Question

When two protons fuse in a star, the product is \({ }^{2} \mathrm{H}\) plus a positron (Equation 21.26). Why do you think the more obvious product of the reaction, \({ }^{2} \mathrm{He}\) is unstable?

Short answer

\({ }^{2} \mathrm{He}\) is unstable because it consists of two protons with no neutrons to balance the electrostatic repulsion between them, causing the nucleus to be unstable and break apart. In contrast, \({ }^{2} \mathrm{H}\) (deuterium) has one proton and one neutron in its nucleus, ensuring its stability. That is why, when two protons fuse in a star, the product is \({ }^{2} \mathrm{H}\) and a positron, not \({ }^{2} \mathrm{He}\).

Step-by-step solution

Understanding the Fusion Reaction

When two protons (hydrogen nuclei) fuse in a star, they undergo nuclear fusion to form a heavier nucleus. The product formed is \({ }^{2} \mathrm{H}\) (deuterium) and a positron (a positive charged electron). The reaction can be represented as: \[ p + p \rightarrow { }^{2} \mathrm{H} + e^{+} \] In this reaction, one of the protons is transformed into a neutron, which combines with the other proton to form a deuterium nucleus. The transformation of a proton into a neutron involves the process of beta-plus decay, which results in the emission of a positron.

Composition of \({ }^{2} \mathrm{He}\) and \({ }^{2} \mathrm{H}\)

\({ }^{2} \mathrm{He}\) (which is not formed) would have two protons and no neutrons in its nucleus. On the other hand, \({ }^{2} \mathrm{H}\) (deuterium) has one proton and one neutron in its nucleus. It's important to note, for the stability of a nucleus, the balance between the numbers of protons and neutrons is crucial. Too many protons will cause an excessive electrostatic repulsion between them, which destabilizes the nucleus.

Why \({ }^{2} \mathrm{He}\) Is Unstable

The reason \({ }^{2} \mathrm{He}\) is unstable is due to the strong electrostatic repulsion between the two protons in its nucleus. Since there are no neutrons in the nucleus to balance this repulsion, the electrostatic force overcomes the strong nuclear force holding the protons together, causing the nucleus to be unstable and break apart. In \({ }^{2} \mathrm{H}\), however, the presence of a neutron with the proton in the nucleus helps in balancing the forces and ensures the stability of the nucleus. This is why the more expected product, \({ }^{2} \mathrm{He}\), is unstable and not formed in the proton-proton fusion reaction in stars. Instead, the reaction results in the formation of the stable \({ }^{2} \mathrm{H}\) and a positron.

Key concepts

Proton-Proton Fusion

The process of proton-proton fusion is at the heart of stars like our Sun, where it provides the energy necessary for the sun to shine. This reaction involves two hydrogen nuclei (protons) coming together under immense temperatures and pressures to form a heavier element.

As the protons get closer, the electrostatic repulsion between their positive charges becomes significant, but the extreme conditions within stars allow the protons to overcome this repulsion. Once fused, a heavier nucleus is created, which is deuterium ({ }^{2} {H}), along with the release of a positron ({e^+}) and a neutrino. This positron emission is a result of one proton being converted into a neutron, a type of reaction known as beta-plus decay.

The balanced forces within the deuterium nucleus make it far more stable than a hypothetical helium-2 ({ }^{2} {He}) nucleus, which would contain two protons without any neutrons to mitigate the repulsive forces between them.

Nucleus Stability

The stability of an atomic nucleus is governed by a delicate balance between the attractive strong nuclear force and the repulsive electrostatic force between protons. The strong nuclear force, effective only at very short distances, binds protons and neutrons together. In contrast, the electrostatic force, governed by Coulomb's law, tends to push the positively charged protons apart.

For lighter elements, a roughly equal number of protons and neutrons generally ensures stability. As elements get heavier, more neutrons are needed to counteract the electrostatic repulsion between the increasing number of protons. Neutrons, being neutral, can be part of the nucleus without contributing to electrostatic repulsion, helping to bind the nucleus more effectively.

A hypothetical isotope of helium, { }^{2} {He}, with two protons and no neutrons, demonstrates a case where nuclear stability is compromised. The absence of neutrons would allow the electrostatic repulsion to dominate, leading to the nucleus's near-immediate disintegration.

Beta-Plus Decay

Beta-plus decay is a type of radioactivity in which a proton inside an atomic nucleus is transformed into a neutron, releasing a positron ({e^+}) and a neutrino. This process is one of the key reactions in the proton-proton fusion process inside stars.

In the confined conditions of a star's core, where proton-proton fusion takes place, a proton can absorb enough energy to undergo this transformation. The emission of a positron is evidence of this decay, as it indicates the conversion of one type of particle into another, assisting in the creation of a more stable nucleus.

Interestingly, beta-plus decay is the inverse of beta-minus decay, where a neutron turns into a proton. Both processes are fundamental in understanding nuclear reactions and how elements evolve in the universe.

Electrostatic Repulsion

Electrostatic repulsion is a fundamental force that plays a critical role in the structure of the atomic nucleus. It emerges due to the Coulomb interaction between charged particles, like protons, which have the same positive charge.

Within a nucleus, electrostatic repulsion is the force that tries to push protons apart from each other. This force increases dramatically as protons get closer, making it increasingly difficult for additional protons to join the nucleus without the neutralizing presence of neutrons.

The electrostatic force competes with the strong nuclear force, which works to hold the nucleus together. In a small nucleus like that of deuterium ({ }^{2} {H}), the strong force is sufficient to maintain stability, despite the electrostatic repulsion. However, in our non-existent helium-2 ({ }^{2} {He}) example, the lack of neutrons would mean that the electrostatic repulsion is unopposed, leading to an unstable nucleus.