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Chapter 4 • Theory & Derivations

Nuclear Reactions, Fission & Thermonuclear Fusion

Exhaustive dynamics of nuclear interactions: reaction kinematics in laboratory and center-of-mass frames, Q-value equations, threshold energies; reaction cross-sections, Breit-Wigner single-level resonance, optical model; direct reactions versus the Bohr compound nucleus hypothesis; nuclear fission physics, liquid drop deformation barrier, mass yield asymmetry, prompt and delayed neutron emissions; four-factor formula and nuclear reactor kinetics; stellar nucleosynthesis (proton-proton chain, CNO cycle) and thermonuclear fusion physics (Lawson criterion, magnetic confinement).

§4.1 Reaction Kinematics: Laboratory vs CM Frames & Threshold Energy

1. General Nuclear Reaction Formulation & Q-Value

A generic two-body nuclear reaction is denoted as $a + X \to Y + b$ or in compact Bethe notation $X(a, b)Y$, where projectile $a$ strikes stationary target nucleus $X$, producing ejectile $b$ and residual nucleus $Y$. The reaction $Q$-value is defined as the invariant rest-mass energy difference:

$$Q = [ (M_a + M_X) - (M_b + M_Y) ] c^2 = (T_b + T_Y) - T_a$$
  • Exothermic Reactions ($Q > 0$): Mass is converted into kinetic energy. The reaction can occur even at zero incident projectile energy ($T_a \to 0$).
  • Endothermic Reactions ($Q < 0$): Kinetic energy is converted into mass. The reaction is forbidden unless the projectile delivers a kinetic energy exceeding a strict threshold energy $E_{\text{th}}$.

2. Laboratory vs Center-of-Mass (CM) Transformation

In the laboratory frame, the total momentum is $p_{\text{lab}} = p_a = \sqrt{2 M_a T_a}$. The velocity of the center-of-mass frame is:

$$v_{\text{cm}} = \frac{M_a}{M_a + M_X} v_a$$

The total kinetic energy available in the center-of-mass frame ($T_{\text{cm}}$) represents the fraction of laboratory energy available to induce nuclear transformations (the remainder $T_{\text{cm, motion}} = \frac{1}{2}(M_a + M_X)v_{\text{cm}}^2$ is locked up in rigid motion of the system):

$$T_{\text{cm}} = \frac{1}{2} \mu v_{\text{rel}}^2 = \frac{M_X}{M_a + M_X} T_a$$

where $\mu = \frac{M_a M_X}{M_a + M_X}$ is the reduced mass.

3. Derivation of Reaction Threshold Energy

For an endothermic reaction ($Q < 0$), the reaction can occur if and only if the center-of-mass kinetic energy is at least sufficient to supply the mass deficit: $T_{\text{cm}} \ge |Q|$. Substituting $T_{\text{cm}}$ gives the threshold laboratory kinetic energy:

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

If Coulomb forces are present (charged projectile and target), the projectile must also overcome or tunnel through the Coulomb barrier $V_C = \frac{1}{4\pi\varepsilon_0}\frac{Z_a Z_X e^2}{R_a + R_X}$, significantly elevating the effective practical threshold.

§4.2 Cross-Sections, Breit-Wigner Resonances & The Optical Model

1. Reaction Cross-Section & Beam Attenuation

The reaction probability is quantified by the cross-section $\sigma$, possessing units of area (standard unit: $1\text{ barn (b)} \equiv 10^{-28}\text{ m}^2 = 100\text{ fm}^2$). For a uniform projectile beam of flux $\Phi = n_a v_a$ (particles per unit area per second) incident on a thin target containing $N_{\text{t}}$ target nuclei per unit area, the reaction rate $R$ is:

$$R = \Phi N_{\text{t}} \sigma$$

For a thick target of mass density $\rho$ and thickness $x$, the unscattered beam intensity decreases exponentially:

$$I(x) = I_0 e^{-\Sigma_t x} = I_0 e^{-n_X \sigma_t x}, \quad \Sigma_t = \frac{\rho N_A}{A} \sigma_t$$

where $\Sigma_t$ is the macroscopic total cross-section ($\text{cm}^{-1}$).

