula: when a neutron is added to odd-$N$ $^{235}\text{U}$, it forms a neutron pair in even-even $^{236}\text{U}$, releasing an extra $\sim 1.2\text{ MeV}$ of pairing energy ($\delta_{\text{pair}} \approx +12 A^{-1/2}\text{ MeV}$). This extra pairing energy pushes $E^*$ well above the critical fission barrier, making $^{235}\text{U}$ readily fissile by zero-energy thermal neutrons!

§1.6Thermal Neutrons: Maxwell-Boltzmann Velocity Distribution & Energy Standards

1. Thermalization and Thermodynamic Equilibrium

In a thermal reactor, fast fission neutrons undergo repeated elastic collisions with light moderator nuclei until they reach approximate thermodynamic equilibrium with the thermal motion of the moderator atoms at core temperature $T$. The velocity distribution of thermal neutrons is given by the Maxwell-Boltzmann distribution: $$n(v) dv = \frac{4 n_0}{\sqrt{\pi}} \left( \frac{m}{2 k_B T} \right)^{3/2} v^2 \exp\left( -\frac{m v^2}{2 k_B T} \right) dv$$ where $n_0$ is the total thermal neutron number density ($\text{neutrons/cm}^3$) and $k_B = 8.61733 \times 10^{-5}\text{ eV/K} = 1.38065 \times 10^{-23}\text{ J/K}$.

2. Characteristic Thermal Parameters

From the Maxwellian distribution, three characteristic velocities and energies arise:

  1. Most Probable Speed ($v_0$): Located at the peak of $n(v)$ where $dn/dv = 0$: $$v_0 = \sqrt{\frac{2 k_B T}{m}}$$ The corresponding kinetic energy is: $$E_0 = \frac{1}{2} m v_0^2 = k_B T$$ At the international standard reference temperature $T_0 = 293.61\text{ K}$ ($20.46^{\circ}\text{C}$): $$E_0 = (8.61733 \times 10^{-5}\text{ eV/K})(293.61\text{ K}) = \mathbf{0.0253\text{ eV}}$$ $$v_0 = \sqrt{\frac{2(0.0253 \times 1.6022 \times 10^{-19}\text{ J})}{1.67493 \times 10^{-27}\text{ kg}}} = \mathbf{2200\text{ m/s}}$$ This defines the universally tabulated $2200\text{ m/s}$ standard cross-section reference state.
  2. Average Speed ($\bar{v}$): $$\bar{v} = \int_0^\infty v \frac{n(v)}{n_0} dv = \sqrt{\frac{8 k_B T}{\pi m}} = \frac{2}{\sqrt{\pi}} v_0 \approx 1.128 v_0$$
  3. Root-Mean-Square Speed ($v_{\text{rms}}$): $$v_{\text{rms}} = \sqrt{\langle v^2 \rangle} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3}{2}} v_0 \approx 1.225 v_0$$ The average kinetic energy of the thermal neutron distribution is: $$\bar{E} = \frac{1}{2} m \langle v^2 \rangle = \frac{3}{2} k_B T = 1.5 E_0$$

§1.7Atom Density Formalism: Elemental, Isotopic & Compound Formulations

1. Atom Density of a Pure Element

In reactor physics, reaction rates are evaluated using the macroscopic cross section $\Sigma = N \sigma$, where $N$ is the atom density (number of target nuclei per unit volume, typically expressed in $\text{atoms/cm}^3$ or $\text{atoms/b}\cdot\text{cm}$, where $1\text{ b}\cdot\text{cm} = 10^{-24}\text{ cm}^3$). For a pure element with mass density $\rho$ ($\text{g/cm}^3$) and atomic weight $M$ ($\text{g/mol}$): $$N = \frac{\rho \cdot N_A}{M}$$ where $N_A = 6.02214076 \times 10^{23}\text{ atoms/mol}$ is Avogadro's number.

