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Chapter 2 • Theory & Derivations

Neutron Interactions, Cross Sections & Nuclear Reaction Rates

Detailed exploration of neutron interactions with matter: classification of neutrons across seven energy decades from relativistic to cold thermal states; laboratory and reactor neutron production mechanisms; physical definitions and units of microscopic cross sections (barns) and macroscopic cross sections (cm⁻¹); the concept of neutron mean free path; exponential beam attenuation through bulk materials; scalar neutron flux φ = nv and vector current density; volumetric nuclear reaction rate calculations R = Σφ; energy-dependent cross section phenomena including the low-energy 1/v absorption law, Doppler-broadened Breit-Wigner resonances, and fission thresholds; and thermal Maxwellian-averaged cross sections with Westcott g-factor corrections.

§2.1 Classification of Neutrons by Kinetic Energy & Nuclear Sources

1. The Energy Spectrum of Neutrons

Neutrons produced in nuclear fission are born with high kinetic energies averaging $\bar{E} \approx 2\text{ MeV}$, but subsequently moderate across nine orders of magnitude down to thermal equilibrium ($E \sim 0.025\text{ eV}$). Reactor physics categorizes neutrons into well-defined energy regimes:

Neutron Classification Energy Range Typical Speed Dominant Interaction Mechanism
Relativistic / High Energy $E > 20\text{ MeV}$ $> 0.2 c$ Nuclear spallation, meson production
Fast Neutrons $0.1\text{ MeV} < E \le 20\text{ MeV}$ $\sim 1.4 \times 10^7\text{ m/s}$ Fission spectrum; elastic & inelastic scattering
Epithermal / Intermediate $1\text{ eV} < E \le 0.1\text{ MeV}$ $\sim 10^5\text{ to }10^6\text{ m/s}$ Elastic moderation ($1/E$ slowing-down spectrum)
Resonance Neutrons $1\text{ eV} \le E \le 1\text{ keV}$ $\sim 10^4\text{ to }10^5\text{ m/s}$ Sharp $(n,\gamma)$ capture resonances ($^{238}\text{U}$)
Thermal Neutrons $0.01\text{ eV} \le E \le 0.5\text{ eV}$ $2200\text{ m/s}$ Thermal equilibrium with moderator; fission in $^{235}\text{U}$
Cold & Ultracold (UCN) $E < 0.005\text{ eV}$ (UCN $< 10^{-7}\text{ eV}$) $< 100\text{ m/s}$ ($< 5\text{ m/s}$) Bragg reflection, total internal reflection in pipes

2. Laboratory and Startup Neutron Sources

To start up a subcritical nuclear reactor safely, external neutron sources are installed to provide an initial detectable neutron flux for nuclear instrumentation:

  • Alpha-Neutron $(\alpha, n)$ Sources: Utilize alpha decay from actinides mixed with beryllium-9: $${}^9_4\text{Be} + \alpha \longrightarrow {}^{12}_6\text{C} + n + 5.70\text{ MeV}$$ Standard commercial sources include $\text{Am-Be}$ ($T_{1/2} = 432\text{ yr}$) and $\text{Pu-Be}$.
  • Photoneutron $(\gamma, n)$ Sources: Photons exceeding the low neutron binding energy of deuterium ($2.22\text{ MeV}$) or beryllium ($1.67\text{ MeV}$): $${}^9_4\text{Be} + \gamma (> 1.67\text{ MeV}) \longrightarrow {}^8_4\text{Be} + n$$ A common source is Antimony-Beryllium (${}^{124}\text{Sb}\text{-Be}$), where $^{124}\text{Sb}$ emits $1.69\text{ MeV}$ gammas.
  • Spontaneous Fission Sources: Californium-252 (${}^{252}_{98}\text{Cf}$, $T_{1/2} = 2.645\text{ yr}$) decays $3.09\%$ by spontaneous fission, emitting an intense stream of $3.76$ neutrons per fission ($2.3 \times 10^{12}\text{ n/s per gram}$).

