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Chapter 21: Problem 31

Using the concept of magic numbers, explain why alpha emission is relatively common, but proton emission is nonexistent.

Short Answer

Expert verified
Using the concept of magic numbers, alpha emission is relatively common because it often results in a more stable daughter nucleus with magic numbers of protons or neutrons. These magic numbers (2, 8, 20, 28, 50, 82, and 126) provide increased stability properties in atomic nuclei. In contrast, proton emission, which hasn't been observed in nature, is nonexistent because it's less likely to produce a daughter nucleus with enhanced stability properties due to magic numbers. This is because proton emission involves the loss of only 1 proton, significantly reducing the potential for achieving a magic number in the daughter nucleus compared to alpha emission, which involves the loss of 2 protons and 2 neutrons.

Step by step solution

01

1. Understand Magic Numbers in Nuclear Physics

Magic numbers are specific numbers of nucleons (protons and neutrons) that have unique stability properties in atomic nuclei. When a nucleus has a magic number of protons or neutrons, it demonstrates increased stability compared to other nuclei. The magic numbers are 2, 8, 20, 28, 50, 82, and 126.
02

2. Determine the Stability of Atomic Nuclei

Atomic nuclei are stable when the number of protons and neutrons results in a strong binding energy that holds the nucleons together. When a nucleus is unstable, it tends to decay by emitting certain particles to become more stable.
03

3. Compare Alpha Emission and Proton Emission

Alpha emission is a type of radioactive decay where an unstable nucleus emits an alpha particle, which consists of 2 protons and 2 neutrons. The alpha emission results in a new nucleus with the atomic number decreased by 2 and the mass number decreased by 4. On the other hand, proton emission is a hypothetical type of decay in which a nucleus would emit a single proton. This emission would result in a new nucleus with an atomic number decreased by 1. Proton emission has not been observed in nature, which is what the exercise asks us to explain.
04

4. Analyze the Impact of Alpha and Proton Emission on Nuclei Stability

We know that nuclei with magic numbers of protons and neutrons demonstrate increased stability. It's important to see how the changes caused by alpha emission and proton emission affect the stability of parent and daughter nuclei. - In alpha emission, the nucleus loses 2 protons and 2 neutrons. This change in composition can result in the daughter nucleus having a magic number of protons or neutrons, making it more stable than the parent nucleus. - In proton emission, the nucleus would lose 1 proton. This change in composition has significantly less potential to result in a daughter nucleus with a magic number of protons or neutrons, making the decay process less favorable.
05

5. Conclusion

Using the concept of magic numbers, we can explain that alpha emission is relatively common because it often results in a more stable daughter nucleus, with magic numbers of protons or neutrons. However, proton emission is nonexistent because it's less likely to produce a daughter nucleus with enhanced stability properties due to magic numbers.

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Most popular questions from this chapter

Complete and balance the following nuclear equations by supplying the missing particle: (a) \({ }_{58}^{252} \mathrm{Cf}+{ }_{5}^{10} \mathrm{~B} \longrightarrow 3{ }_{0}^{1} \mathrm{n}+\) ? (b) \({ }_{1}^{2} \mathrm{H}+{ }_{2}^{3} \mathrm{He} \longrightarrow{ }_{2}^{4} \mathrm{He}+\) ? (c) \({ }_{1}^{1} \mathrm{H}+{ }_{5}^{11} \mathrm{~B} \longrightarrow 3\) ? (d) \({ }_{53}^{122} \mathrm{I} \longrightarrow{ }_{54}^{122} \mathrm{Xe}+\) ? (e) \({ }_{26}^{59} \mathrm{Fe} \longrightarrow{ }_{-1}^{0} \mathrm{e}+\) ?

Based on the following atomic mass values - \({ }^{1} \mathrm{H}, 1.00782\) amu; \({ }^{2} \mathrm{H}, 2.01410 \mathrm{amu} ;{ }^{3} \mathrm{H}, 3.01605 \mathrm{amu} ;{ }^{3} \mathrm{He}, 3.01603 \mathrm{amu} ;\) \({ }^{4} \mathrm{He}, 4.00260 \mathrm{amu}-\) and the mass of the neutron given in the text, calculate the energy released per mole in each of the following nuclear reactions, all of which are possibilities for a controlled fusion process: (a) \({ }_{1}{ }_{1} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} \mathrm{n}\) (b) \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{2}^{3} \mathrm{He}+{ }_{0}^{1} \mathrm{n}\) (c) \({ }_{1}^{2} \mathrm{H}+{ }_{2}^{3} \mathrm{He} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{1}^{1} \mathrm{H}\) \(21.53\) Which of the following nuclei is likely to have the largest mass defect per nucleon: (a) \({ }^{59} \mathrm{Co}\), (b) \({ }^{11} \mathrm{~B}\), (c) \({ }^{118} \mathrm{Sn}\), (d) \({ }^{243} \mathrm{Cm}\) ? Explain your answer.

Decay of which nucleus will lead to the following products: (a) bismuth- 211 by beta decay; (b) chromium 50 by positron emission; (c) tantalum-179 by electron capture; (d) radium-226 by alpha decay?

Cobalt-60 is a strong gamma emitter that has a half-life of \(5.26 \mathrm{yr}\). The cobalt- 60 in a radiotherapy unit must be replaced when its radioactivity falls to \(75 \%\) of the original sample. If an original sample was purchased in June 2013, when will it be necessary to replace the cobalt-60?

(a) What is the function of the moderator in a nuclear reactor? (b) What substance acts as the moderator in a pressurized water generator? (c) What other substances are used as a moderator in nuclear reactor designs?
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