2. Isotopic Atom Densities in an Enriched Mixture

For an element composed of multiple isotopes $i$, if $a_i$ denotes the atomic fraction ($\sum a_i = 1$), the atomic weight of the mixture is $M = \sum a_i M_i$, and the individual isotopic density is: $$N_i = a_i N = a_i \frac{\rho N_A}{M}$$ If the enrichment is given as a weight (mass) fraction $w_i \equiv m_i / m_{\text{total}}$ ($\sum w_i = 1$): $$N_i = \frac{w_i \cdot \rho \cdot N_A}{M_i}$$ The corresponding atomic fraction $a_i$ can be converted via: $$a_i = \frac{w_i / M_i}{\sum_j (w_j / M_j)}$$

3. Atom Densities in Chemical Compounds ($\text{A}_x\text{B}_y$)

For a homogeneous chemical compound with chemical formula $\text{A}_x\text{B}_y$, molecular weight $M_{\text{mol}} = x M_A + y M_B$, and mass density $\rho_{\text{comp}}$: The molecular density $N_{\text{mol}}$ is: $$N_{\text{mol}} = \frac{\rho_{\text{comp}} \cdot N_A}{M_{\text{mol}}}$$ The constituent elemental atom densities are: $$N_A = x \cdot N_{\text{mol}} = x \frac{\rho_{\text{comp}} N_A}{M_{\text{mol}}}, \qquad N_B = y \cdot N_{\text{mol}} = y \frac{\rho_{\text{comp}} N_A}{M_{\text{mol}}}$$ For example, in water ($\text{H}_2\text{O}$, $\rho = 1.0\text{ g/cm}^3$, $M_{\text{mol}} = 18.015\text{ g/mol}$): $$N_{\text{mol}} = \frac{1.0 \times 6.022 \times 10^{23}}{18.015} \approx 3.343 \times 10^{22}\text{ molecules/cm}^3$$ $$N_H = 2 N_{\text{mol}} \approx 6.686 \times 10^{22}\text{ atoms/cm}^3, \qquad N_O = N_{\text{mol}} \approx 3.343 \times 10^{22}\text{ atoms/cm}^3$$

ADVANCED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, quantitative calculations, and step-by-step examination solutions for Unit 1.

SOLVED PROBLEM 1.1

Atom Density and Cross Section Calculation in Enriched UO2 Fuel

A commercial Pressurized Water Reactor uses uranium dioxide fuel ($\text{UO}_2$) enriched to $4.0\text{ wt}\%$ in $^{235}\text{U}$ (with the remaining $96.0\text{ wt}\%$ being $^{238}\text{U}$). The bulk density of the sintered fuel pellets is $\rho = 10.45\text{ g/cm}^3$. Atomic masses: $M({}^{235}\text{U}) = 235.044\text{ g/mol}$, $M({}^{238}\text{U}) = 238.051\text{ g/mol}$, $M({}^{16}\text{O}) = 15.999\text{ g/mol}$. Thermal microscopic fission cross sections: $\sigma_f({}^{235}\text{U}) = 585\text{ b}$, $\sigma_f({}^{238}\text{U}) \approx 0\text{ b}$. Thermal microscopic absorption cross sections: $\sigma_a({}^{235}\text{U}) = 680\text{ b}$, $\sigma_a({}^{238}\text{U}) = 2.70\text{ b}$, $\sigma_a({}^{16}\text{O}) = 0.0002\text{ b}$. (a) Calculate the average atomic weight of the uranium mixture and the molecular weight of the enriched $\text{UO}_2$. (b) Calculate the atom densities $N({}^{235}\text{U})$, $N({}^{238}\text{U})$, and $N({}^{16}\text{O})$ in units of $\text{atoms/cm}^3$ and $\text{atoms/b}\cdot\text{cm}$. (c) Calculate the thermal macroscopic fission cross section $\Sigma_f$ and absorption cross section $\Sigma_a$ of the fuel pellet in $\text{cm}^{-1}$.

