← Part V: Applications
Chapter 13

Nucleosynthesis

Big Bang Nucleosynthesis

In the first few minutes after the Big Bang, the universe was hot and dense enough for nuclear reactions. BBN produced the lightest elements: H, D, He-3, He-4, and traces of Li-7. The key stages:

  • - $t < 1$ s, $T > 1$ MeV: Weak interactions maintain neutron-proton equilibrium: $n/p = e^{-\Delta m c^2/k_BT}$
  • - $t \sim 1$ s, $T \sim 0.8$ MeV: Weak freeze-out. n/p ratio freezes at ~1/6
  • - $1 < t < 180$ s: Free neutron decay reduces n/p to ~1/7
  • - $t \sim 180$ s, $T \sim 0.07$ MeV: Deuterium bottleneck breaks. Nearly all neutrons captured into He-4

He-4 Abundance Prediction

$$Y_p = \frac{2(n/p)}{1 + n/p} = \frac{2 \times 1/7}{1 + 1/7} = \frac{1}{4} = 0.25$$

Observed: $Y_p = 0.245 \pm 0.003$ -- excellent agreement, confirming the standard cosmological model and constraining the baryon density.

Neutron Number (N) β†’Proton Number (Z) β†’N = ZValley of StabilityProton Drip LineNeutron Drip Lines-process(slow n-capture)r-process(rapid n-capture)β⁻ decayStables-processr-processp-dripn-dripβ⁻ decay
Figure 1. Schematic chart of nuclides (N vs Z). The valley of stability (dark band) contains the stable isotopes. The proton drip line (upper boundary) and neutron drip line (lower boundary) mark the limits of nuclear binding. The s-process path (green) zigzags along the valley via slow neutron captures and beta decays. The r-process path (red, dashed) runs far from stability on the neutron-rich side, with subsequent beta decays (purple arrows) bringing nuclei back toward stability.

The Deuterium Bottleneck

Although nuclear reactions could in principle begin when $T \sim 1$ MeV, the low binding energy of deuterium ($B_d = 2.224$ MeV) means that photodissociation destroys deuterium as fast as it forms until $T$ drops well below $B_d$. This is the deuterium bottleneck.

Saha-like Equilibrium

The deuterium abundance relative to baryons in nuclear statistical equilibrium:

$$\frac{n_d}{n_b} \approx \eta \left(\frac{T}{m_N}\right)^{3/2} \exp\!\left(\frac{B_d}{T}\right)$$

where $\eta = n_b/n_\gamma \approx 6.1 \times 10^{-10}$ is the baryon-to-photon ratio. Because $\eta$ is so small, $T$ must drop to $\sim 0.07$ MeV ($\sim 8 \times 10^8$ K) before deuterium survives β€” far below $B_d$.

BBN Reaction Network

Once deuterium survives, a rapid chain of reactions produces heavier nuclei:

$$p + n \to d + \gamma \quad (B_d = 2.224\;\text{MeV})$$

$$d + d \to {}^3\text{He} + n, \quad d + d \to {}^3\text{H} + p$$

$$ {}^3\text{He} + n \to {}^3\text{H} + p, \quad {}^3\text{H} + d \to {}^4\text{He} + n$$

$$ {}^3\text{He} + {}^4\text{He} \to {}^7\text{Be} + \gamma$$

$$ {}^7\text{Be} + n \to {}^7\text{Li} + p$$

No stable nuclei exist at $A = 5$ or $A = 8$, creating mass gaps that prevent BBN from producing significant amounts of elements beyond Li-7.

The Cosmological Lithium Problem

BBN predicts $^7$Li/H $\approx 4.7 \times 10^{-10}$, but observations of metal-poor halo stars (the β€œSpite plateau”) give $^7$Li/H $\approx 1.6 \times 10^{-10}$ β€” a factor of ~3 discrepancy. This is one of the outstanding puzzles in nuclear astrophysics.