§2.2 Microscopic Cross Sections: Geometric Analogy & Probability Units

1. Definition of the Microscopic Cross Section $\sigma$

Consider a monodirectional, uniform beam of neutrons of intensity $I$ ($\text{neutrons/cm}^2\cdot\text{s}$) impinging perpendicularly upon a thin target containing a single isolated nucleus. The probability per unit time that a specific nuclear reaction occurs is directly proportional to the incident beam intensity $I$: $$\text{Interaction Rate} = \sigma \cdot I$$ The constant of proportionality $\sigma$ has dimensions of area ($\text{cm}^2/\text{nucleus}$) and represents the effective target area that the nucleus presents to the passing neutron.

Because nuclear radii are on the order of $R \sim 10^{-12}\text{ cm}$, typical geometric cross sections are: $$\sigma_{\text{geom}} \approx \pi R^2 \approx \pi (10^{-12}\text{ cm})^2 = 3.14 \times 10^{-24}\text{ cm}^2$$ To establish a convenient unit, nuclear physicists in the 1940s Manhattan Project coined the barn ($\text{b}$): $$1\text{ barn (b)} \equiv 10^{-24}\text{ cm}^2 = 10^{-28}\text{ m}^2$$ Sub-units include millibarns ($1\text{ mb} = 10^{-3}\text{ b} = 10^{-27}\text{ cm}^2$) and microbarns ($1\text{ }\mu\text{b} = 10^{-6}\text{ b}$).

2. Partial and Total Cross Sections

A neutron can interact with a nucleus through multiple mutually exclusive reaction channels: $$\sigma_t = \sigma_s + \sigma_a$$

  1. Scattering Cross Section ($\sigma_s$): $$\sigma_s = \sigma_e + \sigma_i$$ where $\sigma_e$ is elastic potential and resonance scattering $(n, n)$ (kinetic energy conserved in CM frame), and $\sigma_i$ is inelastic scattering $(n, n')$ (leaving target nucleus in an excited state).
  2. Absorption Cross Section ($\sigma_a$): $$\sigma_a = \sigma_\gamma + \sigma_f + \sigma_\alpha + \sigma_p$$ where $\sigma_\gamma$ is radiative capture $(n, \gamma)$, $\sigma_f$ is neutron-induced fission $(n, f)$, $\sigma_\alpha$ is alpha emission $(n, \alpha)$, and $\sigma_p$ is proton ejection $(n, p)$.
The total interaction probability is governed by the total microscopic cross section $\sigma_t$.

§2.3 Macroscopic Cross Sections & Neutron Mean Free Path

1. The Macroscopic Cross Section $\Sigma$

In a macroscopic medium containing target nuclei with atom density $N$ ($\text{nuclei/cm}^3$), the macroscopic cross section $\Sigma$ represents the total interaction probability per unit distance traveled by a neutron: $$\Sigma \equiv N \cdot \sigma \quad [\text{cm}^{-1}]$$ Physical significance of dimensions: $$\Sigma = \left( \frac{\text{nuclei}}{\text{cm}^3} \right) \times \left( \frac{\text{cm}^2}{\text{nucleus}} \right) = \mathbf{\text{cm}^{-1}}$$ For a homogeneous mixture of $K$ distinct isotopic species: $$\Sigma = \sum_{i=1}^K N_i \sigma_i = N_1 \sigma_1 + N_2 \sigma_2 + \dots + N_K \sigma_K$$ Total macroscopic cross section: $$\Sigma_t = \Sigma_s + \Sigma_a = \Sigma_e + \Sigma_i + \Sigma_\gamma + \Sigma_f$$