RIGOROUS DERIVATION & EXAM SOLUTION
Full Rigorous Analytical Solution
**(a) Average Atomic and Molecular Weights:** Using weight fractions $w_{235} = 0.040$ and $w_{238} = 0.960$: The average atomic weight of uranium $M_U$ is: $$\frac{1}{M_U} = \frac{w_{235}}{M_{235}} + \frac{w_{238}}{M_{238}} = \frac{0.040}{235.044} + \frac{0.960}{238.051} = 1.7018 \times 10^{-4} + 4.0328 \times 10^{-3} = 4.2030 \times 10^{-3}\text{ mol/g}$$ $$M_U = \frac{1}{4.2030 \times 10^{-3}} \approx \mathbf{237.927\text{ g/mol}}$$ Molecular weight of $\text{UO}_2$: $$M_{\text{UO}_2} = M_U + 2 M_O = 237.927 + 2(15.999) = \mathbf{269.925\text{ g/mol}}$$ **(b) Atom Densities:** Molecular density of $\text{UO}_2$: $$N_{\text{mol}} = \frac{\rho \cdot N_A}{M_{\text{UO}_2}} = \frac{10.45\text{ g/cm}^3 \times 6.02214 \times 10^{23}\text{ molecules/mol}}{269.925\text{ g/mol}} \approx \mathbf{2.3314 \times 10^{22}\text{ molecules/cm}^3}$$ Total Uranium density $N_U = N_{\text{mol}} = 2.3314 \times 10^{22}\text{ atoms/cm}^3$. Oxygen density: $$N_O = 2 N_{\text{mol}} = 2 \times 2.3314 \times 10^{22} \approx \mathbf{4.6628 \times 10^{22}\text{ atoms/cm}^3} = \mathbf{0.04663\text{ atoms/b}\cdot\text{cm}}$$ Using mass fraction formulas directly for individual uranium isotopes: $$N_{235} = \frac{w_{235} \cdot (M_U / M_{\text{UO}_2}) \cdot \rho \cdot N_A}{M_{235}} = \frac{w_{235} \rho_U N_A}{M_{235}}$$ Since the mass fraction of Uranium in $\text{UO}_2$ is $f_U = M_U / M_{\text{UO}_2} = 237.927 / 269.925 \approx 0.88146$: $$\rho_U = 0.88146 \times 10.45\text{ g/cm}^3 \approx 9.2113\text{ g }U/\text{cm}^3$$ $$\rho_{235} = 0.040 \times 9.2113 = 0.36845\text{ g/cm}^3$$ $$N_{235} = \frac{0.36845 \times 6.02214 \times 10^{23}}{235.044} \approx \mathbf{9.440 \times 10^{20}\text{ atoms/cm}^3} = \mathbf{9.440 \times 10^{-4}\text{ atoms/b}\cdot\text{cm}}$$ For ${}^{238}\text{U}$: $$\rho_{238} = 0.960 \times 9.2113 = 8.8428\text{ g/cm}^3$$ $$N_{238} = \frac{8.8428 \times 6.02214 \times 10^{23}}{238.051} \approx \mathbf{2.2370 \times 10^{22}\text{ atoms/cm}^3} = \mathbf{0.02237\text{ atoms/b}\cdot\text{cm}}$$ Check: $N_{235} + N_{238} = 0.0944 \times 10^{22} + 2.2370 \times 10^{22} = 2.3314 \times 10^{22} = N_U$. (Exact match!) **(c) Macroscopic Fission and Absorption Cross Sections:** Recall $1\text{ b} = 10^{-24}\text{ cm}^2$: $$\Sigma_f = N_{235} \sigma_f^{235} + N_{238} \sigma_f^{238} = (9.440 \times 10^{20}\text{ cm}^{-3})(585 \times 10^{-24}\text{ cm}^2) + 0 \approx \mathbf{0.5522\text{ cm}^{-1}}$$ Macroscopic absorption cross section: $$\Sigma_a = N_{235}\sigma_a^{235} + N_{238}\sigma_a^{238} + N_O\sigma_a^O$$ $$\Sigma_a = (9.440 \times 10^{20})(680 \times 10^{-24}) + (2.2370 \times 10^{22})(2.70 \times 10^{-24}) + (4.6628 \times 10^{22})(0.0002 \times 10^{-24})$$ $$\Sigma_a = 0.64192\text{ cm}^{-1} + 0.06040\text{ cm}^{-1} + 9.3 \times 10^{-6}\text{ cm}^{-1} \approx \mathbf{0.7023\text{ cm}^{-1}}$$ The thermal macroscopic fission cross section is **$0.552\text{ cm}^{-1}$** and the total absorption cross section is **$0.702\text{ cm}^{-1}$**.
Final Answer & Verification

Complete rigorous derivation and proof detailed above.