Possible Explanations

  • - Stellar depletion (gravitational settling, rotationally-induced mixing)
  • - New physics (decaying dark matter particles, varying constants)
  • - Nuclear reaction rate uncertainties ($^7$Be destruction channels)
  • - Non-standard BBN (inhomogeneous models)

BBN as a Precision Test

Despite the Li problem, BBN is remarkably successful: it correctly predicts D/H,$^3$He/H, and $Y_p$ (He-4) from a single parameter $\eta$, independently confirmed by the CMB (Planck: $\eta = (6.104 \pm 0.058) \times 10^{-10}$). BBN also constrains the number of neutrino species: $N_\nu = 2.99 \pm 0.17$.

Hydrogen Burning: The pp Chain

The proton-proton chain is the dominant energy source in stars with $M \lesssim 1.3\,M_\odot$. The net reaction is $4p \to {}^4\text{He} + 2e^+ + 2\nu_e$ releasing $Q = 26.73$ MeV.

pp-I Chain (dominant, ~85% in Sun)

$$p + p \to d + e^+ + \nu_e \quad (Q = 1.442\;\text{MeV},\; \langle E_\nu\rangle = 0.267\;\text{MeV})$$

$$d + p \to {}^3\text{He} + \gamma \quad (Q = 5.493\;\text{MeV})$$

$$ {}^3\text{He} + {}^3\text{He} \to {}^4\text{He} + 2p \quad (Q = 12.860\;\text{MeV})$$

The first step is the rate-limiting reaction β€” it requires the weak interaction to convert a proton to a neutron, giving the Sun its ~10 billion year lifetime.

pp-II Chain (~15% in Sun)

$$ {}^3\text{He} + {}^4\text{He} \to {}^7\text{Be} + \gamma$$

$$ {}^7\text{Be} + e^- \to {}^7\text{Li} + \nu_e \quad (E_\nu = 0.862\;\text{MeV})$$

$$ {}^7\text{Li} + p \to 2\,{}^4\text{He}$$

pp-III Chain (~0.02% in Sun)

$$ {}^7\text{Be} + p \to {}^8\text{B} + \gamma$$

$$ {}^8\text{B} \to {}^8\text{Be}^* + e^+ + \nu_e \quad (E_\nu \leq 14.06\;\text{MeV})$$

$$ {}^8\text{Be}^* \to 2\,{}^4\text{He}$$

Though rare, the high-energy $^8$B neutrinos were crucial for the solar neutrino problem (detected by Davis, Kamiokande, SNO) and the discovery of neutrino oscillations.

Thermonuclear Reaction Rates

The reaction rate per unit volume depends on the Gamow peak β€” the overlap between the Maxwell-Boltzmann tail and the tunneling probability through the Coulomb barrier:

$$\langle\sigma v\rangle = \left(\frac{8}{\pi m}\right)^{1/2} \frac{1}{(k_BT)^{3/2}} \int_0^\infty S(E)\, \exp\!\left(-\frac{E}{k_BT} - \frac{E_G^{1/2}}{E^{1/2}}\right) dE$$

where $E_G = 2m(\pi \alpha Z_1 Z_2 c)^2$ is the Gamow energy and $S(E)$ is the astrophysical S-factor (varies slowly with energy). The integrand peaks at the Gamow energy:

$$E_0 = \left(\frac{E_G (k_BT)^2}{4}\right)^{1/3} \approx 5.665\,(Z_1^2 Z_2^2 \mu\, T_6^2)^{1/3}\;\text{keV}$$

For $p + p$ in the Sun ($T_6 \approx 15.7$): $E_0 \approx 5.9$ keV, with a Gamow window width $\Delta \approx 6.4$ keV.

Hydrogen Burning: The CNO Cycle

In stars with $M \gtrsim 1.3\,M_\odot$, the CNO cycle dominates hydrogen burning because its rate scales as $T^{16\text{-}20}$ (vs $T^4$ for pp). Carbon, nitrogen, and oxygen act as catalysts β€” they are not consumed.

