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Chapter 4 β€’ Theory & Derivations

Unit 4: Nuclear Reaction Dynamics, Kinematics & Reaction Cross-Sections

Rigorous theoretical and mathematical treatise on nuclear reactions: conservation laws in nuclear transformations, laboratory versus center-of-mass kinematics, Coulomb potential barrier penetration via Gamow tunneling, exact derivation of reaction Q-values and endoergic threshold energies, the Bohr compound nucleus hypothesis versus direct reaction mechanisms, and cross-section resonance mechanics via the Breit-Wigner formula.

Β§4.1 Fundamentals of Nuclear Reactions: Conservation Laws & Reaction Channels

A nuclear reaction occurs when two subatomic entitiesβ€”typically an incident projectile particle $a$ and a target nucleus $X$β€”collide at close range, resulting in a rearrangement of nucleons to produce one or more product nuclei $Y$ and outgoing ejectile particles $b$:

$$a + X \longrightarrow Y + b \quad \text{or in Bethe shorthand} \quad X(a, b)Y$$

If multiple particles emerge, the reaction is written $X(a, b_1 b_2 \dots)Y$.

Primary Reaction Types

1. Elastic Scattering ($a + X \to X + a$, denoted $X(a, a)X$):

The projectile and target are identical in species and internal quantum state before and after the collision. Kinetic energy is strictly conserved in the center-of-mass frame ($Q = 0$).

2. Inelastic Scattering ($a + X \to X^* + a$, denoted $X(a, a')X^*$):

A portion of the incident kinetic energy is transferred into internal excitation energy of the target nucleus $X^*$, which subsequently de-excites via gamma emission ($Q < 0$).

3. Radiative Capture ($a + X \to Y^* \to Y + \gamma$, denoted $X(a, \gamma)Y$):

The projectile is absorbed into the target nucleus, forming an excited compound state that de-excites solely by gamma photon emission (e.g., $^{115}\text{In}(n, \gamma)^{116}\text{In}$, $^{238}\text{U}(n, \gamma)^{239}\text{U}$).

4. Transfer Reactions:

One or a few nucleons are transferred between projectile and target during a fast peripheral transit:

  • Stripping: Projectile loses nucleons to target (e.g., $(d, p)$, $(d, n)$, $(^3\text{He}, d)$).
  • Pickup: Projectile captures nucleons from target (e.g., $(p, d)$, $(n, d)$, $(\alpha, ^6\text{Li})$).

5. Knockout Reactions:

Incident projectile collides directly with a single constituent nucleon, ejecting it promptly (e.g., $(p, 2p)$, $(p, pn)$).

6. Spallation Reactions:

Ultra-relativistic projectiles ($E > 100\text{ MeV}$) shatter a heavy nucleus into dozens of nucleons and nuclear fragments.

7. Nuclear Fission and Fusion:

Fission splits a heavy nucleus into two intermediate-mass fragments plus neutrons; fusion coalesces light nuclei into a heavier product.

Exact Conservation Laws in Nuclear Reactions

Every nuclear reaction is constrained by fundamental physical conservation laws:

| Conserved Quantity | Mathematical Formalism | Symmetries & Invariances | | :--- | :--- | :--- | | Total Mass-Energy | $E_{\text{total}} = \sum (T_i + m_i c^2) = \text{constant}$ | Invariance under time translation | | Linear Momentum | $\vec{P}_{\text{total}} = \sum \vec{p}_i = \text{constant}$ | Invariance under spatial translation | | Total Angular Momentum | $\vec{J}_{\text{total}} = \sum (\vec{L}_i + \vec{I}_i) = \text{constant}$ | Invariance under spatial rotation | | Electric Charge | $\sum Z_i = \text{constant}$ | $U(1)$ gauge invariance of electrodynamics | | Total Baryon Number | $\sum A_i = \text{constant}$ | Global baryon number conservation | | Parity ($\pi$) | $\pi_{\text{initial}} = \prod \pi_i (-1)^{l_{\text{in}}} = \pi_{\text{final}}$ | Strong & electromagnetic spatial inversion invariance | | Isospin ($T, T_3$) | $\sum T_{3,i} = \text{constant}$; $T$ conserved in strong force | Charge symmetry and charge independence |

Violations of parity or isospin indicate weak force participation (e.g., beta decays or weak nuclear transitions).

Β§4.2 Laboratory Versus Center-of-Mass Kinematics: Coordinate Transformations & Energy Partition

In experimental nuclear physics, measurements are performed in the Laboratory (LAB) frame, where the target nucleus $X$ of mass $M_X$ is initially at rest ($v_X = 0$) and the projectile $a$ of mass $m_a$ strikes it with incident laboratory kinetic energy $T_a^{\text{lab}}$ and momentum $\vec{p}_a = m_a \vec{v}_a$.

Theoretical analyses, however, are vastly simplified in the Center-of-Mass (CM) frame, where the total linear momentum is identically zero:

$$\vec{P}_{\text{CM}} = \vec{p}_a^{\text{CM}} + \vec{p}_X^{\text{CM}} = 0$$

Center-of-Mass Velocity

The velocity of the center of mass in the laboratory frame is:

$$\vec{V}_{\text{CM}} = \frac{m_a \vec{v}_a + M_X \vec{v}_X}{m_a + M_X} = \frac{m_a}{m_a + M_X} \vec{v}_a$$

``` LAB FRAME: CENTER-OF-MASS (CM) FRAME: m_a (v_a) M_X (at rest) m_a (v_a - V_cm) M_X (-V_cm) ───► O ───► ◄─── \ / \ Zero / Total Momentum! ```

Partition of Kinetic Energy

The total laboratory kinetic energy is:

$$T_{\text{total}}^{\text{lab}} = \frac{1}{2} m_a v_a^2 = T_a^{\text{lab}}$$

In the CM frame, the velocities of the particles are:

$$u_a = v_a - V_{\text{CM}} = v_a \left(1 - \frac{m_a}{m_a + M_X}\right) = \frac{M_X}{m_a + M_X} v_a$$
$$u_X = 0 - V_{\text{CM}} = -\frac{m_a}{m_a + M_X} v_a$$

The total kinetic energy available in the Center-of-Mass frame ($T_{\text{total}}^{\text{CM}}$) is:

$$T_{\text{total}}^{\text{CM}} = \frac{1}{2} m_a u_a^2 + \frac{1}{2} M_X u_X^2 = \frac{1}{2} m_a \left(\frac{M_X}{m_a + M_X}\right)^2 v_a^2 + \frac{1}{2} M_X \left(\frac{m_a}{m_a + M_X}\right)^2 v_a^2$$

Factoring:

$$T_{\text{total}}^{\text{CM}} = \frac{1}{2} \left[ \frac{m_a M_X^2 + M_X m_a^2}{(m_a + M_X)^2} \right] v_a^2 = \frac{1}{2} \left[ \frac{m_a M_X (M_X + m_a)}{(m_a + M_X)^2} \right] v_a^2 = \frac{1}{2} \left(\frac{m_a M_X}{m_a + M_X}\right) v_a^2$$

Defining the reduced mass $\mu$:

$$\mu \equiv \frac{m_a M_X}{m_a + M_X}$$

We obtain the fundamental relationship:

$$T_{\text{total}}^{\text{CM}} = \frac{1}{2} \mu v_a^2 = \left(\frac{M_X}{m_a + M_X}\right) T_a^{\text{lab}}$$

The remaining portion of laboratory kinetic energy is tied up in the irreducible forward motion of the center of mass:

$$T_{\text{motion}}^{\text{CM}} = \frac{1}{2} (m_a + M_X) V_{\text{CM}}^2 = \left(\frac{m_a}{m_a + M_X}\right) T_a^{\text{lab}}$$

Only the CM kinetic energy $T_{\text{total}}^{\text{CM}}$ is available to induce internal nuclear excitations, overcome reaction barriers, or satisfy endoergic reaction deficits.

Β§4.3 The Nuclear Coulomb Potential Barrier & Gamow Quantum Tunneling Factor

When a positively charged projectile (such as a proton, deuteron, or alpha particle with charge $z e$) approaches a target nucleus with charge $Z e$, it experiences long-range electrostatic Coulomb repulsion.

Coulomb Potential Barrier Height ($V_C$)

The repulsive Coulomb potential energy at center-to-center separation $r$ is:

$$V(r) = \frac{1}{4\pi\varepsilon_0} \frac{z Z e^2}{r}$$

As the projectile approaches the target, the potential rises monotonically until the surfaces of the two nuclei touch at the contact radius $R_c = R_a + R_X = R_0 (a^{1/3} + X^{1/3})$. At this radius, the attractive short-range strong nuclear force takes over, forming an attractive potential well.

The maximum electrostatic potential energy at contact defines the Coulomb Barrier Height $V_C$:

$$V_C = \frac{1}{4\pi\varepsilon_0} \frac{z Z e^2}{R_a + R_X} = \frac{1.43996\text{ MeV}\cdot\text{fm} \cdot z Z}{R_0 (a^{1/3} + X^{1/3})}$$