SOLVED PROBLEM 1.2

Neutron Separation Energies and Critical Fission Barrier Evaluation

Given the high-precision atomic masses: $M({}^{235}\text{U}) = 235.043930\text{ u}$ $M({}^{236}\text{U}) = 236.045568\text{ u}$ $M({}^{238}\text{U}) = 238.050788\text{ u}$ $M({}^{239}\text{U}) = 239.054293\text{ u}$ $m_n = 1.0086649\text{ u}$, $1\text{ u} = 931.494\text{ MeV}/c^2$. The critical deformation barriers against fission are $E_{\text{crit}}({}^{236}\text{U}) = 5.70\text{ MeV}$ and $E_{\text{crit}}({}^{239}\text{U}) = 5.85\text{ MeV}$. (a) Calculate the neutron separation energy $S_n$ for the compound nucleus $^{236}\text{U}^*$ formed by neutron capture on $^{235}\text{U}$. (b) Calculate the neutron separation energy $S_n$ for the compound nucleus $^{239}\text{U}^*$ formed by neutron capture on $^{238}\text{U}$. (c) Comparing $S_n$ with the critical fission barriers, determine whether zero-energy thermal neutrons can induce fission in $^{235}\text{U}$ and $^{238}\text{U}$, and compute the minimum threshold laboratory kinetic energy $E_{\text{th}}$ required for neutrons to induce fission in $^{238}\text{U}$.

RIGOROUS DERIVATION & EXAM SOLUTION
Full Rigorous Analytical Solution
**(a) Neutron Separation Energy of $^{236}\text{U}^*$:** The capture reaction is $n + {}^{235}\text{U} \to {}^{236}\text{U}^*$. $$\Delta m({}^{236}\text{U}) = \left[ M({}^{235}\text{U}) + m_n \right] - M({}^{236}\text{U})$$ $$\Delta m({}^{236}\text{U}) = [235.043930 + 1.0086649] - 236.045568 = 236.0525949 - 236.045568 = 0.0070269\text{ u}$$ In energy units: $$S_n({}^{236}\text{U}^*) = 0.0070269\text{ u} \times 931.494\text{ MeV/u} \approx \mathbf{6.5455\text{ MeV}}$$ **(b) Neutron Separation Energy of $^{239}\text{U}^*$:** The capture reaction is $n + {}^{238}\text{U} \to {}^{239}\text{U}^*$. $$\Delta m({}^{239}\text{U}) = \left[ M({}^{238}\text{U}) + m_n \right] - M({}^{239}\text{U})$$ $$\Delta m({}^{239}\text{U}) = [238.050788 + 1.0086649] - 239.054293 = 239.0594529 - 239.054293 = 0.0051599\text{ u}$$ In energy units: $$S_n({}^{239}\text{U}^*) = 0.0051599\text{ u} \times 931.494\text{ MeV/u} \approx \mathbf{4.8064\text{ MeV}}$$ **(c) Comparison with Fission Barriers & Threshold Energy:** 1. **For $^{235}\text{U}$:** $$E^* = S_n({}^{236}\text{U}^*) = 6.546\text{ MeV}$$ $$E_{\text{crit}}({}^{236}\text{U}) = 5.70\text{ MeV}$$ $$E^* - E_{\text{crit}} = 6.546 - 5.70 = \mathbf{+0.846\text{ MeV} > 0}$$ Because the excitation energy delivered purely by neutron binding exceeds the fission barrier by $0.85\text{ MeV}$, **thermal neutrons readily induce fission in $^{235}\text{U}$ with colossal cross section ($\sigma_f = 585\text{ b}$)**! 2. **For $^{238}\text{U}$:** $$E^* = S_n({}^{239}\text{U}^*) = 4.806\text{ MeV}$$ $$E_{\text{crit}}({}^{239}\text{U}) = 5.85\text{ MeV}$$ $$E^* - E_{\text{crit}} = 4.806 - 5.85 = \mathbf{-1.044\text{ MeV} < 0}$$ Thermal neutrons leave the compound nucleus $1.04\text{ MeV}$ short of the saddle point. To overcome this deficit, the incident neutron must supply kinetic energy in the center-of-mass frame: $$E_{\text{cm}} \ge E_{\text{crit}} - S_n = 5.85\text{ MeV} - 4.806\text{ MeV} = 1.044\text{ MeV}$$ Converting to laboratory frame kinetic energy ($E_{\text{lab}} = E_{\text{cm}} \frac{m_n + M_{238}}{M_{238}}$): $$E_{\text{th}} = 1.044\text{ MeV} \times \left( \frac{1 + 238}{238} \right) = 1.044 \times 1.0042 \approx \mathbf{1.048\text{ MeV}}$$ Neutrons must have at least **$\approx 1.05\text{ MeV}$** of kinetic energy to induce fission in $^{238}\text{U}$!
Final Answer & Verification

Complete rigorous derivation and proof detailed above.