CNO-I (dominant)

$$ {}^{12}\text{C} + p \to {}^{13}\text{N} + \gamma$$

$$ {}^{13}\text{N} \to {}^{13}\text{C} + e^+ + \nu_e \quad (\tau_{1/2} = 9.97\;\text{min})$$

$$ {}^{13}\text{C} + p \to {}^{14}\text{N} + \gamma$$

$$ {}^{14}\text{N} + p \to {}^{15}\text{O} + \gamma \quad (\text{rate-limiting step})$$

$$ {}^{15}\text{O} \to {}^{15}\text{N} + e^+ + \nu_e \quad (\tau_{1/2} = 2.03\;\text{min})$$

$$ {}^{15}\text{N} + p \to {}^{12}\text{C} + {}^4\text{He}$$

Net result: $4p \to {}^4\text{He} + 2e^+ + 2\nu_e + 3\gamma$ ($Q = 25.03$ MeV). The $^{14}$N(p,$\gamma$)$^{15}$O reaction is the slowest step, so in CNO equilibrium most catalyst nuclei accumulate as $^{14}$N.

Temperature Dependence

The energy generation rate is well approximated by a power law:

$$\epsilon_{\text{CNO}} \approx \epsilon_0\, X_{\text{CNO}}\, X_H\, \rho\, T_6^{16.7}$$

This extreme temperature sensitivity means CNO-dominated stars develop convective cores (energy is generated faster than radiation can transport it). The crossover between pp and CNO dominance occurs at $T \approx 1.7 \times 10^7$ K ($\sim 1.3\,M_\odot$).

Helium Burning: The Triple-Alpha Process

After hydrogen exhaustion, the core contracts and heats to $T \sim 10^8$ K, enabling helium burning. Direct $\alpha + \alpha$ fusion produces $^8$Be, which is unstable ($\tau \sim 10^{-16}$ s). The triple-alpha process overcomes this.

Two-Step Mechanism

$$ {}^4\text{He} + {}^4\text{He} \rightleftharpoons {}^8\text{Be} \quad (Q = -91.78\;\text{keV})$$

$$ {}^8\text{Be} + {}^4\text{He} \to {}^{12}\text{C}^*(0^+_2) \to {}^{12}\text{C} + 2\gamma \quad (Q = 7.367\;\text{MeV})$$

The second step is only possible because of the Hoyle state β€” an excited $0^+$ state of $^{12}$C at 7.654 MeV. Fred Hoyle predicted this resonance in 1953 from the anthropic requirement that carbon must be abundant in the universe. It was experimentally confirmed by Fowler's group at Caltech.

Triple-Alpha Rate

$$r_{3\alpha} = \frac{n_\alpha^3}{6}\langle\sigma v\rangle_{3\alpha} \propto \rho^2 Y^3 T^{41}$$

The extreme temperature sensitivity ($T^{41}$!) leads to the helium flashin low-mass red giants: degenerate cores cannot expand to regulate temperature, causing a thermonuclear runaway that lifts the degeneracy in seconds.

Carbon-to-Oxygen Ratio

After $^{12}$C forms, the crucial reaction $^{12}$C($\alpha,\gamma$)$^{16}$O determines the final C/O ratio, which controls the subsequent evolution and nucleosynthesis products. This reaction rate at stellar energies remains one of the most important unsolved problems in nuclear astrophysics:

$$S_{E1}(300\;\text{keV}) \approx 80 \pm 20\;\text{keV b}$$

If the rate were ~2x higher, stars would produce almost no carbon; if ~2x lower, they would produce very little oxygen. Both elements are essential for life.