2. Breit-Wigner Single-Level Resonance Formula

When the incident projectile energy matches a quasi-bound quantum level of the composite system ($E_0$), the cross-section displays a dramatic, sharp peak called a resonance. Gregory Breit and Eugene Wigner derived the cross-section for isolated s-wave ($\ell = 0$) resonance reactions using quantum scattering theory:

$$\sigma(a, b) = \pi \lambda\hspace{-0.45em}\bar{}^2 g \frac{\Gamma_a \Gamma_b}{(E - E_0)^2 + (\Gamma / 2)^2}$$

where:

  • $\lambda\hspace{-0.45em}\bar{} = \frac{\lambda}{2\pi} = \frac{\hbar}{p_{\text{cm}}}$ is the reduced de Broglie wavelength of the incident channel.
  • $\Gamma = \Gamma_a + \Gamma_b + \Gamma_\gamma + \dots$ is the total resonance energy width, related to the compound state lifetime by $\tau = \hbar / \Gamma$.
  • $\Gamma_a, \Gamma_b$ are partial decay widths for the entrance and exit channels.
  • $g = \frac{2 J + 1}{(2 s_a + 1)(2 I_X + 1)}$ is the spin statistical factor for compound nuclear spin $J$.

At the peak ($E = E_0$), the maximum cross-section is $\sigma_{\text{peak}} = 4\pi \lambda\hspace{-0.45em}\bar{}^2 g \frac{\Gamma_a \Gamma_b}{\Gamma^2}$. For slow neutrons where $\lambda \sim 10^{-10}\text{ m} \gg R$, thermal resonance cross-sections can reach tens of thousands of barns (e.g., $^{113}\text{Cd}$ has $\sigma \approx 20,000\text{ b}$, $^{135}\text{Xe}$ has $\sigma \approx 2.6 \times 10^6\text{ b}$).

3. The Optical Model Potential

To describe both elastic scattering and non-elastic absorption of nucleons by nuclei, Herman Feshbach introduced the optical model, representing the nucleus as a refractive, cloudy crystal ball with a complex potential:

$$U(r) = - V(r) - i W(r) + V_{\text{so}}(r) (\vec{\ell}\cdot\vec{s})$$

The real depth $V(r) \approx 50\text{ MeV}$ governs nuclear refraction (elastic scattering), while the imaginary depth $W(r) \approx 5 - 15\text{ MeV}$ acts as an energy sink, absorbing flux from the incident channel to simulate all possible inelastic, transfer, and compound nuclear reactions.

§4.3 Reaction Mechanisms: Direct Reactions vs Compound Nucleus

1. Niels Bohr's Compound Nucleus Hypothesis (1936)

Niels Bohr proposed that low-energy nuclear reactions proceed through a distinct two-stage process:

$$a + X \longrightarrow C^* \longrightarrow Y + b$$
  1. Formation: The projectile enters the target and undergoes repeated collisions with multiple nucleons, dissipating its kinetic energy across the whole nucleus to form an excited compound nucleus $C^*$. The formation time is $\tau_{\text{form}} \sim 10^{-22}\text{ s}$ (nuclear transit time).
  2. Independence Hypothesis: The compound nucleus survives for a long duration ($\tau_C \sim 10^{-16} - 10^{-18}\text{ s}$, thousands of times longer than transit time) during which all memory of the entrance channel ($a+X$) is completely erased, except for exact conserved quantum numbers (energy $E$, angular momentum $J$, and parity $\pi$).
  3. Decay: De-excitation occurs through statistical evaporation when fluctuations concentrate sufficient energy on a single particle or photon channel:
    $$\sigma(a, b) = \sigma_{\text{form}}(C^*) \times P_{\text{decay}}(b)$$

Compound reactions produce isotropic or forward-backward symmetric angular distributions in the center-of-mass frame ($\frac{d\sigma}{d\Omega}(\theta) = \frac{d\sigma}{d\Omega}(\pi - \theta)$).