SOLVED PROBLEM 1.3

Breeder Reactor Breeding Ratio, Breeding Gain and Doubling Time

A liquid-metal fast breeder reactor (LMFBR) generates a steady thermal power of $P_{\text{th}} = 2500\text{ MWth}$. The reactor core operates with a breeding ratio $BR = 1.25$ and an initial core fissile inventory of $M_{\text{fiss}} = 3200\text{ kg}$ of $^{239}\text{Pu}$. Each fission of $^{239}\text{Pu}$ releases an average recoverable energy of $Q = 205\text{ MeV}$. The ratio of parasitic capture to fission in the core fissile fuel is $\alpha \equiv \sigma_c / \sigma_f = 0.15$. Out-of-pile reprocessing and refabrication losses are $2.0\%$ of the bred fuel, and the reactor capacity factor is $CF = 0.85$. (a) Calculate the daily fission rate $\dot{N}_f$ and the daily mass of $^{239}\text{Pu}$ consumed by fission and absorption. (b) Calculate the net daily production rate of excess fissile $^{239}\text{Pu}$. (c) Determine the simple doubling time $T_d$ of the breeder reactor in calendar years.

RIGOROUS DERIVATION & EXAM SOLUTION
Full Rigorous Analytical Solution
**(a) Fission Rate and Daily Consumption:** Energy per fission in Joules: $$E_f = 205\text{ MeV} \times 1.60218 \times 10^{-13}\text{ J/MeV} = 3.2845 \times 10^{-11}\text{ J/fission}$$ At $P_{\text{th}} = 2500\text{ MWth} = 2.5 \times 10^9\text{ J/s}$: $$\dot{N}_f = \frac{P_{\text{th}}}{E_f} = \frac{2.5 \times 10^9\text{ J/s}}{3.2845 \times 10^{-11}\text{ J/fission}} \approx 7.6115 \times 10^{19}\text{ fissions/sec}$$ Daily fissions ($1\text{ day} = 86400\text{ s}$): $$N_{f, \text{day}} = 7.6115 \times 10^{19} \times 86400 \approx \mathbf{6.576 \times 10^{24}\text{ fissions/day}}$$ Mass consumed by fission alone per day: $$m_{f, \text{day}} = \frac{N_{f, \text{day}} \cdot M_{239}}{N_A} = \frac{(6.576 \times 10^{24})(0.23905\text{ kg/mol})}{6.02214 \times 10^{23}} \approx \mathbf{2.610\text{ kg/day}}$$ Total fissile consumption includes radiative capture ($1 + \alpha = 1 + 0.15 = 1.15$): $$\dot{m}_{\text{consumed}} = (1 + \alpha) \cdot m_{f, \text{day}} = 1.15 \times 2.610\text{ kg/day} \approx \mathbf{3.002\text{ kg } {}^{239}\text{Pu/day}}$$ **(b) Net Daily Production Rate of Excess Fissile $^{239}\text{Pu}$:** By definition of breeding ratio: $$\dot{m}_{\text{produced}} = BR \cdot \dot{m}_{\text{consumed}} = 1.25 \times 3.002\text{ kg/day} \approx 3.7525\text{ kg/day}$$ Gross excess production: $$\Delta\dot{m}_{\text{gross}} = \dot{m}_{\text{produced}} - \dot{m}_{\text{consumed}} = (BR - 1) \cdot \dot{m}_{\text{consumed}} = 0.25 \times 3.002 = 0.7505\text{ kg/day}$$ Accounting for $2.0\%$ reprocessing loss ($\eta_{\text{rep}} = 0.98$): $$\Delta\dot{m}_{\text{net, operating}} = 0.7505\text{ kg/day} - 0.02(3.7525\text{ kg/day}) = 0.7505 - 0.0751 \approx \mathbf{0.6754\text{ kg/day}}$$ Accounting for capacity factor $CF = 0.85$: $$\Delta\dot{m}_{\text{annual}} = 0.6754\text{ kg/day} \times (365.25 \times 0.85\text{ days}) = 0.6754 \times 310.46 \approx \mathbf{209.7\text{ kg/calendar year}}$$ **(c) Simple Doubling Time $T_d$:** The doubling time is the time required to accumulate the initial core inventory $M_{\text{fiss}} = 3200\text{ kg}$: $$T_d = \frac{M_{\text{fiss}}}{\Delta\dot{m}_{\text{annual}}} = \frac{3200\text{ kg}}{209.7\text{ kg/yr}} \approx \mathbf{15.26\text{ calendar years}}$$ The simple doubling time of the breeder reactor is **$\approx 15.3\text{ years}$**.
Final Answer & Verification

Complete rigorous derivation and proof detailed above.