Advanced Burning Stages

Stars with $M \gtrsim 8\,M_\odot$ proceed through additional burning stages, each at higher temperature and shorter duration, forming an onion-shell structure:

StageT (K)$\rho$ (g/cmΒ³)Duration ($25\,M_\odot$)Main Products
H burning$4 \times 10^7$5$7 \times 10^6$ yrHe
He burning$2 \times 10^8$$7 \times 10^2$$5 \times 10^5$ yrC, O
C burning$8 \times 10^8$$2 \times 10^5$600 yrNe, Na, Mg
Ne burning$1.6 \times 10^9$$4 \times 10^6$1 yrO, Mg
O burning$2.0 \times 10^9$$10^7$6 monthsSi, S, Ar, Ca
Si burning$3.5 \times 10^9$$3 \times 10^7$1 dayFe, Ni, Co

Carbon Burning

$$ {}^{12}\text{C} + {}^{12}\text{C} \to {}^{20}\text{Ne} + \alpha \quad (Q = 4.62\;\text{MeV})$$

$$ {}^{12}\text{C} + {}^{12}\text{C} \to {}^{23}\text{Na} + p \quad (Q = 2.24\;\text{MeV})$$

$$ {}^{12}\text{C} + {}^{12}\text{C} \to {}^{23}\text{Mg} + n \quad (Q = -2.60\;\text{MeV})$$

Oxygen Burning

$$ {}^{16}\text{O} + {}^{16}\text{O} \to {}^{28}\text{Si} + \alpha \quad (Q = 9.59\;\text{MeV})$$

$$ {}^{16}\text{O} + {}^{16}\text{O} \to {}^{31}\text{P} + p \quad (Q = 7.68\;\text{MeV})$$

$$ {}^{16}\text{O} + {}^{16}\text{O} \to {}^{31}\text{S} + n \quad (Q = 1.50\;\text{MeV})$$

Silicon Burning & Nuclear Statistical Equilibrium

Silicon burning does not proceed by $^{28}$Si + $^{28}$Si fusion (the Coulomb barrier is too high). Instead, at $T \gtrsim 3 \times 10^9$ K, photodisintegration reactions become fast enough to establish a quasi-equilibrium network of reactions mediated by free protons, neutrons, and alpha particles.

Quasi-Statistical Equilibrium (QSE)

In QSE, abundance ratios within each cluster of nuclei are determined by the nuclear Saha equation:

$$Y(Z,A) \propto G(Z,A)\, A^{3/2}\, \left(\frac{\rho}{\rho_0}\right)^{A-1} \left(\frac{T}{T_0}\right)^{3(1-A)/2} \exp\!\left(\frac{B(Z,A)}{k_BT}\right) Y_p^Z Y_n^{A-Z}$$

where $G$ is the nuclear partition function and $B(Z,A)$ is the binding energy. As the system evolves toward full NSE, abundances favor the most tightly bound nuclei per nucleon β€” the iron-peak elements.

The Iron Peak and the End of Fusion

In NSE at $T \sim 4 \times 10^9$ K, the dominant species is $^{56}$Ni (which decays to $^{56}$Fe via $^{56}$Co). The binding energy per nucleon:

$$B/A: \quad {}^{56}\text{Fe} = 8.790\;\text{MeV}, \quad {}^{62}\text{Ni} = 8.795\;\text{MeV}, \quad {}^{58}\text{Fe} = 8.792\;\text{MeV}$$

Beyond the iron peak, fusion is endothermic. The core can no longer generate thermal pressure to support itself, leading to gravitational collapse and a core-collapse supernova. The collapse happens on a timescale of $\sim 0.1$ s once Si burning ends.

Explosive Nucleosynthesis

Core-collapse supernovae ($M \gtrsim 8\,M_\odot$) and thermonuclear supernovae (Type Ia, white dwarf detonation) produce elements through shock-heated explosive burning.

