Helium and the accounting of light nuclei

A reaction sequence needs both neutrons and protons

Once deuterium can accumulate, a connected set of reactions can build other light nuclei, including helium-3, tritium and helium-4. Competing forward and reverse reactions and the time available determine which nuclei remain.

A helium-4 nucleus contains two protons and two neutrons. A hydrogen-1 nucleus is one proton. These are nuclear identities; the early ionized population should not be pictured as neutral atoms with permanent electron orbits.

The inventory of available neutrons gives a useful approximate handle on helium production because each helium-4 nucleus requires two. It does not eliminate the need to calculate capture rates or the residual lighter and trace nuclei.

The supplied account assumes that every available neutron ends in helium-4 and that enough protons are present. It is a limiting H/He inventory rather than a prediction of actual primordial composition.

Conserve nucleons and nuclear charge

Take separate initial compositions with the same total of 240 nucleons. Let their neutron counts be 24, 30 or 40, with the remaining nucleons initially free protons. Ignore neutron decay and all residual nuclides for this chosen counting exercise.

If N_n ≤ N_p and every neutron is captured into helium-4, the helium count is N_He = N_n/2. That consumes N_n protons as well, leaving N_H = N_p − N_n hydrogen-1 nuclei.

The final nucleon count is N_H + 4N_He = N_p + N_n. The final nuclear charge is N_H + 2N_He = N_p, the same as the initial proton count. Nuclear capture has not turned proton charge into nothing.

The chosen even neutron counts make integer nuclei in the table. The rule is an inventory assumption, not a microscopic account of reaction paths or the energy emitted in binding.

The denominator changes the meaning of an abundance

The helium-to-hydrogen number ratio is N_He/N_H. The helium fraction among final nuclei is N_He/(N_H + N_He). The helium nucleon fraction is Y_He^(b) = 4N_He/(N_H + 4N_He). These answer different questions.

For 30 neutrons and 210 protons, the assumption gives 15 helium-4 nuclei and 180 hydrogen-1 nuclei. The number ratio He/H is 1/12, helium is 1/13 of the final nuclei, and helium contains 1/4 of all nucleons.

Thus a helium nucleon fraction of one quarter does not mean one quarter of the nuclei are helium. Each helium nucleus contributes four nucleons to the third denominator but only one nucleus to the second.

The other rows change the initial neutron inventory while holding the nucleon total fixed. They are separate compositions, rather than successive time stages of one reaction calculation.

A conventional mass approximation needs a qualification

If r = N_n/N_p, the assumed helium nucleon fraction becomes Y_He^(b) = 2r/(1 + r). The He/H number ratio is r/[2(1 − r)] when hydrogen remains. These formulas retain their distinct denominators.

Treating all nucleons as equal-mass and neglecting binding and electron corrections makes the nucleon fraction an approximation to a rest-mass fraction. A precise rest-mass fraction instead needs the actual constituent masses and consistent conventions for nuclei, atoms and electrons.

At N_n = N_p this inventory leaves only helium-4 and no hydrogen-1. The He/H number ratio is then undefined, with a divergent limiting ratio as hydrogen tends to zero. The helium nucleon fraction remains well defined and equals 1.

If neutrons outnumber protons, the assumed complete capture into helium-4 with only residual hydrogen is impossible. The proton supply limits that construction, so its formulas must not be continued beyond their domain.

Residual light elements require reaction dynamics

A full nucleosynthesis calculation follows connected abundances and rate equations, including reverse reactions and decay channels. The small residual deuterium and other light-nucleus abundances are not supplied by the helium inventory alone.

Reaction coefficients depend on cross sections averaged over the relevant relative velocities. Densities and the expansion rate control how many reactions can occur before conditions change. Those inputs can affect several nuclides together.

The idealized table contains only H and He because that is its declared scope, not because actual primordial material contains no trace nuclei. Its fractions should not be labelled measured abundances.

Before comparing a predicted or inferred helium result, identify whether it is a number ratio, a nucleus fraction, a nucleon-weighted fraction or a precise mass fraction. Matching units without matching denominators is insufficient.

Worked reasoning

Try the reasoning

1. Does Y_He^(b) = 1/4 mean one quarter of the final nuclei are helium in the worked case?

   Show the answer and reasoning

   No; only 1/13 of the final nuclei are helium.

   The nucleon fraction weights each helium nucleus by four.

2. Does the inventory calculate residual deuterium or lithium abundances?

   Show the answer and reasoning

   No.

   Those require the connected reaction network and its rates.

3. Is the nucleon fraction exactly the rest-mass fraction without further qualification?

   Show the answer and reasoning

   No.

   The conventional identification neglects nucleon-mass differences, binding and consistent electron or atomic mass corrections.

What to carry forward

• A helium-4 nucleus uses two neutrons and two protons.
• Number ratios, nucleus fractions and nucleon fractions have different denominators.
• The ideal inventory checks conservation but does not replace a full nuclear reaction calculation.