``` Potential Energy V(r) β–² V_C β”‚ /\ Coulomb Barrier Peak β”‚ / \ β”‚ / \____ V(r) = (1/4πΡ₀) * (zZeΒ²/r) β”‚ / 0 ┼──────/────────────────────────► Separation r β”‚ β”‚ R_c (Contact Radius) -Vβ‚€ β”‚_____β”‚ Attractive Strong Well ```

For an alpha particle ($z = 2, a = 4$) striking an uranium-238 nucleus ($Z = 92, X = 238$) with $R_0 = 1.25\text{ fm}$:

$$R_c = 1.25(4^{1/3} + 238^{1/3}) = 1.25(1.587 + 6.197) = 1.25(7.784) \approx 9.73\text{ fm}$$
$$V_C = \frac{(1.44\text{ MeV}\cdot\text{fm})(2)(92)}{9.73\text{ fm}} = \frac{264.96}{9.73} \approx 27.2\text{ MeV}$$

Quantum Mechanical Gamow Tunneling Factor

Under classical mechanics, a projectile with kinetic energy $T < V_C$ is strictly forbidden from reaching the nucleus and undergoing a nuclear reaction. In 1928, George Gamow (and independently Condon and Gurney) demonstrated that a quantum wavepacket has a non-zero probability of tunneling through the classically forbidden barrier.

Using the WKB (Wentzel-Kramers-Brillouin) approximation, the transmission probability $P$ through a barrier $V(r)$ from classical turning point $r_0$ to contact radius $R_c$ is:

$$P \approx \exp\left( -2 \int_{R_c}^{r_0} k(r) dr \right) = \exp\left( -\frac{2}{\hbar} \int_{R_c}^{r_0} \sqrt{2\mu [V(r) - E]} \, dr \right)$$

where $E$ is the center-of-mass energy and $r_0$ satisfies $V(r_0) = E \implies r_0 = \frac{z Z e^2}{4\pi\varepsilon_0 E}$.

Evaluating the integral analytically for $R_c \ll r_0$:

$$P \approx \exp(-2\pi \eta)$$

where $\eta$ is the dimensionless Sommerfeld parameter:

$$\eta = \frac{z Z e^2}{4\pi\varepsilon_0 \hbar v} = \frac{z Z \alpha}{v / c} = \sqrt{\frac{\mu c^2}{2E}} z Z \alpha$$

where $\alpha = \frac{e^2}{4\pi\varepsilon_0 \hbar c} \approx \frac{1}{137.036}$ is the fine-structure constant.

The factor $\exp(-2\pi \eta)$ is the famous Gamow factor. It explains:

  1. Why low-energy alpha particles ($4 - 8\text{ MeV}$) can tunnel out of heavy nuclei despite the $25 - 30\text{ MeV}$ barrier (Geiger-Nuttall law).
  2. Why thermonuclear fusion in stellar interiors (e.g., the Sun's core at $T \approx 1.5 \times 10^7\text{ K}$, where thermal kinetic energies are a mere $\sim 1.3\text{ keV}$) occurs at measurable rates through the Gamow window.

Β§4.4 The Q-Value Derivation, Exoergic Dynamics & Threshold Energy of Endoergic Reactions

The energetics of any nuclear reaction $X(a, b)Y$ are governed by its reaction energy, designated as the $Q$-value.

Derivation of the $Q$-Value from Mass-Energy Equivalence

By relativistic conservation of total mass-energy:

$$E_{\text{reactants}} = E_{\text{products}}$$
$$(T_a + m_a c^2) + (T_X + M_X c^2) = (T_b + m_b c^2) + (T_Y + M_Y c^2)$$

Rearranging terms:

$$(T_b + T_Y) - (T_a + T_X) = (m_a + M_X)c^2 - (m_b + M_Y)c^2$$

The $Q$-value is formally defined as the net change in kinetic energy, which equals the net difference in rest masses:

$$Q \equiv (T_b + T_Y) - (T_a + T_X) = \left[ (m_a + M_X) - (m_b + M_Y) \right] c^2$$

Because binding energy $B$ is defined as $B = [Z m_p + N m_n - M] c^2$, the nucleon numbers $Z$ and $N$ cancel identically, yielding:

$$Q = \left[ B(Y) + B(b) \right] - \left[ B(X) + B(a) \right] = \sum B_{\text{products}} - \sum B_{\text{reactants}}$$

Energetic Classifications

1. Exoergic (Exothermic) Reactions ($Q > 0$):

Rest mass is converted into kinetic energy ($\Delta m > 0$). The reaction can occur at arbitrarily low projectile kinetic energies (even at thermal energies $T_a \approx 0$ for uncharged neutrons).

2. Endoergic (Endothermic) Reactions ($Q < 0$):

Kinetic energy is converted into rest mass ($\Delta m < 0$). The reaction is energetically impossible unless the incident projectile supplies sufficient kinetic energy to cover the mass deficit plus the unavoidable center-of-mass recoil motion.

Exact Derivation of Threshold Energy ($E_{\text{th}}$)

For an endoergic reaction ($Q < 0$) in the laboratory frame with stationary target ($T_X = 0$), what is the minimum kinetic energy $T_a^{\text{lab}} = E_{\text{th}}$ required for the reaction to occur?