Core-Collapse Supernovae (Type II/Ib/Ic)

  • - Shock passes through onion-shell structure
  • - Explosive Si burning: $^{56}$Ni (decays to $^{56}$Fe)
  • - Explosive O burning: Si, S, Ar, Ca
  • - Neutrino-driven wind: may produce light r-process elements
  • - $\nu$-process: $^{11}$B, $^{19}$F, $^{138}$La

Type Ia Supernovae

  • - Thermonuclear detonation of C/O white dwarf
  • - Primary source of iron-peak elements
  • - Produces $\sim 0.6\,M_\odot$ of $^{56}$Ni per event
  • - Yields Si, S, Ca, Fe, Ni, Cr, Mn
  • - Standard candles for cosmological distance measurements

The s-Process (Slow Neutron Capture)

Elements heavier than iron are produced primarily by neutron capture. In the s-process, neutron capture is slow compared to beta decay ($\tau_{\text{capture}} \gg \tau_\beta$), so the capture path follows the valley of beta stability.

s-Process Sites and Neutron Sources

The s-process operates in AGB stars (thermally-pulsing asymptotic giant branch) with two main neutron sources:

$$ {}^{13}\text{C}(\alpha,n){}^{16}\text{O} \quad (T \sim 10^8\;\text{K}, \text{main source in low-mass AGB})$$

$$ {}^{22}\text{Ne}(\alpha,n){}^{25}\text{Mg} \quad (T \sim 3 \times 10^8\;\text{K}, \text{in massive AGB})$$

Typical neutron density: $n_n \sim 10^7\text{-}10^8$ cm$^{-3}$. Neutron exposure: $\tau = \int n_n v_T\, dt \sim 0.1\text{-}0.4$ mb$^{-1}$.

Classical s-Process Model

In the steady-flow approximation, the product $\sigma N$ (cross section times abundance) is approximately constant along the s-process path:

$$\sigma_A N_A \approx \text{constant} \quad (\text{local approximation})$$

More precisely, assuming an exponential distribution of neutron exposures:

$$\sigma_A N_s(A) = \frac{f N_{56}}{\tau_0} \prod_{i=56}^{A} \frac{1}{1 + 1/(\sigma_i \tau_0)}$$

The s-process produces abundance peaks at magic neutron numbers(N = 50, 82, 126) where neutron capture cross sections are small, creating bottlenecks: $^{88}$Sr, $^{138}$Ba, $^{208}$Pb.

Branch Points

At certain unstable isotopes, the beta-decay and neutron-capture timescales become comparable, creating branch points where the s-process path splits. Key examples include$^{85}$Kr ($\tau_{1/2} = 10.76$ yr), $^{151}$Sm, and$^{176}$Lu. Branch point ratios serve as thermometers and clocks for the s-process environment.

The r-Process (Rapid Neutron Capture)

In the r-process, neutron capture is much faster than beta decay ($\tau_{\text{capture}} \ll \tau_\beta$). The capture path runs far from stability along lines of constant neutron separation energy$S_n \approx 2\text{-}3$ MeV, until beta decay becomes energetically favored.

r-Process Sites

Neutron Star Mergers (confirmed)

  • - GW170817: gravitational waves + kilonova AT2017gfo
  • - Infrared afterglow from lanthanide-rich ejecta
  • - $n_n \sim 10^{24}\text{-}10^{30}$ cm$^{-3}$
  • - Estimated $\sim 0.01\text{-}0.05\,M_\odot$ of r-process material per event

Core-Collapse Supernovae (debated)

  • - Neutrino-driven wind from proto-neutron star
  • - May produce light r-process (A < 130)
  • - Magneto-rotational jets in special cases
  • - Chemical evolution models suggest multiple sites

Waiting-Point Approximation

In the equilibrium r-process, for each element Z, an equilibrium abundance distribution is established among isotopes via (n,$\gamma$) $\rightleftharpoons$($\gamma$,n):

$$\frac{Y(Z,A+1)}{Y(Z,A)} = n_n \frac{G(Z,A+1)}{2G(Z,A)} \left(\frac{2\pi\hbar^2}{m_n k_BT}\right)^{3/2} \exp\!\left(\frac{S_n(Z,A+1)}{k_BT}\right)$$

The abundance peak for each Z occurs where $S_n \approx 2\text{-}3$ MeV. The overall flow is governed by beta-decay rates of these waiting-point nuclei. The r-process abundance peaks are shifted to lower A compared to s-process peaks: A ~ 80, 130, 195 (vs s-process: 88, 138, 208).