2. The Neutron Mean Free Path $\lambda$

The mean free path $\lambda$ is the average distance a neutron travels through a medium between two successive collisions. The probability that a neutron travels distance $x$ without interaction is $e^{-\Sigma_t x}$, and the probability of colliding in differential slice $dx$ is $\Sigma_t dx$. Thus, the collision probability density is: $$p(x) dx = \Sigma_t e^{-\Sigma_t x} dx$$ The expectation value of travel distance is: $$\lambda \equiv \langle x \rangle = \int_0^\infty x \, p(x) dx = \Sigma_t \int_0^\infty x e^{-\Sigma_t x} dx = \Sigma_t \left[ \frac{1}{\Sigma_t^2} \right] = \mathbf{\frac{1}{\Sigma_t}}$$ Partial mean free paths: $$\lambda_s = \frac{1}{\Sigma_s} \quad \text{(Mean free path for scattering)}, \qquad \lambda_a = \frac{1}{\Sigma_a} \quad \text{(Mean free path for absorption)}$$ Because $\Sigma_t = \Sigma_s + \Sigma_a$: $$\frac{1}{\lambda_t} = \frac{1}{\lambda_s} + \frac{1}{\lambda_a}$$ In light water, thermal neutrons have $\lambda_s \approx 0.3\text{ cm}$ and $\lambda_a \approx 45\text{ cm}$, demonstrating that a neutron scatters over a hundred times before undergoing absorption!

§2.4 Neutron Beam Attenuation & Transmission Probability

1. Derivation of the Exponential Attenuation Law

Consider a collimated, monoenergetic neutron beam of initial intensity $I_0$ ($\text{neutrons/cm}^2\cdot\text{s}$) entering a slab of material at normal incidence ($x = 0$). Let $I(x)$ denote the uncollided beam intensity at depth $x$. In traversing a differential layer of thickness $dx$: The number of collisions occurring in unit area per second is: $$-dI(x) = I(x) \cdot \Sigma_t \cdot dx$$ Rearranging into a differential equation: $$\frac{dI(x)}{dx} = -\Sigma_t I(x)$$ Integrating with boundary condition $I(0) = I_0$: $$\int_{I_0}^{I(x)} \frac{dI}{I} = -\Sigma_t \int_0^x dx \implies \ln\left( \frac{I(x)}{I_0} \right) = -\Sigma_t x$$ $$I(x) = I_0 \exp(-\Sigma_t x)$$ This is the fundamental exponential attenuation law for uncollided neutrons.

2. Transmission, Absorption, and Half-Value Thickness

For a shield slab of total physical thickness $t$:

  • Transmission Probability ($T$): The fraction of the beam penetrating without undergoing any interaction: $$T \equiv \frac{I(t)}{I_0} = e^{-\Sigma_t t}$$
  • Interaction Probability ($P_{\text{int}}$): The fraction undergoing at least one collision: $$P_{\text{int}} = 1 - T = 1 - e^{-\Sigma_t t}$$
  • Half-Value Layer ($HVL$): The thickness of material required to reduce the uncollided beam intensity by exactly $50\%$ ($I/I_0 = 1/2$): $$e^{-\Sigma_t \cdot HVL} = \frac{1}{2} \implies HVL = \frac{\ln 2}{\Sigma_t} = \frac{0.69315}{\Sigma_t} \approx 0.693 \lambda_t$$
  • Tenth-Value Layer ($TVL$): The thickness required to attenuate the uncollided beam by a factor of 10: $$TVL = \frac{\ln 10}{\Sigma_t} = \frac{2.3026}{\Sigma_t}$$

§2.5 Neutron Flux, Current Density & Total Reaction Rates

1. Scalar Neutron Flux $\phi(\vec{r}, E)$

In a reactor core, neutrons are not collimated into a single beam; they travel isotropically in all directions. If $n(\vec{r}, E) dE$ is the density of neutrons ($\text{neutrons/cm}^3$) at position $\vec{r}$ with energy in $dE$, and $v$ is their speed: The scalar neutron flux $\phi(\vec{r}, E)$ is defined as: $$\phi(\vec{r}, E) \equiv n(\vec{r}, E) \cdot v \quad [\text{neutrons/cm}^2\cdot\text{s}]$$ Physical interpretation: $\phi$ is the total track length swept out by all neutrons contained within a unit volume per second ($\text{cm of track} / \text{cm}^3\cdot\text{s} = \text{cm}^{-2}\text{s}^{-1}$).