2. Direct Reactions (Stripping & Pickup)

At higher incident energies ($E \gtrsim 20 - 100\text{ MeV}$), the projectile traverses the nucleus in a single passage ($\tau \sim 10^{-22}\text{ s}$) and interacts with only one or two valence nucleons at the nuclear surface:

  • Stripping Reactions: The projectile deposits one or more nucleons into a single-particle orbit of the target nucleus, while the remainder continues forward (e.g., $(d, p)$ stripping deposits a neutron).
  • Pickup Reactions: The projectile scoops up a nucleon from the target (e.g., $(p, d)$ picks up a neutron).

Direct reactions display strong forward-peaked angular distributions whose diffraction peaks depend uniquely on the transferred orbital angular momentum $\ell$, providing a powerful experimental probe of single-particle shell structures.

§4.4 Nuclear Fission: Deformation Barrier, Mass Yield & Neutrons

1. Liquid Drop Model Explanation of Nuclear Fission

In 1939, Otto Frisch and Lise Meitner interpreted Otto Hahn and Fritz Strassmann's discovery of barium in neutron-irradiated uranium as the binary splitting—nuclear fission—of the heavy uranium nucleus. Niels Bohr and John Wheeler parameterized the stability of a spherical drop deformed into a prolate spheroid with eccentricity $\varepsilon$:

$$R(\theta) = R_0 [1 + \alpha_2 P_2(\cos\theta)], \quad E_{\text{def}} = \Delta E_S + \Delta E_C \approx E_S^{(0)} \left( \frac{2}{5} \alpha_2^2 \right) + E_C^{(0)} \left( - \frac{1}{5} \alpha_2^2 \right)$$

The deformed droplet is stable against spontaneous fission if $E_{\text{def}} > 0$. The threshold for spontaneous instability occurs when Coulomb repulsion exceeds twice surface tension:

$$\frac{E_C^{(0)}}{2 E_S^{(0)}} \ge 1 \implies \frac{a_c Z^2 / A^{1/3}}{2 a_s A^{2/3}} \ge 1 \implies \frac{Z^2}{A} \ge \frac{2 a_s}{a_c} \approx \frac{2 \times 17.8}{0.711} \approx 50$$

The ratio $x = \frac{Z^2 / A}{(Z^2 / A)_{\text{crit}}} \approx \frac{Z^2 / A}{48}$ is the fissility parameter. For $^{238}_{92}\text{U}$, $Z^2/A \approx 35.6$, resulting in a finite fission barrier height $E_f \approx 5.8\text{ MeV}$.

2. Energy Release in Fission

For a typical heavy nucleus ($A \approx 236$), $B/A \approx 7.6\text{ MeV}$, while the medium-mass fission fragments have $B/A \approx 8.5\text{ MeV}$. The energy released per fission event is:

$$Q_{\text{fission}} \approx A [ (B/A)_{\text{fragments}} - (B/A)_{\text{parent}} ] \approx 236 \times (8.5 - 7.6)\text{ MeV} \approx 200\text{ MeV}$$

The $200\text{ MeV}$ is partitioned as follows:

  • Fragment Kinetic Energy: $\approx 168\text{ MeV}$ ($84\%$, deposited immediately as heat within a few micrometers).
  • Prompt Fission Neutrons: $\approx 5\text{ MeV}$ (average $2 - 3$ neutrons per fission, mean kinetic energy $\sim 2\text{ MeV}$).
  • Prompt Gamma Rays: $\approx 7\text{ MeV}$.
  • Delayed Beta Particles: $\approx 8\text{ MeV}$ from fragment radioactive decay chains.
  • Delayed Antineutrinos: $\approx 12\text{ MeV}$ (escapes the reactor without depositing heat).

3. Mass Yield Asymmetry & Fission Neutrons

Low-energy thermal neutron fission of $^{235}\text{U}$ produces an asymmetric two-humped mass yield curve with peaks at light mass $A_L \approx 95$ and heavy mass $A_H \approx 140$. Symmetric fission ($A_1 = A_2 \approx 118$) is suppressed by a factor of 600 due to shell closures in the nascent fragments ($Z=50, N=82$). At high excitation energies ($E_n > 50\text{ MeV}$), shell effects wash out and symmetric fission dominates.