Fission Recycling

In very neutron-rich environments, the r-process path reaches nuclei with $A \gtrsim 260$that undergo neutron-induced or beta-delayed fission. The fission fragments ($A \sim 100\text{-}150$) are recaptured, creating a fission recyclingloop that enhances abundances near $A \sim 130$ and contributes to the robustness of the r-process abundance pattern. This also limits the r-process to producing elements up to $A \sim 270$.

The p-Process and Other Processes

About 35 proton-rich stable isotopes cannot be produced by neutron capture. These p-nuclei require different mechanisms.

$\gamma$-Process (Photodisintegration)

In the O/Ne-rich layers of core-collapse supernovae, shock-heated material reaches$T \sim 2\text{-}3 \times 10^9$ K. Photodisintegration reactions strip neutrons and protons from pre-existing s-process and r-process seed nuclei:

$$(\gamma,n), \; (\gamma,p), \; (\gamma,\alpha)$$

This is the dominant mechanism for most p-nuclei, though it underproduces the lightest ones ($^{92,94}$Mo, $^{96,98}$Ru).

$\nu p$-Process

In the proton-rich neutrino-driven wind from a proto-neutron star, antineutrino captures on protons create neutrons: $\bar\nu_e + p \to n + e^+$. These neutrons allow (n,p) reactions that bypass slow beta decays, enabling proton-rich nucleosynthesis up to $A \sim 100$. This may explain the light p-nuclei underproduced by the $\gamma$-process.

i-Process (Intermediate)

Recently identified process with neutron densities $n_n \sim 10^{13}\text{-}10^{15}$ cm$^{-3}$, intermediate between s and r. May occur during proton ingestion events in low-metallicity AGB stars or rapidly accreting white dwarfs. Produces elements with distinct isotopic patterns that explain some observed anomalies in carbon-enhanced metal-poor (CEMP) stars.

Nuclear Cosmochronology

Long-lived radioactive isotopes serve as clocks for dating nucleosynthesis events and constraining the age of the elements and the Galaxy.

Key Chronometers

Isotope$\tau_{1/2}$ (Gyr)DaughterProcessApplication
$^{232}$Th14.05$^{208}$PbrAge of Galaxy
$^{238}$U4.468$^{206}$PbrAge of Solar System
$^{235}$U0.704$^{207}$PbrMeteorite dating
$^{187}$Re43.3$^{187}$Oss/rGalactic chemical evolution

Th/U Cosmochronometry in Metal-Poor Stars

By measuring the Th/Eu and U/Th ratios in ultra-metal-poor halo stars and comparing with r-process production ratios, we can estimate the age of the r-process event that enriched the star's birth cloud:

$$t = \frac{\tau}{\ln 2} \ln\!\left(\frac{({}^{232}\text{Th}/{}^{238}\text{U})_{\text{prod}}}{({}^{232}\text{Th}/{}^{238}\text{U})_{\text{obs}}}\right) \cdot \frac{\tau_{232}\tau_{238}}{\tau_{232}-\tau_{238}}$$

The star CS 31082-001 yields an age of $\sim 14 \pm 3$ Gyr, consistent with the age of the universe from the CMB ($13.8$ Gyr).

The Solar System Abundance Pattern

The solar abundance distribution encodes the integrated nucleosynthesis history of the Galaxy over ~9 billion years before the Solar System formed. Key features:

  • - H and He dominance: 98.5% by mass β€” from BBN and stellar H-burning
  • - Li-Be-B gap: extremely low abundances (fragile, destroyed in stars; produced by cosmic-ray spallation)
  • - Alpha-element peaks: C, O, Ne, Mg, Si, S, Ca (multiples of $\alpha$)
  • - Iron peak: local maximum at Fe-Ni group (NSE products)
  • - Odd-even staggering: even-Z and even-N nuclei are more abundant (pairing energy)
  • - s-process peaks: Sr-Y-Zr (N=50), Ba-La-Ce (N=82), Pb (N=126)
  • - r-process peaks: Se-Br-Kr (A~80), Te-I-Xe (A~130), Os-Ir-Pt (A~195)
  • - Exponential decline: abundances fall ~10 orders of magnitude from Fe to U

Python Simulation: BBN Abundances

Simplified Big Bang nucleosynthesis showing light element abundance evolution and dependence on the baryon-to-photon ratio.