At the absolute threshold, all reaction products $Y$ and $b$ emerge with zero relative velocity in the center-of-mass frame; they move forward as a single consolidated mass $(M_Y + m_b)$ at the velocity of the center of mass $V_{\text{CM}}$. By conservation of linear momentum:

$$m_a v_{\text{th}} = (M_Y + m_b) V_{\text{CM}} \approx (m_a + M_X) V_{\text{CM}} \implies V_{\text{CM}} = \frac{m_a}{m_a + M_X} v_{\text{th}}$$

The total laboratory kinetic energy of the products at threshold is:

$$T_{\text{products}}^{\text{lab}} = \frac{1}{2} (M_Y + m_b) V_{\text{CM}}^2 \approx \frac{1}{2} (m_a + M_X) \left(\frac{m_a}{m_a + M_X} v_{\text{th}}\right)^2 = \left(\frac{m_a}{m_a + M_X}\right) \left(\frac{1}{2} m_a v_{\text{th}}^2\right) = \left(\frac{m_a}{m_a + M_X}\right) E_{\text{th}}$$

From the definition of $Q$:

$$Q = T_{\text{products}}^{\text{lab}} - T_{\text{reactants}}^{\text{lab}} = \left(\frac{m_a}{m_a + M_X}\right) E_{\text{th}} - E_{\text{th}} = -E_{\text{th}} \left( 1 - \frac{m_a}{m_a + M_X} \right) = -E_{\text{th}} \left(\frac{M_X}{m_a + M_X}\right)$$

Solving for $E_{\text{th}}$ (with $Q < 0$):

$$E_{\text{th}} = -Q \left(\frac{m_a + M_X}{M_X}\right) = |Q| \left(1 + \frac{m_a}{M_X}\right)$$

This elegant formula shows that the threshold energy in the lab is always strictly greater than $|Q|$. The factor $(m_a / M_X) |Q|$ represents the inescapable kinetic energy locked in the forward recoil of the center of mass.

Β§4.5 Reaction Mechanisms: Bohr Compound Nucleus Hypothesis Versus Direct Nuclear Reactions

In 1936, Niels Bohr proposed the Compound Nucleus Hypothesis to explain the narrow, intense resonance peaks observed in low-energy neutron capture reactions.

The Two-Stage Compound Nucleus Model

Bohr posited that a nuclear reaction proceeds in two completely independent, decoupled stages:

1. Formation Stage: The incident projectile $a$ is absorbed by target $X$, merging into an intermediate excited compound state $C^*$:

$$a + X \longrightarrow C^*$$

The incident kinetic energy and binding energy are rapidly distributed among all nucleons through hundreds of stochastic nucleon-nucleon collisions. The compound nucleus reaches thermodynamic quasi-equilibrium within $\sim 10^{-16}\text{ to }10^{-18}\text{ seconds}$β€”an eternity compared to the nuclear transit time $\tau_{\text{transit}} \approx 2R/v \sim 10^{-22}\text{ seconds}$.

2. Decay (De-excitation) Stage: The compound nucleus "forgets" the specific manner in which it was formed (the Bohr Independence Hypothesis). It de-excites purely statistically through whatever open decay channel stochastic fluctuations concentrate enough energy into:

$$C^* \longrightarrow \begin{cases} X + a & \text{(Elastic / Shape Resonant Scattering)} \\ X^* + a' & \text{(Inelastic Scattering)} \\ Y_1 + b_1 & \text{(Particle Emission, e.g., } (n, p), (n, \alpha)) \\ C + \gamma & \text{(Radiative Capture)} \\ F_1 + F_2 & \text{(Nuclear Fission)} \end{cases}$$

``` FORMATION (Fast: ~10⁻²² s) EQUILIBRATION (~10⁻¹⁢ s) DECAY (Statistical) a + X ────────────────────────► [ Compound Nucleus C* ] ───► Y₁ + b₁ (Energy shared among all) ───► Yβ‚‚ + bβ‚‚ ───► C + Ξ³ ```

Mathematically, the cross-section for reaction channel $a \to b$ factorizes:

$$\sigma(a, b) = \sigma_{\text{formation}}(C^*) \cdot P_{\text{decay}}(b)$$

where $P_{\text{decay}}(b) = \Gamma_b / \Gamma_{\text{total}}$ is the branching fraction for decay into mode $b$.

Experimental proof was provided by Ghoshal (1950), who produced the identical compound nucleus $^{64}\text{Zn}^*$ via two completely different entrances:

$$p + ^{63}_{29}\text{Cu} \longrightarrow [^{64}_{30}\text{Zn}^*] \quad \text{and} \quad \alpha + ^{60}_{28}\text{Ni} \longrightarrow [^{64}_{30}\text{Zn}^*]$$

At identical excitation energies, the cross-section ratios for de-excitation via $(n)$, $(2n)$, and $(pn)$ channels were experimentally identical, proving the independence hypothesis.

Direct Nuclear Reactions

In contrast, when the projectile energy is high ($E > 20\text{ MeV}$) or the impact parameter corresponds to the nuclear surface, the reaction bypasses compound nucleus formation:

  • Timescale: Ultra-fast, $\tau \sim 10^{-22}\text{ s}$ (single transit).
  • Mechanism: Projectile interacts directly with a single valence nucleon without exciting the bulk core.
  • Angular Distribution: Strongly forward-peaked ($\theta \approx 0^\circ$).
  • Excitation Functions: Smooth and monotonic with energy, showing no sharp resonances.

Β§4.6 Reaction Cross-Section Formalism: The Barn, Differential Cross-Sections & Excitation Functions

The nuclear reaction cross-section $\sigma$ is an effective target area quantifying the intrinsic probability that a given nuclear reaction will take place between a projectile and a target nucleus.

Physical Formulation of Cross-Section

Consider a uniform beam of incident particles with flux $\Phi$ (particles per unit area per unit time, $\text{cm}^{-2}\cdot\text{s}^{-1}$) striking a thin target foil of area $A$, thickness $dx$, and atomic number density $n$ ($\text{nuclei/cm}^3$). The total number of target nuclei exposed to the beam is $N_T = n \cdot A \cdot dx$.

The reaction rate $R$ (number of nuclear events occurring per second) is experimentally found to be directly proportional to the incident flux and the total number of exposed target nuclei:

$$R = \sigma \cdot \Phi \cdot N_T = \sigma \cdot \Phi \cdot (n \cdot A \cdot dx)$$

Solving for $\sigma$:

$$\sigma = \frac{R}{\Phi \cdot N_T} = \frac{\text{Events / second}}{(\text{Incident particles / cm}^2\cdot\text{s}) \cdot (\text{Target nuclei})}$$

Cross-section has physical dimensions of area ($[\text{L}]^2$).

The Unit of Cross-Section: The Barn

Because typical nuclear radii are $R \sim 5\text{ fm} = 5 \times 10^{-13}\text{ cm}$, the geometric cross-sectional area of a nucleus is:

$$\sigma_{\text{geom}} \approx \pi R^2 \approx \pi (5 \times 10^{-13}\text{ cm})^2 \approx 8 \times 10^{-25}\text{ cm}^2$$

During the Manhattan Project in 1942, physicists Purdue and Oppenheimer coined the term barn ($\text{b}$) to indicate an area "as big as a barn" for nuclear collisions:

$$1\text{ barn (b)} \equiv 10^{-24}\text{ cm}^2 = 10^{-28}\text{ m}^2 = 100\text{ fm}^2$$

Subunits:

  • Millibarn ($\text{mb} = 10^{-3}\text{ b} = 10^{-27}\text{ cm}^2$)
  • Microbarn ($\mu\text{b} = 10^{-6}\text{ b} = 10^{-30}\text{ cm}^2$)
  • Kilobarn ($\text{kb} = 10^3\text{ b} = 10^{-21}\text{ cm}^2$)

Differential Cross-Section ($d\sigma/d\Omega$)

To describe the angular distribution of reaction products, we define the differential cross-section:

$$\frac{d\sigma}{d\Omega}(\theta, \phi) = \frac{1}{\Phi \cdot N_T} \frac{dR(\theta, \phi)}{d\Omega}$$

where $d\Omega = \sin\theta d\theta d\phi$ is the element of solid angle (steradians, $\text{sr}$). The total cross-section is obtained by integrating over all solid angles:

$$\sigma_{\text{total}} = \int_{4\pi} \left(\frac{d\sigma}{d\Omega}\right) d\Omega = \int_0^{2\pi} d\phi \int_0^\pi \left(\frac{d\sigma}{d\Omega}\right) \sin\theta d\theta$$

Macroscopic Cross-Section ($\Sigma$) and Mean Free Path ($\lambda_{\text{mfp}}$)

In reactor physics and radiation shielding, the probability of interaction per unit path length in a bulk material is described by the macroscopic cross-section $\Sigma$:

$$\Sigma = n \cdot \sigma = \frac{\rho N_A}{M} \sigma \quad (\text{dimensions: } \text{cm}^{-1})$$

A beam of intensity $I(0)$ traversing distance $x$ through the medium is attenuated according to:

$$I(x) = I(0) e^{-\Sigma x}$$

The nuclear mean free path $\lambda_{\text{mfp}}$ between collisions is:

$$\lambda_{\text{mfp}} = \frac{1}{\Sigma} = \frac{1}{n \sigma}$$

Β§4.7 The Breit-Wigner Single-Level Resonance Formula & Neutron Absorption Cross-Sections

When the incident kinetic energy of a projectile corresponds precisely to a discrete quasi-stationary quantum energy state of the compound nucleus ($E_{\text{inc}} \approx E_0$), the cross-section exhibits a sharp, dramatic spike known as a nuclear resonance.

Quantum Derivation of the Breit-Wigner Single-Level Formula

In 1936, Gregory Breit and Eugene Wigner formulated the dispersion theory for nuclear resonances, analogizing the process to an optical oscillator or damped harmonic wave. The wave function of an unstable compound state decaying with lifetime $\tau$ has the time dependence:

$$\psi(t) = \psi(0) \exp\left( -i \frac{E_0}{\hbar} t - \frac{t}{2\tau} \right) = \psi(0) \exp\left( -i \frac{E_0 - i\Gamma/2}{\hbar} t \right)$$

where the total resonance energy width $\Gamma$ is related to mean lifetime by the Heisenberg uncertainty principle:

$$\Gamma = \frac{\hbar}{\tau} = \sum_i \Gamma_i = \Gamma_n + \Gamma_\gamma + \Gamma_\alpha + \dots$$

Here, each $\Gamma_i$ is the partial width corresponding to the probability of decay into channel $i$.

Taking the Fourier transform of $\psi(t)$ into the energy domain yields the probability amplitude:

$$f(E) \propto \frac{1}{E - E_0 + i\Gamma/2}$$

The cross-section for reaction channel $a \to b$ is proportional to $|f(E)|^2$:

$$\sigma(a, b) = \pi \lambdabar^2 g_J \frac{\Gamma_a \Gamma_b}{(E - E_0)^2 + (\Gamma/2)^2}$$

where:

  • $\lambdabar = \frac{\lambda}{2\pi} = \frac{\hbar}{p}$ is the reduced de Broglie wavelength of the incident particle.
  • $g_J$ is the statistical spin factor accounting for angular momentum coupling between projectile spin $s_a$ and target nuclear spin $I_X$ to form compound spin $J$:
$$g_J = \frac{2J + 1}{(2s_a + 1)(2I_X + 1)}$$