Fission releases an average of $\bar{\nu} \approx 2.43$ neutrons per event for $^{235}\text{U}$. Crucially, approximately $0.65\%$ of these neutrons are delayed neutrons emitted seconds to minutes later by beta-decay precursors (e.g., $^{87}\text{Br} \to {^{87}\text{Kr}} \to {^{86}\text{Kr}} + n$), providing the indispensable time delay required for mechanical control rods to safely stabilize nuclear power reactors.

§4.5 Controlled Fission: Four-Factor Formula & Reactor Kinetics

1. The Neutron Multiplication Factor (k)

The continuity of a nuclear fission chain reaction is dictated by the effective multiplication factor $k_{\text{eff}}$, defined as the ratio of neutrons in generation $n+1$ to neutrons in generation $n$:

  • $k_{\text{eff}} < 1$: Subcritical (neutron population and fission power die away exponentially).
  • $k_{\text{eff}} = 1$: Critical (steady, self-sustaining stationary power generation).
  • $k_{\text{eff}} > 1$: Supercritical (neutron flux grows exponentially).

2. Fermi's Four-Factor Formula ($k_\infty$)

In an infinitely extended homogeneous or heterogeneous reactor core (ignoring surface neutron leakage), the multiplication factor is governed by the Four-Factor Formula:

$$k_\infty = \varepsilon \cdot p \cdot \eta \cdot f$$
  1. Fast Fission Factor ($\varepsilon \approx 1.03 - 1.08$): Ratio of total neutrons produced by both fast and thermal fissions to those produced solely by thermal fissions.
  2. Resonance Escape Probability ($p \approx 0.85 - 0.92$): The probability that a fast fission neutron slows down through the dangerous intermediate resonance capture region of $^{238}\text{U}$ ($1 - 1000\text{ eV}$) without being absorbed. Heterogeneous fuel lump arrangements maximize $p$ through spatial self-shielding.
  3. Thermal Utilization Factor ($f \approx 0.70 - 0.90$): The probability that a completely moderated thermal neutron is absorbed in the nuclear fuel rather than in the moderator, structural cladding, or control poisons:
    $$f = \frac{\Sigma_a^{\text{fuel}}}{\Sigma_a^{\text{fuel}} + \Sigma_a^{\text{other}}}$$
  4. Neutron Reproduction Factor ($\eta \approx 1.3 - 2.1$): The average number of fission neutrons emitted per thermal neutron absorbed in the fuel:
    $$\eta = \nu \frac{\Sigma_f^{\text{fuel}}}{\Sigma_a^{\text{fuel}}}$$

3. Reactor Kinetics and Prompt Criticality

If all fission neutrons were prompt, the average neutron lifetime would be $\ell \sim 10^{-4}\text{ s}$ (in a thermal reactor). The reactor power would evolve as $P(t) = P_0 e^{(k - 1)t / \ell}$. Even a $0.1\%$ excess ($k = 1.001$) would cause power to escalate by $e^{10} \approx 22,000$ in one second, rendering reactor control physically impossible.

With delayed neutron fraction $\beta \approx 0.0065$, the effective mean lifetime is extended to $\bar{\ell} = (1 - \beta)\ell + \sum \beta_i \tau_i \approx 0.08 - 0.1\text{ seconds}$. Provided $k_{\text{eff}} < 1 + \beta$, the reactor is delayed critical and responds smoothly over human and mechanical timescales.

§4.6 Thermonuclear Fusion: Stellar Cycles & The Lawson Criterion

1. Stellar Nucleosynthesis: The Proton-Proton Chain

In main-sequence stars like our Sun ($T_{\text{core}} \approx 1.5 \times 10^7\text{ K}$, $k_B T \approx 1.3\text{ keV}$), stellar energy generation is powered by the fusion of four protons into helium-4 ($4 p \to {^4\text{He}} + 2 e^+ + 2 \nu_e + 26.73\text{ MeV}$). The dominant sequence is the $p$-$p$ chain:

  1. $p + p \to {^2\text{H}} + e^+ + \nu_e$ ($Q = 1.442\text{ MeV}$): Weak interaction bottle-neck process ($p \to n + e^+ + \nu_e$) with an extraordinarily tiny cross-section ($\sigma \sim 10^{-47}\text{ cm}^2$). A proton in the solar core waits an average of $10^9\text{ years}$ to undergo this reaction, ensuring the Sun's multi-billion-year stability.
  2. $^2\text{H} + p \to {^3\text{He}} + \gamma$ ($Q = 5.493\text{ MeV}$): Rapid electromagnetic capture ($\sim 1\text{ second}$).
  3. $^3\text{He} + {^3\text{He}} \to {^4\text{He}} + 2 p$ ($Q = 12.86\text{ MeV}$, $pp\text{-I}$ branch): Completes the synthesis of $^4\text{He}$.

In heavier, hotter stars ($T > 2 \times 10^7\text{ K}$), the catalytic CNO cycle ($^{12}\text{C} \to {^{13}\text{N}} \to {^{13}\text{C}} \to {^{14}\text{N}} \to {^{15}\text{O}} \to {^{15}\text{N}} \to {^{12}\text{C}} + {^4\text{He}}$) dominates due to its steeper temperature dependence ($\epsilon_{\text{CNO}} \propto T^{17}$ vs $\epsilon_{pp} \propto T^4$).

2. Controlled Terrestrial Fusion: The D-T Reaction

For magnetic confinement fusion (tokamaks, stellarators), the most accessible reaction is Deuterium-Tritium fusion:

$$^2_1\text{H} + {^3_1\text{H}} \longrightarrow {^4_2\text{He}} (3.5\text{ MeV}) + n (14.1\text{ MeV}) \quad (Q = 17.59\text{ MeV})$$

The D-T reaction possesses the lowest Coulomb barrier and highest cross-section ($\sigma_{\text{peak}} \approx 5.0\text{ b}$ at $E_{\text{cm}} \approx 64\text{ keV}$, accessible at thermal plasma temperatures $T \sim 15\text{ keV} \approx 1.7 \times 10^8\text{ K}$). The Gamow window represents the convolution of the Maxwell-Boltzmann tail $e^{-E / k_B T}$ with the quantum tunneling transmission $e^{-b / \sqrt{E}}$.

3. The Lawson Criterion & Triple Product

In 1957, J. D. Lawson formulated the ignition condition where thermonuclear self-heating by alpha particles ($E_\alpha = 3.5\text{ MeV}$) exceeds plasma Bremsstrahlung radiation and conduction losses without external heating:

$$n \cdot \tau_E \ge \frac{12 k_B T}{\langle \sigma v \rangle Q_\alpha}$$

where $n$ is fuel ion density and $\tau_E$ is the energy confinement time. For D-T fusion at the optimum temperature $T \approx 15\text{ keV}$, this requires the famous fusion triple product:

$$n \cdot T \cdot \tau_E \ge 3 \times 10^{21}\text{ keV}\cdot\text{s}\cdot\text{m}^{-3} \approx 5 \times 10^{28}\text{ K}\cdot\text{s}\cdot\text{m}^{-3}$$

Magnetic confinement achieves this at low density and high confinement ($n \sim 10^{20}\text{ m}^{-3}, \tau_E \sim 3\text{ s}$), while inertial confinement uses extreme compression ($n \sim 10^{31}\text{ m}^{-3}, \tau_E \sim 10^{-10}\text{ s}$).

Solved Problem Example 4.1: Threshold Kinetic Energy and Kinematics of Nitrogen-14 Alpha Reaction

The historical Rutherford nuclear reaction is $^{14}_7\text{N}(\alpha, p)^{17}_8\text{O}$. The atomic masses are: $M(^{14}\text{N}) = 14.003074 \text{ u}$, $M(^4\text{He}) = 4.002603 \text{ u}$, $M(^1\text{H}) = 1.007825 \text{ u}$, and $M(^{17}\text{O}) = 16.999131 \text{ u}$. (a) Calculate the reaction $Q$-value in $\text{MeV}$. Is the reaction endothermic or exothermic? (b) Determine the minimum threshold kinetic energy $E_{\text{th}}$ of the incident alpha particle in the laboratory frame. (c) If the incident alpha particle has laboratory energy $T_\alpha = 7.70 \text{ MeV}$, find the total kinetic energy available in the center-of-mass frame.

