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Chapter 21: Problem 72

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

Expert verified
The more obvious product of proton fusion, \({ }^{2}\mathrm{He}\), is unstable due to its N/Z ratio of 0, which indicates that the repulsive force between the two protons in the nucleus goes unbalanced and leads to instability. In contrast, the actual product \({ }^{2}\mathrm{H}\) has a stable N/Z ratio of 1, with the neutron helping to balance the repulsive force, making it the preferred and more stable product of proton fusion in a star.

Step by step solution

01

Understand Proton Fusion

Proton fusion is a nuclear reaction that takes place in the core of stars. During this process, two protons (hydrogen nuclei) come together under extreme temperatures and pressures to form a new nucleus. In the specific case we're investigating, the fusion reaction produces deuterium (\({ }^{2}\mathrm{H}\)), which is a stable isotope of hydrogen consisting of one proton, one neutron, and one electron, and a positron (a positively charged electron). The potential product we are questioning, \({ }^{2}\mathrm{He}\), is an unstable isotope of helium. It consists of two protons and two electrons, but no neutrons.
02

Understand Nuclear Stability

Nuclear stability is determined by the balance between the attractive strong nuclear force, which holds protons and neutrons together in the nucleus, and the repulsive electromagnetic force, which causes protons to repel each other. The stability of a nucleus can be predicted by examining the neutron-to-proton ratio (N/Z). If the number of neutrons relative to protons falls within a certain range, usually close to 1 for lighter elements, the nucleus is more likely to be stable. As elements become heavier, the stable N/Z ratio tends to increase slightly.
03

Examine the N/Z Ratio for \({ }^{2}\mathrm{He}\) and \({ }^{2}\mathrm{H}\)

For the potential product \({ }^{2}\mathrm{He}\), the nucleus contains two protons and no neutrons, which gives us an N/Z ratio of 0/2 or 0. This ratio is far from 1, indicating that the nucleus is unstable due to the overwhelming repulsive force among the protons. On the other hand, for the actual product \({ }^{2}\mathrm{H}\), the nucleus contains one proton and one neutron, which gives us an N/Z ratio of 1/1 or 1. This ratio is close to 1, indicating that the nucleus is stable with a balanced attractive and repulsive forces in the nucleus.
04

Explain the Stability of \({ }^{2}\mathrm{H}\)

The reason why \({ }^{2}\mathrm{H}\) is more stable than \({ }^{2}\mathrm{He}\) can be explained by the strong nuclear force and the neutron-to-proton ratio. The deuterium nucleus (\({ }^{2}\mathrm{H}\)) has one proton and one neutron, giving it a stable N/Z ratio of 1. This means that the neutron helps to balance the repulsive force between the protons, making the nucleus stable.
05

Conclude the Reason for \({ }^{2}\mathrm{He}\) Instability

The more obvious product of proton fusion, \({ }^{2}\mathrm{He}\), is unstable because its nucleus has an N/Z ratio of 0, far away from the ideal N/Z ratio for stability. This means that the repulsive force between the two protons in the \({ }^{2}\mathrm{He}\) nucleus goes unbalanced, leading the nucleus to be unstable. On the other hand, the actual product \({ }^{2}\mathrm{H}\) has a stable N/Z ratio, making it the preferred and more stable product of proton fusion in a star.

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Most popular questions from this chapter

What particle is produced during the following decay processes: (a) sodium- 24 decays to magnesium- \(24 ;\) (b) mercury188 decays to gold-188; (c) iodine-122 decays to xenon-122; (d) plutonium-242 decays to uranium-238?

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