``` Cross Section Οƒ(E) β–² Οƒβ‚€ β”‚ /\ Resonance Peak Eβ‚€ β”‚ / \ β”‚ / \ Οƒβ‚€/2β”‚----------/------\---------- FWHM = Ξ“ (Total Width) β”‚ / \ β”‚_______/ \_______ └───────┴─────┴──────┴───────► Projectile Energy E Eβ‚€-Ξ“/2 Eβ‚€+Ξ“/2 ```

The $1/v$ Law for Low-Energy Neutron Capture

For slow (thermal) s-wave neutrons ($l = 0$), the incident energy is far below the first resonance ($E \ll E_0$). The de Broglie wavelength squared scales as:

$$\lambdabar^2 = \frac{\hbar^2}{p^2} = \frac{\hbar^2}{2 m_n E} \propto \frac{1}{E} \propto \frac{1}{v^2}$$

For s-wave neutrons, the neutron emission partial width $\Gamma_n$ is proportional to the outgoing neutron velocity (density of final states):

$$\Gamma_n \propto v$$

While the radiative capture width $\Gamma_\gamma$ is constant (independent of neutron speed). Substituting these into the Breit-Wigner formula with $(E - E_0)^2 \approx E_0^2$:

$$\sigma(n, \gamma) \propto \lambdabar^2 \Gamma_n \Gamma_\gamma \propto \left(\frac{1}{v^2}\right) (v) (\text{constant}) \propto \frac{1}{v}$$

This yields the fundamental $1/v$ Law of Slow Neutron Capture:

$$\sigma(v) = \sigma_0 \left(\frac{v_0}{v}\right) = \sigma_0 \sqrt{\frac{E_0}{E}}$$

where standard thermal neutron cross-sections are tabulated at reference velocity $v_0 = 2200\text{ m/s}$ ($E_0 = 0.0253\text{ eV}$ at $T = 293.6\text{ K}$). This explains why thermal neutrons have enormous capture cross sections (thousands of barns) compared to fast neutrons (a few barns).

Β§4.8 Direct Reaction Mechanisms: Optical Model, DWBA & Transfer Spectroscopy

At incident projectile energies exceeding $10 - 20\text{ MeV}$, nuclear reactions bypass compound nucleus formation, proceeding via direct reactions where the projectile interacts with only one or two valence nucleons during a single transit time ($\tau \sim 10^{-22}\text{ s}$).

The Optical Model of Elastic Nuclear Scattering

To describe direct elastic scattering and total reaction cross-sections, Herman Feshbach, Charles Porter, and Victor Weisskopf (1954) introduced the Optical Model. The nucleus is treated as a partially transparent, refractive, and absorbing cloudy crystal sphere for projectile matter waves, modeled by a complex phenomenological potential:

$$U(r) = V(r) + i W(r)$$
  • Real Part $V(r)$: Refracts the incoming matter wave, describing elastic scattering. Modeled using a Woods-Saxon volume potential:
$$V(r) = -V_0 f(r, R_v, a_v) \quad \text{where } f(r, R, a) = \frac{1}{1 + \exp\left(\frac{r - R}{a}\right)}$$
  • Imaginary Part $W(r)$: Absorbs flux from the elastic channel, describing all non-elastic reaction processes (inelastic scattering, capture, transfer, fission):
$$W(r) = -W_v f(r, R_w, a_w) + 4 a_s W_s \frac{d}{dr}f(r, R_s, a_s)$$
  • Spin-Orbit Term $V_{so}(r)$: Describes polarization and spin-flip scattering:
$$V_{so}(r) = V_{so} \left(\frac{\hbar}{m_\pi c}\right)^2 \frac{1}{r} \frac{df}{dr} (\vec{L} \cdot \vec{S})$$

The Distorted Wave Born Approximation (DWBA) for Single-Nucleon Transfer

In a transfer reaction such as $(d, p)$ stripping:

$$d + X \longrightarrow p + Y \quad \text{where } Y = X + n$$

The deuteron breaks up; the neutron is captured into an unoccupied single-particle shell-model orbital with orbital angular momentum $l$ and total angular momentum $j$, while the proton escapes.

