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Chapter 21: Problem 12

Why is it preferable to use nuclear binding energy per nucleon for a comparison of the stabilitiesof different nuclei?

Short Answer

Expert verified
Using binding energy per nucleon allows for consistent comparison of nuclear stability across different nucleus sizes.

Step by step solution

01

Understanding Binding Energy

Nuclear binding energy is the energy required to disassemble a nucleus into its individual protons and neutrons. It represents how strongly the nucleus is held together.
02

Importance of Per Nucleon

When comparing the stability of different nuclei, it is important to use binding energy per nucleon, not just total binding energy. This is because the total binding energy depends on the number of nucleons and does not accurately compare nuclei of different sizes.
03

Comparison of Stability

Binding energy per nucleon allows us to compare the stability of different nuclei regardless of their size. A higher binding energy per nucleon indicates a more stable nucleus because more energy is required to break it apart.
04

Conclusion

Using nuclear binding energy per nucleon normalizes the energy across different nuclei sizes, providing a consistent and comparable measure of nuclear stability.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binding Energy per Nucleon
Binding energy per nucleon is a crucial concept in nuclear physics. It refers to the average energy that binds each proton and neutron (collectively called nucleons) in a nucleus. By considering energy on a per nucleon basis, we achieve a more accurate measure of nuclear stability, especially when comparing nuclei of different sizes. For example, a small nucleus might have a lower total binding energy compared to a large one. However, its binding energy per nucleon might be higher, indicating greater stability. This measure normalizes the binding energy, making it independent of the number of nucleons and allowing for a fair comparison.
Nuclear Stability
Nuclear stability is fundamentally about how resistant a nucleus is to disassembly. A stable nucleus requires a significant amount of energy to be broken down into its component protons and neutrons. The binding energy per nucleon helps quantify this stability. Higher values mean more energy is needed for nuclear disassembly, reflecting greater stability. Understanding nuclear stability is important for several applications, including nuclear power generation, where stable nuclei release energy through processes like fission.
Nucleus Disassembly
Disassembling a nucleus involves breaking it apart into its individual nucleons. This process requires substantial energy, equivalent to the binding energy of the nucleus. Energy must be supplied to overcome the strong nuclear forces holding the nucleons together. When nucleons are tightly bound, more energy is needed. The difficulty of nucleus disassembly varies with the binding energy per nucleon. Nuclei with higher binding energy per nucleon resist disassembly more strongly, highlighting their stability.
Comparison of Stability
When comparing the stability of different nuclei, binding energy per nucleon provides a standardized metric. It levels the playing field, as it accounts for both size differences and nucleon count. For instance, to compare two different elements' nuclei, looking at their total binding energy alone could be misleading due to size discrepancies. The binding energy per nucleon, however, removes these disparities. Nuclei with a higher binding energy per nucleon indicate greater structural integrity and stability. This allows scientists to predict which nuclei are more likely to remain intact without undergoing spontaneous decay.

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Most popular questions from this chapter

Outline the principle for dating materials using radioactive isotopes.

Cobalt- 60 is an isotope used in diagnostic medicine and cancer treatment. It decays with \(\gamma\) ray emission. Calculate the wavelength of the radiation in nanometers if the energy of the \(\gamma\) ray is \(2.4 \times 10^{-13}\) \(\mathrm{J} /\) photon.

In the thorium decay series, thorium -232 loses a total of \(6 \alpha\) particles and \(4 \beta\) particles in a 10 -stage process. What is the final isotope produced?

These cquations are for nuclear reactions that are known to occur in the explosion of an atomic bomb. Identify \(X\). (a) \({ }^{235} \mathrm{U}+{ }^{1} \mathrm{n} \longrightarrow{ }_{56}^{140} \mathrm{Ba}+3_{0}^{1} \mathrm{n}+\mathrm{X}\) (b) \({ }^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{53}^{144} \mathrm{Cs}+{ }_{37}^{90} \mathrm{Rb}+2 \mathrm{X}\) (c) \({ }^{235} \mathrm{U}+{ }^{1} \mathrm{n} \longrightarrow{ }^{\mathrm{s} 7} \mathrm{Br}+3{ }^{1} \mathrm{n}+\mathrm{X}\) (d) \({ }^{235} \mathrm{U}+{ }^{1} \mathrm{n} \longrightarrow{ }^{160} \mathrm{Sm}+\frac{72}{30} \mathrm{Zn}+4 \mathrm{X}\)

What is the difference between radioactive decay and nuclear transmutation?
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