2. Neutron Current Density Vector $\vec{J}(\vec{r}, E)$

While scalar flux $\phi$ is an orientation-independent scalar measure of neutron activity, the neutron current density $\vec{J}$ is a vector describing the net rate of flow of neutrons across a unit surface oriented perpendicular to $\vec{J}$: $$\vec{J}(\vec{r}, E) \equiv \int_{4\pi} \vec{\Omega} \, \psi(\vec{r}, \vec{\Omega}, E) \, d\Omega$$ where $\psi(\vec{r}, \vec{\Omega}, E) = v \cdot n(\vec{r}, \vec{\Omega}, E)$ is the angular flux. In an isotropic neutron field, the current vanishes ($\vec{J} = 0$), but the scalar flux remains non-zero ($\phi > 0$).

3. Volumetric Reaction Rate Density

The volumetric reaction rate density $R_x(\vec{r})$ (number of reactions of type $x$ occurring per unit volume per second) is given by: $$R_x(\vec{r}) = \int_0^\infty \Sigma_x(\vec{r}, E) \, \phi(\vec{r}, E) \, dE \quad [\text{reactions/cm}^3\cdot\text{s}]$$ For a monoenergetic or one-group thermal flux $\phi_{\text{th}}$: $$R_x = \Sigma_x \phi_{\text{th}} = N \sigma_x \phi_{\text{th}}$$ Integrating over the entire reactor core volume $V_{\text{core}}$ gives the total core reaction rate: $$\mathcal{R}_x = \int_{V_{\text{core}}} \Sigma_x(\vec{r}) \phi(\vec{r}) d^3r \quad [\text{reactions/s}]$$ For nuclear fission, the total thermal power $P$ released is: $$P = Q_f \cdot \mathcal{R}_f = Q_f \int_{V_{\text{core}}} \Sigma_f(\vec{r}) \phi(\vec{r}) d^3r \quad [\text{Watts}]$$

§2.6 Energy Dependence of Cross Sections: 1/v Law, Resonances & Fission

1. The $1/v$ Low-Energy Law

In the low-energy region below isolated resonances ($E \lesssim 0.1\text{ eV}$), quantum perturbation theory predicts that the probability of absorbing an $s$-wave ($l = 0$) neutron is proportional to the time the neutron spends in the vicinity of the nuclear potential well: $$\text{Interaction Time} \Delta t \propto \frac{R_{\text{nuc}}}{v} \propto \frac{1}{\sqrt{E}}$$ Consequently, for almost all non-threshold absorption reactions, the microscopic cross section follows the universal $1/v$ law: $$\sigma_a(v) = \sigma_a(v_0) \frac{v_0}{v} = \sigma_a(E_0) \sqrt{\frac{E_0}{E}}$$ where $v_0 = 2200\text{ m/s}$ and $E_0 = 0.0253\text{ eV}$. Plotted on a $\log\sigma_a$ versus $\log E$ graph, the $1/v$ cross section is a straight line with a characteristic slope of $-1/2$: $$\log\sigma_a = \text{const} - \frac{1}{2}\log E$$

2. Breit-Wigner Resonances and Doppler Broadening

In the intermediate energy regime ($1\text{ eV} \le E \le 10\text{ keV}$), incident neutron energies match discrete compound nuclear quasi-bound states. The cross section exhibits colossal resonance peaks accurately parameterized by the single-level Breit-Wigner dispersion formula: $$\sigma_\gamma(E) = \frac{\pi}{k^2} g_J \frac{\Gamma_n \Gamma_\gamma}{(E - E_0)^2 + (\Gamma/2)^2}$$ where $E_0$ is the resonance energy, $\Gamma = \Gamma_n + \Gamma_\gamma + \Gamma_f$ is the total resonance width, and $g_J$ is the statistical spin factor.