In the Distorted Wave Born Approximation (DWBA), the transition matrix element is:

$$T_{fi} = \int d^3r_i \int d^3r_f \, \chi_f^{(-)*}(\vec{k}_f, \vec{r}_f) \langle \psi_Y \psi_p | V_{\text{trans}} | \psi_X \psi_d \rangle \chi_i^{(+)}(\vec{k}_i, \vec{r}_i)$$

where $\chi_i$ and $\chi_f$ are distorted waves calculated using Optical Model potentials. The experimental differential cross-section factorizes into:

$$\left(\frac{d\sigma}{d\Omega}\right)_{\text{exp}} = S_{l j} \cdot \left(\frac{d\sigma}{d\Omega}\right)_{\text{DWBA}}$$

where $S_{l j}$ is the Spectroscopic Factor, quantifying the purity of the single-particle configuration in the residual nuclear state. DWBA transfer reactions serve as the primary experimental tool for mapping single-particle energy levels across the chart of nuclides.

Giant Resonance Parameters of Slow-Neutron Absorber Isotopes

| Nuclide | Natural Abundance | Target Spin $I^\pi$ | Thermal Cross-Section $\sigma_{\text{th}}$ | Resonance Energy $E_0$ | Peak Cross-Section $\sigma_0$ | Primary Application | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | $^{10}\text{B}$ | $19.9\%$ | $3^+$ | $3,840\text{ b}$ | $1/v$ up to $100\text{ keV}$ | β€” | Reactor control rods, neutron shielding | | $^{113}\text{Cd}$ | $12.22\%$ | $1/2^+$ | $20,600\text{ b}$ | $0.178\text{ eV}$ | $60,000\text{ b}$ | Thermal neutron cut-off filter ($E_{\text{Cd}} = 0.55\text{ eV}$) | | $^{115}\text{In}$ | $95.71\%$ | $9/2^+$ | $202\text{ b}$ | $1.457\text{ eV}$ | $35,100\text{ b}$ | Neutron flux activation foil | | $^{135}\text{Xe}$ | $0.0\%$ (Fission) | $3/2^+$ | $2,650,000\text{ b}$ | $0.084\text{ eV}$ | $3.5 \times 10^6\text{ b}$ | Strongest known reactor poison | | $^{149}\text{Sm}$ | $13.82\%$ | $7/2^-$ | $41,000\text{ b}$ | $0.097\text{ eV}$ | $120,000\text{ b}$ | Permanent non-decaying reactor poison | | $^{157}\text{Gd}$ | $15.65\%$ | $3/2^-$ | $254,000\text{ b}$ | $0.031\text{ eV}$ | $2.6 \times 10^5\text{ b}$ | Burnable poison in nuclear fuel assemblies |

Intermediate Example 4.1: Exact Q-Value and Threshold Energy for C-12(alpha, n)O-15

Consider the endoergic nuclear reaction:

$$^{12}_{6}\text{C} + ^{4}_{2}\alpha \longrightarrow ^{15}_{8}\text{O} + ^{1}_{0}n$$

Given the atomic masses:

  • $M(^{12}\text{C}) = 12.000000\text{ u}$
  • $m(\alpha) = 4.001506\text{ u}$
  • $M(^{15}\text{O}) = 15.003065\text{ u}$
  • $m_n = 1.008665\text{ u}$
  • $1\text{ u} = 931.4941\text{ MeV}/c^2$
  1. Calculate the reaction $Q$-value in $\text{MeV}$.
  2. Determine the minimum threshold kinetic energy $E_{\text{th}}$ in the laboratory frame required for incident alpha particles striking a stationary $^{12}\text{C}$ target.

Step 1: Calculate $Q$-Value

Sum of reactant masses:

$$m_{\text{reactants}} = M(^{12}\text{C}) + m(\alpha) = 12.000000 + 4.001506 = 16.001506\text{ u}$$

Sum of product masses:

$$m_{\text{products}} = M(^{15}\text{O}) + m_n = 15.003065 + 1.008665 = 16.011730\text{ u}$$

Mass difference $\Delta m$:

$$\Delta m = m_{\text{reactants}} - m_{\text{products}} = 16.001506 - 16.011730 = -0.010224\text{ u}$$

$Q$-value:

$$Q = -0.010224\text{ u} \times 931.4941\text{ MeV/u} \approx -9.5236\text{ MeV}$$

Because $Q < 0$, the reaction is endoergic.

Step 2: Threshold Energy $E_{\text{th}}$

Laboratory threshold formula:

$$E_{\text{th}} = |Q| \left(1 + \frac{m_\alpha}{M_C}\right)$$

Using masses $m_\alpha \approx 4.0015\text{ u}$ and $M_C = 12.0000\text{ u}$:

$$\frac{m_\alpha}{M_C} = \frac{4.001506}{12.000000} \approx 0.333459$$
$$E_{\text{th}} = 9.5236\text{ MeV} \times (1 + 0.333459) = 9.5236 \times 1.333459 \approx 12.6995\text{ MeV}$$

The incident alpha particle must have a laboratory kinetic energy of at least $12.70\text{ MeV}$ to initiate the reaction.

Easy Example 4.2: Coulomb Barrier and Gamow Tunneling for D-T Fusion

In deuterium-tritium thermonuclear fusion:

$$^2_1\text{H} + ^3_1\text{H} \longrightarrow ^4_2\text{He} + ^1_0n + 17.59\text{ MeV}$$

Given the nuclear radius constant $R_0 = 1.30\text{ fm}$:

  1. Calculate the contact radius $R_c = R_D + R_T$ in $\text{fm}$.
  2. Determine the classical Coulomb barrier height $V_C$ in $\text{keV}$.
  3. Explain why magnetically confined fusion reactors operate at plasma temperatures of $T \approx 15\text{ keV}$ ($150\text{ million K}$), far below $V_C$.

Step 1: Contact Radius $R_c$

$$R_D = R_0 (2)^{1/3} = 1.30 \times 1.2599 \approx 1.638\text{ fm}$$
$$R_T = R_0 (3)^{1/3} = 1.30 \times 1.4422 \approx 1.875\text{ fm}$$
$$R_c = R_D + R_T = 1.638 + 1.875 = 3.513\text{ fm}$$

Step 2: Classical Coulomb Barrier $V_C$

Both deuteron and triton have $z_1 = 1, z_2 = 1$:

$$V_C = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{R_c} = \frac{1.43996\text{ MeV}\cdot\text{fm}}{3.513\text{ fm}} \approx 0.4099\text{ MeV} = 410\text{ keV}$$

Step 3: Physical Explanation of Plasma Temperature

Although the classical electrostatic barrier is $410\text{ keV}$:

1. Gamow Quantum Tunneling: Nuclei do not need to scale the crest of the barrier; quantum tunneling enables penetration with substantial probability at energies well below $V_C$.

2. Maxwell-Boltzmann High-Energy Tail: In a thermal plasma at $15\text{ keV}$, the Maxwellian distribution has an exponential tail where particles with $E \sim 4 - 5 k_B T \approx 60 - 80\text{ keV}$ exist in significant numbers.

  1. The convolution of the rising tunneling probability $\exp(-2\pi \eta)$ with the falling Maxwellian distribution $\exp(-E/k_B T)$ forms the Gamow peak centered at $\approx 65\text{ keV}$, yielding immense reaction rates at an operating temperature of only $15\text{ keV}$.
Intermediate Example 4.3: Breit-Wigner Peak Cross-Section for Indium-115 Resonance

Indium-115 ($^{115}\text{In}$, ground spin $I_X = 9/2^+$) exhibits a famous thermal/epithermal neutron capture resonance at laboratory energy $E_0 = 1.457\text{ eV}$. The resonance parameters are:

  • Total width $\Gamma = 0.089\text{ eV}$
  • Neutron partial width $\Gamma_n = 0.0033\text{ eV}$
  • Radiative capture partial width $\Gamma_\gamma = 0.0857\text{ eV}$
  • Compound nucleus resonance spin $J = 5$
  • Neutron spin $s_a = 1/2$
  1. Calculate the statistical spin factor $g_J$.
  2. Compute the reduced de Broglie wavelength $\lambdabar$ of the neutron at $E_0$.
  3. Calculate the peak radiative capture cross-section $\sigma(n, \gamma)$ at resonance in barns.