BBN Light Element Abundances

Python

Big Bang nucleosynthesis abundance evolution and eta dependence

script.py130 lines

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Python Simulation: Gamow Peak & Stellar Energy Generation

Visualization of the Gamow peak for thermonuclear reactions and energy generation rates for pp chain vs CNO cycle as functions of temperature.

Gamow Peak & Stellar Energy Generation

Python

Thermonuclear reaction rates, pp vs CNO, binding energy curve, burning timescales

script.py122 lines

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Python Simulation: Solar Abundance Pattern

The solar system elemental abundances as a function of atomic number, showing the key features produced by different nucleosynthesis processes.

Solar System Abundance Pattern

Python

Elemental abundances showing signatures of different nucleosynthesis processes

script.py77 lines

Click Run to execute the Python code

Code will be executed with Python 3 on the server

Fortran: Thermonuclear Reaction Rates

Computes the Gamow peak energy, window width, and thermally-averaged cross section for key stellar reactions at various temperatures.

Thermonuclear Reaction Rates

Fortran

Gamow peak energies and reaction rate parameters for stellar burning reactions

thermonuclear_rates.f9069 lines

Click Run to execute the Fortran code

Code will be compiled with gfortran and executed on the server

Fortran: s-Process Neutron Capture Chain

Simulates slow neutron capture nucleosynthesis from iron seed nuclei, showing how magic-number bottlenecks create the s-process abundance peaks.

s-Process Neutron Capture Chain

Fortran

Simulates slow neutron capture nucleosynthesis from iron seed nuclei

s_process_chain.f9077 lines

Click Run to execute the Fortran code

Code will be compiled with gfortran and executed on the server

Practice Problems

Problem 1:Calculate the Gamow peak energy $E_0$ for the $p + p$ reaction at the solar core temperature $T = 1.5 \times 10^7$ K.

Solution:

Step 1: The Gamow peak energy is $E_0 = \left(\frac{b\,k_BT}{2}\right)^{2/3}$ where the Gamow energy parameter is $b = \sqrt{E_G} = \pi\alpha Z_1 Z_2\sqrt{2\mu c^2}$.

Step 2: For $p + p$: $Z_1 = Z_2 = 1$, $\mu = m_p/2 = 469.5$ MeV/$c^2$. The Gamow energy: $E_G = (2\pi\alpha)^2 \cdot 2\mu c^2/2 = 493$ keV.

Step 3: Solar core: $k_BT = 1.293$ keV. So $E_0 = \left(\frac{\sqrt{493} \times 1.293}{2}\right)^{2/3}$ keV.

Step 4: $E_0 = \left(\frac{22.2 \times 1.293}{2}\right)^{2/3} = (14.36)^{2/3} \approx 5.9$ keV.

Answer: $E_0 \approx 6$ keV, which is far below the Coulomb barrier ($\sim 550$ keV) but far above $k_BT = 1.3$ keV, illustrating why quantum tunneling is essential for stellar fusion.

Problem 2:Calculate the total energy released in the pp chain: $4p \to \,^4\text{He} + 2e^+ + 2\nu_e$.

Solution:

Step 1: Mass of 4 protons: $4 \times 938.272 = 3753.088$ MeV/$c^2$.

Step 2: Mass of products: $m(^4\text{He}) = 3727.379$ MeV/$c^2$, $2m_e = 2 \times 0.511 = 1.022$ MeV/$c^2$.

Step 3: $Q = 3753.088 - 3727.379 - 1.022 = 24.687$ MeV.