Step 1: Compute Mass Difference and Q-Value
$$\Delta m = [M(^{14}\text{N}) + M(^4\text{He})] - [M(^1\text{H}) + M(^{17}\text{O})] = [14.003074 + 4.002603] - [1.007825 + 16.999131]\text{ u} = -0.001279\text{ u}$$

Subtract product masses from reactant masses.

Step 2: Convert Q-Value to MeV
$$Q = (-0.001279\text{ u}) \times 931.494\text{ MeV/u} \approx -1.1914\text{ MeV}$$

Because Q < 0, the reaction is endothermic.

Step 3: Calculate Threshold Laboratory Kinetic Energy
$$E_{\text{th}} = |Q| \left( 1 + \frac{M_\alpha}{M_N} \right) = 1.1914\text{ MeV} \times \left( 1 + \frac{4.0026}{14.0031} \right) \approx 1.1914 \times (1 + 0.2858) \approx 1.532\text{ MeV}$$

Calculate minimum projectile energy required in laboratory frame.

Step 4: Center-of-Mass Kinetic Energy at 7.70 MeV
$$T_{\text{cm}} = T_\alpha \left( \frac{M_N}{M_\alpha + M_N} \right) = 7.70\text{ MeV} \times \left( \frac{14.0031}{18.0057} \right) \approx 7.70 \times 0.7777 \approx 5.988\text{ MeV}$$

Compute available CM energy at 7.70 MeV.

Final Answer & Physical Insight

Q = -1.191 \text{ MeV} \quad (\text{Endothermic}), \quad E_{\text{th}} = 1.532 \text{ MeV}, \quad T_{\text{cm}} = 5.988 \text{ MeV}

Solved Problem Example 4.2: Nuclear Energy Release in Complete Fission of Uranium-235

A commercial nuclear reactor operates at a thermal power output of $P_{\text{th}} = 3000 \text{ MW}$ ($3.0 \times 10^9 \text{ J/s}$). Assuming an average usable energy release of $Q = 200 \text{ MeV}$ per fission event of $^{235}_{92}\text{U}$: (a) Calculate the fission rate (number of fissions per second). (b) Determine the rate of mass consumption of $^{235}\text{U}$ in kilograms per day. (c) Compare this daily fuel consumption to that of a coal-fired power plant of identical thermal capacity burning coal with a heat of combustion of $29.0 \text{ MJ/kg}$.

Step 1: Convert Energy per Fission to Joules
$$E_{\text{fission}} = 200\text{ MeV} \times (1.6022 \times 10^{-13}\text{ J/MeV}) = 3.2044 \times 10^{-11}\text{ J}$$

Convert 200 MeV to Joules.

Step 2: Calculate Fission Rate per Second
$$R_{\text{fiss}} = \frac{P_{\text{th}}}{E_{\text{fission}}} = \frac{3.0 \times 10^9\text{ J/s}}{3.2044 \times 10^{-11}\text{ J}} \approx 9.362 \times 10^{19}\text{ fissions/s}$$

Divide power by energy per fission.

Step 3: Calculate Daily Mass Consumption of Uranium-235
$$m_{\text{day}} = \frac{R_{\text{fiss}} \times 86400\text{ s} \times 0.23504\text{ kg/mol}}{6.0221 \times 10^{23}\text{ mol}^{-1}} = \frac{(8.089 \times 10^{24}) \times 0.23504}{6.0221 \times 10^{23}}\text{ kg} \approx 3.157\text{ kg/day}$$

Multiply daily fissions by molar mass divided by Avogadro's number: approx 3.16 kg/day.

Step 4: Calculate Coal Consumption for Comparison
$$m_{\text{coal}} = \frac{3.0 \times 10^9\text{ J/s} \times 86400\text{ s}}{29.0 \times 10^6\text{ J/kg}} = \frac{2.592 \times 10^{14}\text{ J}}{2.90 \times 10^7\text{ J/kg}} \approx 8.938 \times 10^6\text{ kg/day} \approx 8940\text{ metric tons/day}$$

Compute coal consumption: 8,940 tonnes per day, demonstrating a ~3,000,000:1 fuel mass density advantage.