Step 1: Statistical Spin Factor $g_J$

$$g_J = \frac{2J + 1}{(2s_a + 1)(2I_X + 1)} = \frac{2(5) + 1}{(2(1/2) + 1)(2(9/2) + 1)} = \frac{11}{(2)(10)} = \frac{11}{20} = 0.550$$

Step 2: Reduced de Broglie Wavelength $\lambdabar$

Neutron kinetic energy $E_0 = 1.457\text{ eV} = 1.457 \times 1.60218 \times 10^{-19}\text{ J} = 2.3344 \times 10^{-19}\text{ J}$. Momentum:

$$p = \sqrt{2 m_n E} = \sqrt{2 (1.67493 \times 10^{-27}\text{ kg})(2.3344 \times 10^{-19}\text{ J})} = \sqrt{7.820 \times 10^{-46}} = 2.7964 \times 10^{-23}\text{ kg}\cdot\text{m/s}$$
$$\lambdabar = \frac{\hbar}{p} = \frac{1.05457 \times 10^{-34}\text{ J}\cdot\text{s}}{2.7964 \times 10^{-23}\text{ kg}\cdot\text{m/s}} \approx 3.7712 \times 10^{-12}\text{ m} = 3771.2\text{ fm}$$
$$\pi \lambdabar^2 = \pi (3.7712 \times 10^{-12}\text{ m})^2 = \pi (1.4222 \times 10^{-23}\text{ m}^2) = 4.468 \times 10^{-23}\text{ m}^2$$

In barns ($1\text{ b} = 10^{-28}\text{ m}^2$):

$$\pi \lambdabar^2 = \frac{4.468 \times 10^{-23}\text{ m}^2}{10^{-28}\text{ m}^2/\text{b}} = 446,800\text{ barns}$$

Step 3: Peak Capture Cross-Section

At resonance peak ($E = E_0$), the energy denominator $(E - E_0)^2 + (\Gamma/2)^2$ reduces to $(\Gamma/2)^2 = \Gamma^2 / 4$:

$$\sigma_{\text{peak}} = 4\pi \lambdabar^2 g_J \frac{\Gamma_n \Gamma_\gamma}{\Gamma^2}$$

Substitute values:

$$\frac{\Gamma_n \Gamma_\gamma}{\Gamma^2} = \frac{(0.0033\text{ eV})(0.0857\text{ eV})}{(0.089\text{ eV})^2} = \frac{0.0002828}{0.007921} \approx 0.03570$$
$$\sigma_{\text{peak}} = 4 \times (446,800\text{ b}) \times (0.550) \times (0.03570) = 1,787,200 \times 0.019636 \approx 35,095\text{ barns}$$

The peak cross section is an astounding $\approx 35,100\text{ barns}$ (explaining why indium foils are standard neutron detectors).

Easy Example 4.4: Macroscopic Cross-Section and Neutron Mean Free Path in Natural Uranium

Pure metallic uranium fuel has a mass density $\rho = 19.1\text{ g/cm}^3$ and atomic weight $M = 238.03\text{ g/mol}$. For thermal neutrons ($v = 2200\text{ m/s}$):

  • Total microscopic scattering cross-section $\sigma_s = 8.3\text{ barns}$
  • Total microscopic absorption cross-section $\sigma_a = 7.6\text{ barns}$
  1. Calculate the atomic number density $n$ of uranium in $\text{atoms/cm}^3$.
  2. Calculate the macroscopic scattering cross-section $\Sigma_s$ and absorption cross-section $\Sigma_a$ in $\text{cm}^{-1}$.
  3. Determine the total macroscopic cross-section $\Sigma_t$ and the mean free path $\lambda_{\text{mfp}}$ in centimeters.

Step 1: Number Density $n$

$$n = \frac{\rho N_A}{M} = \frac{(19.1\text{ g/cm}^3)(6.02214 \times 10^{23}\text{ atoms/mol})}{238.03\text{ g/mol}} \approx 4.832 \times 10^{22}\text{ atoms/cm}^3$$

Step 2: Macroscopic Cross-Sections

Convert barns to $\text{cm}^2$: $1\text{ b} = 10^{-24}\text{ cm}^2$.

  1. Scattering:
$$\sigma_s = 8.3 \times 10^{-24}\text{ cm}^2$$
$$\Sigma_s = n \sigma_s = (4.832 \times 10^{22}\text{ cm}^{-3})(8.3 \times 10^{-24}\text{ cm}^2) \approx 0.4011\text{ cm}^{-1}$$
  1. Absorption:
$$\sigma_a = 7.6 \times 10^{-24}\text{ cm}^2$$
$$\Sigma_a = n \sigma_a = (4.832 \times 10^{22}\text{ cm}^{-3})(7.6 \times 10^{-24}\text{ cm}^2) \approx 0.3672\text{ cm}^{-1}$$

Step 3: Total Macroscopic Cross-Section and Mean Free Path

$$\Sigma_t = \Sigma_s + \Sigma_a = 0.4011 + 0.3672 = 0.7683\text{ cm}^{-1}$$

The mean free path between collisions is:

$$\lambda_{\text{mfp}} = \frac{1}{\Sigma_t} = \frac{1}{0.7683\text{ cm}^{-1}} \approx 1.302\text{ cm}$$

A thermal neutron travels on average only $1.30\text{ cm}$ in metallic uranium before undergoing a nuclear interaction.

Intermediate Example 4.5: Thermal 1/v Cross-Section Scaling to Reactor Operating Temperatures

Boron-10 is widely used as a neutron absorber in reactor control rods via the $^{10}\text{B}(n, \alpha)^7\text{Li}$ reaction. At room temperature ($T_0 = 293.6\text{ K}$, thermal energy $E_0 = 0.0253\text{ eV}$), its capture cross-section is $\sigma_0 = 3,840\text{ barns}$. Assuming the cross-section strictly obeys the $1/v$ law:

  1. Formulate the relationship between effective cross-section $\sigma(T)$ and absolute temperature $T$.
  2. Calculate the absorption cross-section of $^{10}\text{B}$ in a pressurized water reactor operating at core temperature $T = 310^\circ\text{C}$ ($583.15\text{ K}$).

Step 1: Temperature Scaling Formulation

For thermal neutrons in Maxwellian equilibrium with a moderator at temperature $T$, the average kinetic energy scales linearly with temperature:

$$\bar{E} = \frac{3}{2} k_B T \implies v_{\text{thermal}} \propto \sqrt{T}$$

According to the $1/v$ law:

$$\sigma(v) \propto \frac{1}{v} \propto \frac{1}{\sqrt{T}}$$

Therefore:

$$\sigma(T) = \sigma_0 \sqrt{\frac{T_0}{T}} = \sigma_0 \sqrt{\frac{E_0}{k_B T}}$$

Step 2: Numerical Calculation at Reactor Operating Temperature

Reference temperature: $T_0 = 293.6\text{ K}$ ($20.45^\circ\text{C}$). Operating temperature: $T = 310 + 273.15 = 583.15\text{ K}$. Temperature ratio:

$$\frac{T_0}{T} = \frac{293.6\text{ K}}{583.15\text{ K}} \approx 0.50347$$

Square root factor:

$$\sqrt{\frac{T_0}{T}} = \sqrt{0.50347} \approx 0.70956$$

Effective absorption cross-section:

$$\sigma(583.15\text{ K}) = 3,840\text{ b} \times 0.70956 \approx 2,725\text{ barns}$$

At $310^\circ\text{C}$, the cross-section decreases by nearly $30\%$ due to thermal spectral hardening.