As the reactor fuel temperature $T$ increases, thermal vibrations of the actinide lattice nuclei broaden the relative velocity distribution between target nuclei and incident neutrons. This phenomenon—Doppler Broadening—flattens the peak height but broadens the resonance wings without altering the total area: $$\Psi(\theta, x) = \frac{\theta}{2\sqrt{\pi}} \int_{-\infty}^\infty \frac{\exp\left( -\frac{\theta^2}{4}(x - y)^2 \right)}{1 + y^2} dy, \quad \theta \equiv \frac{\Gamma}{\Delta_{\text{Doppler}}}$$ In thick fuel rods, Doppler broadening exposes more neutrons to resonance capture (reducing self-shielding), providing an instantaneous, inherently safe negative temperature reactivity feedback!

§2.7 Thermal Maxwellian-Averaged Cross Sections & The Westcott g-Factor

1. Maxwellian Average of $1/v$ Cross Sections

Because thermal neutrons are distributed across a continuous Maxwellian spectrum $n(v)$, the effective reaction rate in a thermal reactor is: $$R_a = \int_0^\infty N \sigma_a(v) v \, n(v) \, dv$$ If the absorber obeys the ideal $1/v$ law ($\sigma_a(v) = \sigma_0 v_0 / v$): $$R_a = N \sigma_0 v_0 \int_0^\infty n(v) dv = N \sigma_0 v_0 n_0$$ Defining the conventional 2200 m/s thermal flux $\phi_0 \equiv n_0 v_0$: $$R_a = N \sigma_0 \phi_0 = \Sigma_0 \phi_0$$ Remarkably, for a pure $1/v$ absorber, the reaction rate is independent of the moderator temperature $T$ when evaluated using the 2200 m/s cross section $\sigma_0$ and the standard flux $\phi_0 = n_0 v_0$!

2. The Westcott $g$-Factor for Non-$1/v$ Absorbers

For nuclides that possess low-lying resonances near thermal energies (such as $^{235}\text{U}$, $^{239}\text{Pu}$, $^{241}\text{Pu}$, and $^{113}\text{Cd}$), the absorption cross section deviates significantly from $1/v$. Carl Westcott introduced the dimensionless Westcott $g$-factor $g(T)$ to correct the 2200 m/s cross section: $$g(T) \equiv \frac{1}{\sigma_0 v_0} \frac{\int_0^\infty \sigma_a(v) v \, n(v) \, dv}{\int_0^\infty n(v) \, dv} = \frac{\bar{\sigma}_a(T)}{\sigma_0} \frac{\bar{v}}{v_0} \frac{\sqrt{\pi}}{2}$$ The true thermal reaction rate is then calculated as: $$R_a = g(T) \cdot \Sigma_0 \cdot \phi_0$$

  • For an ideal $1/v$ absorber (such as Boron-10): $g(T) \equiv 1.000$ at all temperatures.
  • For Uranium-235: $g_a(293.6\text{ K}) = 0.9780$, $g_f(293.6\text{ K}) = 0.9766$.
  • For Plutonium-239: Due to a massive resonance at $0.296\text{ eV}$, $g_a(293.6\text{ K}) = 1.072$, rising above $1.4$ as coolant temperature rises to $600\text{ K}$!

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 Example 2.1: Neutron Beam Attenuation and Half-Value Thickness in Borated Polyethylene

A narrow, collimated beam of thermal neutrons ($E = 0.0253\text{ eV}$) with initial intensity $I_0 = 1.5 \times 10^7\text{ neutrons/cm}^2\cdot\text{s}$ impinges perpendicularly upon a shielding slab of borated polyethylene ($5.0\text{ wt}\%$ natural Boron). The mass density of the borated polyethylene is $\rho = 0.95\text{ g/cm}^3$. Composition by weight: $5.0\%$ Boron ($M_B = 10.811\text{ g/mol}$), $13.6\%$ Hydrogen ($M_H = 1.008\text{ g/mol}$), and $81.4\%$ Carbon ($M_C = 12.011\text{ g/mol}$). Microscopic cross sections at $0.0253\text{ eV}$:

  • Boron: $\sigma_a = 767\text{ b}$, $\sigma_s = 4.0\text{ b}$
  • Hydrogen: $\sigma_a = 0.332\text{ b}$, $\sigma_s = 21.0\text{ b}$
  • Carbon: $\sigma_a = 0.0035\text{ b}$, $\sigma_s = 4.8\text{ b}$

