research32 min studyLesson 598 in Physics

Proton-rich nuclei and photodisintegration networks

Track competing gamma-induced particle emissions and relate stellar reverse rates to capture without inventing a single universal p-process.

Before you begin

On this page

The question for today

What you’ll understand

Balance neutron, proton, and alpha photodisintegration channels.
Compute competing channel fluxes from rates per nucleus and abundances.
Distinguish gamma-process mechanisms, stellar excited-state rates, and a complete origin model for proton-rich isotopes.

Some naturally occurring proton-rich isotopes are bypassed by the dominant neutron-capture paths. Their production is often discussed under the broad p-process label, which should not be read as a claim that direct proton capture makes every one of them. A gamma process can move pre-existing heavy seeds toward proton-rich isotopes through photodisintegration. Competing neutron, proton, and alpha emission redirect the flow, while the temperature history and later captures determine which products survive.

Photons open several distinct exits

A thermal photon bath has a high-energy tail capable of breaking a nucleus apart when the threshold and transition strength permit. Neutron emission reduces mass number by one without changing proton number. Proton emission reduces both by one, and alpha emission reduces mass number by four and proton number by two. These changes must also be recorded in the free-particle inventories. A network that follows only the heavy residue but forgets emitted particles cannot check baryon or charge conservation.

For a prescribed thermal radiation field, each photodisintegration channel is a rate per target nucleus. Its volumetric flow multiplies that rate by the target abundance or density. A plot of large rate coefficients alone can therefore be misleading when the associated targets are scarce. As the path removes neutrons, separation energies and charged-particle barriers can change which emission channel wins. Rates depend on temperature, nuclear structure, and the thermally populated target states, not simply on the count of available photons above a threshold.

(A,Z)\xrightarrow{\gamma,n}(A-1,Z),\quad (A,Z)\xrightarrow{\gamma,p}(A-1,Z-1),\quad (A,Z)\xrightarrow{\gamma,\alpha}(A-4,Z-2),\qquad r_{\gamma x}=n_{A,Z}\lambda_{\gamma x}

The emitted particle carries the missing baryon number and charge. Lambda is inverse time, while r is events per volume per time; the photon bath is prescribed and not depleted by these reactions.↔ Scroll wide equations horizontally.

Capture and photodisintegration are linked

For a capture reaction and its inverse, microscopic reciprocity connects the stellar rates through masses, statistical factors, and thermal partition functions. Schematically, the inverse rate contains a thermal phase-space factor times the capture reactivity and an exponential of minus the capture energy release over thermal energy. This makes small separation-energy differences important. The relation is not an invitation to replace a stellar rate with a laboratory ground-state cross section without accounting for thermally populated target states.

The usual compact reciprocity expression also uses assumptions about the photon distribution and reaction regime. Replacing the Planck denominator by its high-energy Maxwell tail is justified when the contributing photon energies make the omitted stimulated factor negligible; near low thresholds that step deserves checking. At fixed temperature, the photodisintegration hazard per target has no additional projectile-density factor, whereas the opposing particle capture hazard does. Changing density can therefore move the balance even when the photon temperature stays unchanged.

n_A n_a\langle\sigma v\rangle_{Aa}=n_B\lambda_{\gamma a}\quad\text{at equilibrium},\qquad \lambda_{\gamma a}\propto T^{3/2}e^{-Q/(k_BT)}\langle\sigma v\rangle_{Aa}

This schematic reverse relation omits mass, spin, and nuclear partition-function prefactors and assumes a valid high-energy photon-tail approximation. The exact equilibrium flow equation shown is for distinguishable A and a with consistently defined stellar rates.↔ Scroll wide equations horizontally.

Seeds and trajectories limit origin claims

Photodisintegration rearranges an existing heavy-nucleus inventory; it does not create heavy nuclei from an empty parcel simply because the radiation is hot. Seed abundances inherited or produced before heating thus matter. Material exposed too weakly can retain its seeds, while material heated too strongly can destroy the desired products. Ejecta from different zones can sample different histories, and their mass-weighted combination is a separate step from the reaction calculation in a single parcel.

The broad class of proton-rich isotopes can receive contributions from several mechanisms, including gamma-induced flows and other proton- or neutrino-assisted paths under different conditions. Identifying one successful local channel does not establish a unique astrophysical origin for the entire class. Likewise, the branch probability of one nucleus is not its final abundance if products are subsequently destroyed or replenished. A defensible model states its seed composition, thermal evolution, reaction network, and uncertainty treatment before comparing final isotope ratios with observations.