Final Answer & Physical Insight

R_{\text{fiss}} = 9.36 \times 10^{19} \text{ s}^{-1}, \quad m_{\text{U}} = 3.16 \text{ kg/day}, \quad m_{\text{coal}} = 8940 \text{ tonnes/day} \quad (2.83 \times 10^6 \times)

Solved Problem Example 4.3: Thermonuclear D-T Fusion Power Density and Lawson Parameter

In a D-T magnetic confinement fusion reactor, the deuterium and tritium ion densities are equal: $n_D = n_T = \frac{1}{2} n_i = 1.0 \times 10^{20} \text{ m}^{-3}$ ($n_i = 2.0 \times 10^{20} \text{ m}^{-3}$). At a plasma temperature of $T = 15.0 \text{ keV}$ ($1.74 \times 10^8 \text{ K}$), the reaction rate parameter is $\langle \sigma v \rangle = 2.80 \times 10^{-22} \text{ m}^3/\text{s}$, and each reaction produces $Q = 17.6 \text{ MeV}$ ($2.82 \times 10^{-12} \text{ J}$), including an alpha particle of $E_\alpha = 3.52 \text{ MeV}$ ($5.64 \times 10^{-13} \text{ J}$). (a) Calculate the total fusion power density $P_f$ in $\text{MW/m}^3$. (b) Calculate the alpha particle self-heating power density $P_\alpha$. (c) If the plasma energy confinement time is $\tau_E = 3.50 \text{ s}$, evaluate the Lawson ignition parameter and determine if ignition is achieved.

Step 1: Calculate Total Fusion Power Density
$$P_f = n_D n_T \langle \sigma v \rangle Q = (1.0 \times 10^{20})^2 (2.80 \times 10^{-22}\text{ m}^3/\text{s})(2.82 \times 10^{-12}\text{ J}) \approx 7.896 \times 10^6\text{ W/m}^3 = 7.90\text{ MW/m}^3$$

Evaluate fusion power density P_f.

Step 2: Calculate Alpha Self-Heating Power Density
$$P_\alpha = n_D n_T \langle \sigma v \rangle E_\alpha = (1.0 \times 10^{40})(2.80 \times 10^{-22})(5.64 \times 10^{-13}\text{ J}) \approx 1.579 \times 10^6\text{ W/m}^3 = 1.58\text{ MW/m}^3$$

Evaluate alpha heating power density P_alpha.

Step 3: Evaluate Plasma Energy Loss Rate
$$P_{\text{loss}} = \frac{3 n_e k_B T}{\tau_E} = \frac{3 (2.0 \times 10^{20}\text{ m}^{-3})(15.0 \times 1.6022 \times 10^{-16}\text{ J})}{3.50\text{ s}} = \frac{1.442 \times 10^6}{3.50}\text{ W/m}^3 \approx 0.412\text{ MW/m}^3$$

Calculate plasma thermal energy loss per unit volume.

Step 4: Check Ignition Criterion
$$\frac{P_\alpha}{P_{\text{loss}}} = \frac{1.579\text{ MW/m}^3}{0.412\text{ MW/m}^3} \approx 3.83 > 1. \quad \text{Triple Product: } n_i T \tau_E = (2.0 \times 10^{20})(15)(3.5) = 1.05 \times 10^{22}\text{ keV}\cdot\text{s}\cdot\text{m}^{-3} > 3 \times 10^{21}$$

Because P_alpha exceeds P_loss by a factor of 3.8, alpha self-heating sustains the temperature without auxiliary heating: ignition is robustly achieved.

Final Answer & Physical Insight

P_f = 7.90 \text{ MW/m}^3, \quad P_\alpha = 1.58 \text{ MW/m}^3, \quad n T \tau_E = 1.05 \times 10^{22} \text{ keV}\cdot\text{s}\cdot\text{m}^{-3} \quad (\text{Ignition Achieved})

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