Advanced Example 4.6: Two-Body Reaction Ejectile Energy as Function of Emission Angle

In the exoergic reaction $^7_3\text{Li}(p, \alpha)^4_2\text{He}$ ($Q = +17.347\text{ MeV}$), incident protons with kinetic energy $T_p = 3.00\text{ MeV}$ strike stationary lithium-7.

  1. Using non-relativistic kinematics, derive the exact formula for ejectile kinetic energy $T_\alpha(\theta)$ as a function of laboratory emission angle $\theta$.
  2. Calculate the kinetic energy of the alpha particle emitted at $\theta = 0^\circ$ (forward) and $\theta = 90^\circ$ (perpendicular).

Step 1: Kinematic Derivation

Let projectile $a$ ($p$, mass $m_a$), target $X$ ($^7\text{Li}$, mass $M_X$), ejectile $b$ ($\alpha$, mass $m_b$), and recoil $Y$ ($^4\text{He}$, mass $M_Y$). By conservation of momentum:

$$\vec{p}_Y = \vec{p}_a - \vec{p}_b$$
$$p_Y^2 = p_a^2 + p_b^2 - 2 p_a p_b \cos\theta$$

Dividing by $2 M_Y$:

$$T_Y = \frac{p_Y^2}{2 M_Y} = \frac{m_a}{M_Y} T_a + \frac{m_b}{M_Y} T_b - \frac{2 \sqrt{m_a m_b}}{M_Y} \sqrt{T_a T_b} \cos\theta$$

Substitute into $Q = T_b + T_Y - T_a$:

$$Q = T_b + \left[ \frac{m_a}{M_Y} T_a + \frac{m_b}{M_Y} T_b - \frac{2 \sqrt{m_a m_b}}{M_Y} \sqrt{T_a T_b} \cos\theta \right] - T_a$$

Group powers of $\sqrt{T_b}$:

$$\left(1 + \frac{m_b}{M_Y}\right) T_b - \left(\frac{2 \sqrt{m_a m_b T_a}}{M_Y} \cos\theta\right) \sqrt{T_b} - \left[ Q + T_a\left(1 - \frac{m_a}{M_Y}\right) \right] = 0$$

Let:

$$A = \frac{M_Y + m_b}{M_Y}, \quad B = \frac{2 \sqrt{m_a m_b T_a} \cos\theta}{M_Y}, \quad C = Q + T_a\left(\frac{M_Y - m_a}{M_Y}\right)$$

The quadratic equation $A (\sqrt{T_b})^2 - B \sqrt{T_b} - C = 0$ yields:

$$\sqrt{T_b} = \frac{B + \sqrt{B^2 + 4 A C}}{2 A}$$

Step 2: Numerical Calculation

Masses: $m_a \approx 1$, $M_X \approx 7$, $m_b \approx 4$, $M_Y \approx 4$. $T_a = 3.00\text{ MeV}$, $Q = 17.347\text{ MeV}$.

  • $A = \frac{4 + 4}{4} = 2.00$
  • $C = 17.347 + 3.00\left(\frac{4 - 1}{4}\right) = 17.347 + 3.00(0.75) = 17.347 + 2.250 = 19.597\text{ MeV}$

1. At $\theta = 0^\circ$ ($\cos 0^\circ = 1$):

$$B = \frac{2 \sqrt{1 \times 4 \times 3.00}}{4} (1) = \frac{2 \sqrt{12}}{4} = \frac{\sqrt{12}}{2} = \sqrt{3} \approx 1.73205$$
$$B^2 + 4 A C = 3 + 4(2.00)(19.597) = 3 + 156.776 = 159.776$$
$$\sqrt{B^2 + 4 A C} = \sqrt{159.776} \approx 12.64025$$
$$\sqrt{T_\alpha} = \frac{1.73205 + 12.64025}{4.00} = \frac{14.3723}{4.00} \approx 3.59308$$
$$T_\alpha(0^\circ) = (3.59308)^2 \approx 12.91\text{ MeV}$$

2. At $\theta = 90^\circ$ ($\cos 90^\circ = 0 \implies B = 0$):

$$\sqrt{T_\alpha} = \frac{\sqrt{4 A C}}{2 A} = \sqrt{\frac{C}{A}} = \sqrt{\frac{19.597}{2.00}} = \sqrt{9.7985} \approx 3.13025$$
$$T_\alpha(90^\circ) = 9.80\text{ MeV}$$
Intermediate Example 4.7: Ghoshal Compound Nucleus Verification Calculation

In Ghoshal's classic 1950 test of the Bohr independence hypothesis, the compound nucleus $^{64}_{30}\text{Zn}^*$ was formed via: Channel A: $p + ^{63}_{29}\text{Cu} \to [^{64}_{30}\text{Zn}^*]$ ($Q_A = +7.71\text{ MeV}$) Channel B: $\alpha + ^{60}_{28}\text{Ni} \to [^{64}_{30}\text{Zn}^*]$ ($Q_B = +3.98\text{ MeV}$)

  1. For an incident proton kinetic energy $T_p^{\text{lab}} = 12.00\text{ MeV}$, calculate the excitation energy $E^$ of the compound nucleus $^{64}\text{Zn}^$.
  2. Determine the laboratory alpha particle energy $T_\alpha^{\text{lab}}$ required to produce $^{64}\text{Zn}^*$ at the exact same excitation energy.

Step 1: Compound Nucleus Excitation Energy from Channel A

In channel A, the center-of-mass kinetic energy is:

$$T_{\text{CM}, A} = \left(\frac{M_{\text{Cu}}}{m_p + M_{\text{Cu}}}\right) T_p^{\text{lab}} = \left(\frac{63}{1 + 63}\right) 12.00\text{ MeV} = \frac{63}{64} \times 12.00 = 11.8125\text{ MeV}$$

The excitation energy $E^*$ equals the center-of-mass kinetic energy plus the reaction $Q$-value:

$$E^* = T_{\text{CM}, A} + Q_A = 11.8125\text{ MeV} + 7.71\text{ MeV} = 19.5225\text{ MeV}$$

Step 2: Required Incident Alpha Energy from Channel B

For channel B to create the identical compound nucleus:

$$E^* = T_{\text{CM}, B} + Q_B = 19.5225\text{ MeV}$$
$$T_{\text{CM}, B} = E^* - Q_B = 19.5225\text{ MeV} - 3.98\text{ MeV} = 15.5425\text{ MeV}$$

Convert CM energy to laboratory frame for incident alpha:

$$T_{\text{CM}, B} = \left(\frac{M_{\text{Ni}}}{m_\alpha + M_{\text{Ni}}}\right) T_\alpha^{\text{lab}} = \left(\frac{60}{4 + 60}\right) T_\alpha^{\text{lab}} = \frac{60}{64} T_\alpha^{\text{lab}} = \frac{15}{16} T_\alpha^{\text{lab}}$$
$$T_\alpha^{\text{lab}} = \frac{16}{15} T_{\text{CM}, B} = \frac{16}{15} \times 15.5425\text{ MeV} \approx 16.5787\text{ MeV}$$

Incident alphas at $16.58\text{ MeV}$ create the compound state at the identical excitation energy as protons at $12.00\text{ MeV}$.