(a) Calculate the atom densities $N_B, N_H, N_C$ in $\text{atoms/cm}^3$. (b) Calculate the total macroscopic cross section $\Sigma_t$ in $\text{cm}^{-1}$ and the neutron mean free path $\lambda_t$ in $\text{cm}$. (c) Determine the Half-Value Layer ($HVL$) and Tenth-Value Layer ($TVL$) of the shield. (d) Calculate the thickness $t$ of the slab required to attenuate the uncollided beam intensity down to $I(t) = 100\text{ neutrons/cm}^2\cdot\text{s}$.

(a) Atom Densities: Using $N_i = \frac{w_i \cdot \rho \cdot N_A}{M_i}$:

$$N_B = \frac{0.050 \times 0.95\text{ g/cm}^3 \times 6.02214 \times 10^{23}}{10.811\text{ g/mol}} \approx \mathbf{2.6457 \times 10^{21}\text{ atoms/cm}^3} = 2.6457 \times 10^{-3}\text{ b}^{-1}\text{cm}^{-1}$$
$$N_H = \frac{0.136 \times 0.95 \times 6.02214 \times 10^{23}}{1.008} \approx \mathbf{7.7214 \times 10^{22}\text{ atoms/cm}^3} = 0.07721\text{ b}^{-1}\text{cm}^{-1}$$
$$N_C = \frac{0.814 \times 0.95 \times 6.02214 \times 10^{23}}{12.011} \approx \mathbf{3.8757 \times 10^{22}\text{ atoms/cm}^3} = 0.03876\text{ b}^{-1}\text{cm}^{-1}$$

(b) Total Macroscopic Cross Section $\Sigma_t$ and Mean Free Path $\lambda_t$: Total microscopic cross sections ($\sigma_t = \sigma_a + \sigma_s$):

$$\sigma_t(B) = 767 + 4.0 = 771\text{ b} = 771 \times 10^{-24}\text{ cm}^2$$
$$\sigma_t(H) = 0.332 + 21.0 = 21.332\text{ b} = 21.332 \times 10^{-24}\text{ cm}^2$$
$$\sigma_t(C) = 0.0035 + 4.8 = 4.8035\text{ b} = 4.8035 \times 10^{-24}\text{ cm}^2$$

Macroscopic cross section components:

$$\Sigma_t(B) = (2.6457 \times 10^{-3}\text{ b}^{-1}\text{cm}^{-1})(771\text{ b}) = 2.0398\text{ cm}^{-1}$$
$$\Sigma_t(H) = (0.07721)(21.332) = 1.6470\text{ cm}^{-1}$$
$$\Sigma_t(C) = (0.03876)(4.8035) = 0.1862\text{ cm}^{-1}$$

Total macroscopic cross section:

$$\Sigma_t = 2.0398 + 1.6470 + 0.1862 = \mathbf{3.873\text{ cm}^{-1}}$$

Mean free path:

$$\lambda_t = \frac{1}{\Sigma_t} = \frac{1}{3.873\text{ cm}^{-1}} \approx \mathbf{0.2582\text{ cm}} = \mathbf{2.58\text{ mm}}$$

(c) Half-Value Layer ($HVL$) and Tenth-Value Layer ($TVL$):

$$HVL = \frac{\ln 2}{\Sigma_t} = \frac{0.69315}{3.873\text{ cm}^{-1}} \approx \mathbf{0.1790\text{ cm}} \approx \mathbf{1.79\text{ mm}}$$
$$TVL = \frac{\ln 10}{\Sigma_t} = \frac{2.30259}{3.873\text{ cm}^{-1}} \approx \mathbf{0.5945\text{ cm}} \approx \mathbf{5.95\text{ mm}}$$