\lambda_{\rm tot}=\lambda_{\gamma n}+\lambda_{\gamma p}+\lambda_{\gamma\alpha},\qquad f_x=\frac{\lambda_{\gamma x}}{\lambda_{\rm tot}},\qquad F_x(t)=f_x\left(1-e^{-\lambda_{\rm tot}t}\right)

For constant independent competing hazards and no feeding, fx is the eventual exit fraction and Fx is the fraction exited through x by time t. Daughter reactions and additional channels require an expanded network.↔ Scroll wide equations horizontally.

Work it through

A three-way emission branch after one half-depletion time

A schematic target with $(A,Z)=(100,44)$ has rates $(\lambda_{\gamma n},\lambda_{\gamma p},\lambda_{\gamma\alpha})=(6,3,1)\,\mathrm{s}^{-1}$ . Start with 1000 targets and neglect feeding and daughter reactions. At $t=\ln2/(10\,\mathrm{s}^{-1})$ , find remaining parents and mean daughter counts with their nuclear labels.

The total rate is $10\,\mathrm{s}^{-1}$ , so the chosen time leaves survival $e^{-\ln2}=1/2$ . There are 500 expected parents remaining and 500 that have taken an exit.
The three eventual channel fractions are $0.6$ , $0.3$ , and $0.1$ . Because the hazards are constant, apply them to the 500 exited parents, giving 300 neutron-emission, 150 proton-emission, and 50 alpha-emission daughters.
The daughter labels are respectively $(99,44)$ , $(99,43)$ , and $(96,42)$ . Recording 300 free neutrons, 150 free protons, and 50 alpha particles restores the baryon and charge removed from the heavy residues.

The result

Mean counts are 500 parents and 300, 150, and 50 daughters in the neutron, proton, and alpha channels, with the emitted particles retained in the bookkeeping.

The living notebook

Explore the model

Interactive lab

Interactive lab · probability

Conditional counts for one selected exit

Reset and set success probability to 0.3 and trials to 10. Predict mean three proton-emission labels and variance 2.1, conditional on ten independently exited parents in the stated constant-branch model. The instrument shows exact binomial probabilities. It does not include survival, the other two channels jointly, changing temperature, seed production, or a gamma-process abundance network.

Exact probabilityHighlighted outcomeMean np

Mean · np: 5
Variance · np(1−p): 2.5
Total probability: 1

Success probability p0.5Trials n10

t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forConditional branching statistics do not replace parent survival and subsequent daughter evolution.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Particle capture requires projectiles and has a density factor per target. Thermal photodisintegration already incorporates the prescribed radiation field into its temperature-dependent hazard; volumetric flow still multiplies by target density.

Your turn

Try it yourself

For a schematic distinguishable capture pair at one fixed temperature, take projectile density $n_a=10^{20}\,\mathrm{cm}^{-3}$ , capture reactivity $3\times10^{-20}\,\mathrm{cm}^3\mathrm{s}^{-1}$ , and inverse photodisintegration hazard $6\,\mathrm{s}^{-1}$ . Find the equilibrium ratio $n_B/n_A$ , then double only projectile density.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Equate the two volumetric flows. The given forward and inverse coefficients are held at the same temperature; only the reservoir density changes.

Compare with the solution

The initial capture hazard per A is $n_a\langle\sigma v\rangle=3\,\mathrm{s}^{-1}$ , giving $n_B/n_A=3/6=1/2$ . Doubling projectile density doubles that hazard to six and makes the equilibrium ratio one. This illustrates density dependence of the equilibrium composition without treating the forward and reverse nuclear coefficients as independently adjustable thermal physics.

Keep the idea

Take it with you

Photodisintegration channels must conserve baryons and charge including emitted particles.
Stellar forward and reverse rates are linked and need thermal-state conventions.
A local gamma-process branch does not establish universal proton-rich isotope yields or origins.

Read further

Original sources and deeper explanations to follow at your own pace.

T. Rauscher et al.: Constraining the Astrophysical Origin of the p-Nuclei through Nuclear Physics and Meteoritic Data, sections 3, 5.2, and 6.1

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