Advanced Example 4.8: Center-of-Mass Transformation of Differential Cross-Section

A nuclear reaction $X(a, b)Y$ has a differential cross-section $(d\sigma/d\Omega)_{\text{CM}}$ in the center-of-mass frame that is isotropic:

$$\left(\frac{d\sigma}{d\Omega}\right)_{\text{CM}} = \frac{\sigma_{\text{total}}}{4\pi} = \text{constant}$$
  1. Derive the kinematic transformation relating the laboratory scattering angle $\theta_{\text{lab}}$ to the center-of-mass angle $\theta_{\text{CM}}$:
$$\tan\theta_{\text{lab}} = \frac{\sin\theta_{\text{CM}}}{\cos\theta_{\text{CM}} + \gamma}$$

where $\gamma = V_{\text{CM}} / v_b^{\text{CM}}$.

  1. Derive the transformation formula for the laboratory differential cross-section $(d\sigma/d\Omega)_{\text{lab}}$.
  2. Show that forward scattering ($\theta_{\text{lab}} \to 0$) is kinematically enhanced in the laboratory frame.

Step 1: Angular Transformation Derivation

In the laboratory frame, the velocity components of particle $b$ are:

$$v_{b,\parallel}^{\text{lab}} = v_b^{\text{CM}} \cos\theta_{\text{CM}} + V_{\text{CM}}$$
$$v_{b,\perp}^{\text{lab}} = v_b^{\text{CM}} \sin\theta_{\text{CM}}$$

The laboratory emission angle $\theta_{\text{lab}}$ satisfies:

$$\tan\theta_{\text{lab}} = \frac{v_{b,\perp}^{\text{lab}}}{v_{b,\parallel}^{\text{lab}}} = \frac{v_b^{\text{CM}} \sin\theta_{\text{CM}}}{v_b^{\text{CM}} \cos\theta_{\text{CM}} + V_{\text{CM}}} = \frac{\sin\theta_{\text{CM}}}{\cos\theta_{\text{CM}} + \gamma}$$

where $\gamma = \frac{V_{\text{CM}}}{v_b^{\text{CM}}}$.

Step 2: Cross-Section Transformation

By definition of total particle conservation into solid angle:

$$\left(\frac{d\sigma}{d\Omega}\right)_{\text{lab}} d\Omega_{\text{lab}} = \left(\frac{d\sigma}{d\Omega}\right)_{\text{CM}} d\Omega_{\text{CM}}$$
$$\left(\frac{d\sigma}{d\Omega}\right)_{\text{lab}} = \left(\frac{d\sigma}{d\Omega}\right)_{\text{CM}} \frac{\sin\theta_{\text{CM}} d\theta_{\text{CM}}}{\sin\theta_{\text{lab}} d\theta_{\text{lab}}} = \left(\frac{d\sigma}{d\Omega}\right)_{\text{CM}} \left| \frac{d(\cos\theta_{\text{CM}})}{d(\cos\theta_{\text{lab}})} \right|$$

Evaluating the derivative:

$$\left(\frac{d\sigma}{d\Omega}\right)_{\text{lab}} = \left(\frac{d\sigma}{d\Omega}\right)_{\text{CM}} \frac{(1 + 2\gamma\cos\theta_{\text{CM}} + \gamma^2)^{3/2}}{|1 + \gamma\cos\theta_{\text{CM}}|}$$

Step 3: Forward Kinematic Enhancement

At forward angle $\theta_{\text{CM}} = 0^\circ$ ($\cos 0^\circ = 1$):

$$\left(\frac{d\sigma}{d\Omega}\right)_{\text{lab}} = \left(\frac{d\sigma}{d\Omega}\right)_{\text{CM}} \frac{(1 + 2\gamma + \gamma^2)^{3/2}}{1 + \gamma} = \left(\frac{d\sigma}{d\Omega}\right)_{\text{CM}} \frac{[(1 + \gamma)^2]^{3/2}}{1 + \gamma} = \left(\frac{d\sigma}{d\Omega}\right)_{\text{CM}} (1 + \gamma)^2$$

Because $\gamma > 0$, $(1 + \gamma)^2 > 1$. The laboratory cross-section in the forward direction is enhanced by the factor $(1 + \gamma)^2$ due to center-of-mass forward focusing!

Advanced Example 4.9: Breit-Wigner Interference Between Resonant and Potential Elastic Scattering

In low-energy neutron elastic scattering $(n, n)$, the total scattering amplitude $f(\theta)$ is the coherent quantum sum of two amplitudes:

  1. Hard-sphere potential scattering amplitude: $f_{\text{pot}} = -R$ (where $R$ is the nuclear radius).
  2. Resonance Breit-Wigner scattering amplitude: $f_{\text{res}} = -\frac{\lambdabar \Gamma_n / 2}{E - E_0 + i\Gamma/2}$.
  3. Formulate the total elastic cross-section $\sigma_{\text{sc}}(E) = 4\pi |f_{\text{pot}} + f_{\text{res}}|^2$.
  4. Show that quantum interference produces an asymmetric cross-section profile (Fano resonance) with a deep destructive interference minimum on the low-energy side of the resonance ($E < E_0$).

Step 1: Coherent Scattering Cross-Section

The total scattering amplitude is:

$$f(E) = -R - \frac{\lambdabar \Gamma_n / 2}{E - E_0 + i\Gamma/2}$$

Multiplying numerator and denominator of the resonant term by $(E - E_0 - i\Gamma/2)$:

$$f(E) = -R - \frac{\lambdabar \Gamma_n (E - E_0)}{2 [(E - E_0)^2 + \Gamma^2/4]} + i \frac{\lambdabar \Gamma_n \Gamma / 4}{(E - E_0)^2 + \Gamma^2/4}$$

Total scattering cross-section $\sigma_{\text{sc}} = 4\pi |f(E)|^2$:

$$\sigma_{\text{sc}}(E) = 4\pi \left[ \left( R + \frac{\lambdabar \Gamma_n (E - E_0) / 2}{(E - E_0)^2 + \Gamma^2/4} \right)^2 + \left( \frac{\lambdabar \Gamma_n \Gamma / 4}{(E - E_0)^2 + \Gamma^2/4} \right)^2 \right]$$

Expanding the square:

$$\sigma_{\text{sc}}(E) = 4\pi R^2 + \frac{\pi \lambdabar^2 \Gamma_n^2}{(E - E_0)^2 + \Gamma^2/4} + \frac{4\pi R \lambdabar \Gamma_n (E - E_0)}{(E - E_0)^2 + \Gamma^2/4}$$
  • The first term is the constant potential scattering cross-section $\sigma_{\text{pot}} = 4\pi R^2$.
  • The second term is the symmetric Breit-Wigner resonance peak.
  • The third term is the Quantum Interference Term!

Step 2: Destructive Interference Minimum

Notice the sign of the interference term:

  • For $E > E_0$: $(E - E_0) > 0$, constructive interference enhances the cross-section.
  • For $E < E_0$: $(E - E_0) < 0$, destructive interference depresses the cross-section.

At an energy just below resonance:

$$E_{\min} \approx E_0 - \frac{\lambdabar \Gamma_n}{2 R}$$

The potential scattering amplitude and resonance amplitude have opposite signs and cancel each other destructively! The cross-section drops to a deep minimum near zero (the "resonance dip"). This asymmetry is ubiquitous in neutron transmission spectra.

Solved Honors Problems & Derivations

Step-by-step rigorous solutions with full physical, thermodynamic, and nuclear kinematic validation.