(d) Required Shield Thickness for $I(t) = 100\text{ n/cm}^2\cdot\text{s}$: Using $I(t) = I_0 e^{-\Sigma_t t}$:

$$\frac{I(t)}{I_0} = \frac{100}{1.5 \times 10^7} = 6.6667 \times 10^{-6}$$
$$\ln\left( \frac{I(t)}{I_0} \right) = \ln(6.6667 \times 10^{-6}) = -11.9184$$
$$t = \frac{11.9184}{\Sigma_t} = \frac{11.9184}{3.873\text{ cm}^{-1}} \approx \mathbf{3.077\text{ cm}} \approx \mathbf{30.8\text{ mm}}$$

A shield plate of only $3.08\text{ cm}$ ($1.2\text{ inches}$) thickness reduces the beam intensity by over five orders of magnitude!

Solved Problem Example 2.2: Thermal Reaction Rates and Westcott g-Factor Correction for Plutonium-239

A research reactor core contains a high-purity foil of Plutonium-239 (${}^{239}\text{Pu}$) placed in a thermal neutron flux. The 2200 m/s standard microscopic cross sections are $\sigma_0(f) = 748\text{ b}$ and $\sigma_0(a) = 1017\text{ b}$. The neutron density in the thermal column is $n_0 = 5.0 \times 10^7\text{ neutrons/cm}^3$. (a) Calculate the conventional 2200 m/s thermal flux $\phi_0$ in $\text{neutrons/cm}^2\cdot\text{s}$. (b) At room temperature ($T_1 = 293.6\text{ K}$), the Westcott $g$-factors for $^{239}\text{Pu}$ are $g_f(T_1) = 1.055$ and $g_a(T_1) = 1.072$. Calculate the true effective thermal cross sections $\hat{\sigma}_f$ and $\hat{\sigma}_a$, and compute the microscopic fission rate per $^{239}\text{Pu}$ nucleus. (c) When the core heats up to operating temperature $T_2 = 600\text{ K}$, the $0.296\text{ eV}$ resonance increases the Westcott factors to $g_f(T_2) = 1.340$ and $g_a(T_2) = 1.425$. For the same neutron density $n_0$, calculate the percentage increase in the fission rate per nucleus due to thermal spectrum hardening.

(a) Conventional 2200 m/s Flux $\phi_0$: With standard reference velocity $v_0 = 2200\text{ m/s} = 2.2 \times 10^5\text{ cm/s}$:

$$\phi_0 = n_0 \cdot v_0 = (5.0 \times 10^7\text{ cm}^{-3})(2.2 \times 10^5\text{ cm/s}) = \mathbf{1.10 \times 10^{13}\text{ neutrons/cm}^2\cdot\text{s}}$$

(b) Effective Cross Sections and Reaction Rates at $T_1 = 293.6\text{ K}$: The Westcott effective cross sections are $\hat{\sigma} = g(T) \cdot \sigma_0$:

$$\hat{\sigma}_f(T_1) = g_f(T_1) \cdot \sigma_0(f) = 1.055 \times 748\text{ b} \approx \mathbf{789.14\text{ b}} = 7.8914 \times 10^{-21}\text{ cm}^2$$
$$\hat{\sigma}_a(T_1) = g_a(T_1) \cdot \sigma_0(a) = 1.072 \times 1017\text{ b} \approx \mathbf{1090.22\text{ b}} = 1.0902 \times 10^{-20}\text{ cm}^2$$

Fission reaction rate per nucleus:

$$R_{f, 1} = \hat{\sigma}_f(T_1) \cdot \phi_0 = (7.8914 \times 10^{-21}\text{ cm}^2)(1.10 \times 10^{13}\text{ cm}^{-2}\text{s}^{-1}) \approx \mathbf{8.681 \times 10^{-8}\text{ fissions/nucleus}\cdot\text{s}}$$

(c) Spectrum Hardening Effect at $T_2 = 600\text{ K}$: At $T_2 = 600\text{ K}$:

$$\hat{\sigma}_f(T_2) = g_f(T_2) \cdot \sigma_0(f) = 1.340 \times 748\text{ b} \approx \mathbf{1002.32\text{ b}} = 1.0023 \times 10^{-20}\text{ cm}^2$$