Step 4: Including positron annihilation ($2e^+ + 2e^- \to 4\gamma$), add $2 \times 1.022 = 2.044$ MeV. Total deposited energy: $\approx 26.73$ MeV. Neutrinos carry away $\sim 0.59$ MeV on average.

Answer: $Q = 26.73$ MeV total, of which $\sim 26.1$ MeV is deposited as thermal energy and $\sim 0.59$ MeV is lost to neutrinos.

Problem 3:The CNO cycle rate scales as $\epsilon_{\text{CNO}} \propto T^{16.7}$ near $T = 1.5 \times 10^7$ K. If the core temperature increases by 10%, by what factor does the CNO energy generation rate increase?

Solution:

Step 1: The power-law approximation gives $\epsilon \propto T^\nu$ with $\nu \approx 16.7$ for the CNO cycle.

Step 2: If $T \to 1.1\,T$: $\epsilon'/\epsilon = (1.1)^{16.7}$.

Step 3: Compute: $\ln(1.1^{16.7}) = 16.7 \times \ln(1.1) = 16.7 \times 0.0953 = 1.592$.

Step 4: $\epsilon'/\epsilon = e^{1.592} \approx 4.91$.

Answer: A 10% temperature increase multiplies the CNO rate by a factor of $\approx 4.9$. This extreme sensitivity explains why the CNO cycle dominates only in stars more massive than the Sun ($T_c > 1.7 \times 10^7$ K).

Problem 4:Calculate the binding energy per nucleon for $^{56}$Fe (total binding energy $B = 492.26$ MeV) and $^{4}$He ($B = 28.30$ MeV). Explain the significance.

Solution:

Step 1: For $^{56}$Fe: $B/A = 492.26/56 = 8.790$ MeV/nucleon.

Step 2: For $^{4}$He: $B/A = 28.30/4 = 7.075$ MeV/nucleon.

Step 3: $^{56}$Fe has the highest $B/A$ of any nuclide (actually $^{62}$Ni is slightly higher at 8.795 MeV/nucleon).

Step 4: Fusion releases energy for $A < 56$ (moving up the curve), while fission releases energy for $A > 56$ (also moving toward the peak).

Answer: $B/A(^{56}\text{Fe}) = 8.79$ MeV/nucleon, $B/A(^{4}\text{He}) = 7.07$ MeV/nucleon. Iron sits at the peak of the binding energy curve, which is why stellar nucleosynthesis halts at iron β€” no further energy can be extracted from fusion.

Problem 5:Estimate the width of the Gamow window $\Delta E$ for the $^{12}\text{C}(\alpha, \gamma)\,^{16}\text{O}$ reaction at $T = 2 \times 10^8$ K.

Solution:

Step 1: The Gamow window width is $\Delta E = \frac{4}{\sqrt{3}}\sqrt{E_0\,k_BT}$.

Step 2: For $^{12}\text{C} + \alpha$: $Z_1 Z_2 = 12$, $\mu = 12 \times 4/(12 + 4) = 3$ u. The Gamow energy: $E_G = 2\mu c^2(\pi\alpha Z_1 Z_2)^2 \approx 32.1$ MeV.

Step 3: At $T = 2 \times 10^8$ K: $k_BT = 17.2$ keV. Gamow peak: $E_0 = (E_G(k_BT)^2/4)^{1/3} = (32100 \times 17.2^2/4)^{1/3} \approx 300$ keV.

Step 4: Width: $\Delta E = \frac{4}{\sqrt{3}}\sqrt{300 \times 17.2} \approx 2.31 \times 71.8 \approx 166$ keV.

Answer: $\Delta E \approx 170$ keV. This narrow window centered at $E_0 \approx 300$ keV is where the reaction rate is dominated, making the $^{12}\text{C}(\alpha,\gamma)\,^{16}\text{O}$ cross section at these energies one of the most important and difficult measurements in nuclear astrophysics.

Share:XRedditLinkedIn
← Nuclear ReactorsNuclear Medicine β†’
Rate this chapter: