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Chapter 10: Problem 17

Why does the fusion of light nuclei into heavier nuclei release energy?

Short Answer

Expert verified
In conclusion, the fusion of light nuclei into heavier nuclei releases energy because the final nucleus has a larger mass defect and subsequently higher binding energy per nucleon than the initial nuclei. This difference in binding energies is converted into energy, consistent with Einstein's mass-energy equivalence principle, expressed as \(E = mc^2\). The released energy appears as kinetic energy or light, depending on the fusion reaction.

Step by step solution

01

Understand the concept of binding energy

Binding energy is the energy required to disassemble a nucleus into its constituent protons and neutrons. It is the measure of the work required to overcome the strong nuclear force that holds the protons and neutrons together. The higher the binding energy, the stronger the force holding the nucleus together, making it more stable.
02

Mass defect and its relationship to binding energy

Mass defect refers to the difference between the mass of a nucleus and the sum of the masses of its individual nucleons (protons and neutrons). The mass defect occurs because the binding energy needed to hold the nucleus together is converted to mass according to Einstein's mass-energy equivalence principle, expressed as \(E = mc^2\), where E is energy, m is mass, and c is the speed of light. This transformation of energy into mass decreases the overall mass of the nucleus; hence the energy required to overcome the strong nuclear force is equal to the mass defect multiplied by the speed of light squared.
03

Binding energy per nucleon and nuclear stability

Comparing different nuclei for their stability requires calculating the binding energy per nucleon, which is calculated by dividing the total binding energy of the nucleus by the number of nucleons (protons + neutrons). Nuclei with higher binding energy per nucleon are more stable.
04

Fusion process and the release of energy

In the fusion of light nuclei into heavier nuclei, the binding energy per nucleon of the product nucleus is greater than that of the initial nuclei. This difference in binding energy per nucleon means that the mass defect of the final nucleus is larger than the sum of the mass defects of the initial nuclei. Now, recalling Einstein's mass-energy equivalence, this difference in mass defects corresponds to a release of energy, calculated as \( \Delta E = \Delta m \times c^2 \), where \( \Delta E \) is the energy released, \( \Delta m \) is the difference in mass defects, and c is the speed of light. This released energy shows up as kinetic energy or light, depending on the fusion reaction.
05

Conclusion

In conclusion, the fusion of light nuclei into heavier nuclei releases energy because the final nucleus has a larger mass defect and subsequently higher binding energy per nucleon than the initial nuclei. This difference in binding energies is converted into energy, consistent with Einstein's mass-energy equivalence principle.

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Most popular questions from this chapter

The power output of the Sun is \(4 \times 10^{26} \mathrm{W}\). (a) If \(90 \%\) of this energy is supplied by the proton-proton chain, how many protons are consumed per second? (b) How many neutrinos per second should there be per square meter at the surface of Earth from this process?

Why is the number of neutrons greater than the number of protons in stable nuclei that have an \(A\) greater than about 40? Why is this effect more pronounced for the heaviest nuclei?

Verify that the total number of nucleons, and total charge are conserved for each of the following fusion reactions in the proton-proton chain. (i) \(^{1} \mathrm{H}+^{1} \mathrm{H} \rightarrow^{2} \mathrm{H}+e^{+}+v_{\mathrm{e}},\) (ii) \(^{1} \mathrm{H}+^{2} \mathrm{H} \rightarrow^{3} \mathrm{He}+\gamma,\) and (iii) \(^{3} \mathrm{He}+^{3} \mathrm{He} \rightarrow^{4} \mathrm{He}+^{1} \mathrm{H}+^{1} \mathrm{H}\). (List the value of each of the conserved quantities before and after each of the reactions.)

Why is Earth's core molten?

If a 1.50 -cm-thick piece of lead can absorb \(90.0 \%\) of the rays from a radioactive source, how many centimeters of lead are needed to absorb all but \(0.100 \%\) of the rays?
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