Since neutron density $n_0$ is held constant, the reference flux $\phi_0 = n_0 v_0 = 1.10 \times 10^{13}\text{ cm}^{-2}\text{s}^{-1}$ remains unchanged:

$$R_{f, 2} = \hat{\sigma}_f(T_2) \cdot \phi_0 = (1.0023 \times 10^{-20})(1.10 \times 10^{13}) \approx \mathbf{1.1025 \times 10^{-7}\text{ fissions/nucleus}\cdot\text{s}}$$

Percentage increase in fission rate:

$$\text{Increase} = \frac{R_{f, 2} - R_{f, 1}}{R_{f, 1}} \times 100\% = \frac{g_f(T_2) - g_f(T_1)}{g_f(T_1)} \times 100\%$$
$$\text{Increase} = \frac{1.340 - 1.055}{1.055} \times 100\% = \frac{0.285}{1.055} \times 100\% \approx \mathbf{+27.01\%}$$

The fission rate per nucleus jumps by $+27.0\%$ simply due to thermal spectrum hardening moving toward the $0.296\text{ eV}$ resonance of Plutonium-239!

Solved Problem Example 2.3: Volumetric Fission Rate, Power Density and Thermal Neutron Flux in a PWR Fuel Pin

A cylindrical $\text{UO}_2$ fuel pin in a commercial Pressurized Water Reactor has fuel pellet radius $R = 0.41\text{ cm}$ and active fuel length $H = 366\text{ cm}$. The linear heat generation rate at the axial core midplane is $q' = 18.5\text{ kW/m} = 185\text{ W/cm}$. The average energy released per fission is $Q = 200\text{ MeV} = 3.204 \times 10^{-11}\text{ J}$. The macroscopic thermal fission cross section of the fuel is $\Sigma_f = 0.355\text{ cm}^{-1}$. (a) Calculate the volumetric power density $q'''$ in $\text{W/cm}^3$ and $\text{MW/m}^3$ within the fuel pellet. (b) Calculate the volumetric fission rate $R_f$ in $\text{fissions/cm}^3\cdot\text{s}$. (c) Determine the required average thermal neutron flux $\phi_{\text{th}}$ inside the fuel pellet in $\text{neutrons/cm}^2\cdot\text{s}$.

(a) Volumetric Power Density $q'''$: The cross-sectional area of the cylindrical fuel pellet is:

$$A_{\text{pellet}} = \pi R^2 = \pi (0.41\text{ cm})^2 \approx 0.5281\text{ cm}^2$$

The volumetric heat generation rate is the linear power divided by cross-sectional area:

$$q''' = \frac{q'}{A_{\text{pellet}}} = \frac{185\text{ W/cm}}{0.5281\text{ cm}^2} \approx \mathbf{350.3\text{ W/cm}^3}$$

In standard engineering units:

$$q''' = 350.3 \times 10^6\text{ W/m}^3 = \mathbf{350.3\text{ MW/m}^3}$$

(b) Volumetric Fission Rate $R_f$: Since power density is $q''' = Q \cdot R_f$:

$$R_f = \frac{q'''}{Q} = \frac{350.3\text{ J/s}\cdot\text{cm}^3}{3.20436 \times 10^{-11}\text{ J/fission}} \approx \mathbf{1.0932 \times 10^{13}\text{ fissions/cm}^3\cdot\text{s}}$$

(c) Required Average Thermal Neutron Flux $\phi_{\text{th}}$: Using $R_f = \Sigma_f \cdot \phi_{\text{th}}$:

$$\phi_{\text{th}} = \frac{R_f}{\Sigma_f} = \frac{1.0932 \times 10^{13}\text{ fissions/cm}^3\cdot\text{s}}{0.355\text{ cm}^{-1}} \approx \mathbf{3.079 \times 10^{13}\text{ neutrons/cm}^2\cdot\text{s}}$$

The required local thermal neutron flux inside the fuel pellet is $3.08 \times 10^{13}\text{ neutrons/cm}^2\cdot\text{s}$.