Question

which statement describes why energy is released in a nuclear fusion reaction based on mass-energy equivalence?
○ for large nuclei, the mass of the original nucleus is greater than the mass of the products.
○ for large nuclei, the mass of the original nucleus is less than the mass of the products.
○ for small nuclei, the binding energy of the lighter nuclei is greater than the binding energy of the heavier nucleus.
○ for small nuclei, the binding energy of the lighter nuclei is less than the binding energy of the heavier nucleus.

Explanation:

1. Recall mass - energy equivalence ($E = mc^{2}$) and nuclear fusion: Nuclear fusion involves combining small nuclei (like hydrogen isotopes) to form a larger nucleus. The binding energy per nucleon increases as we move from very light nuclei to nuclei around the mass of iron.
2. Analyze each option:

• Option 1 and 2: Nuclear fusion involves small nuclei, not large nuclei. So these options are about nuclear fission (which involves large nuclei splitting), so they are incorrect.
• Option 3: If the binding energy of lighter (small) nuclei was greater than the heavier nucleus formed, energy would be absorbed, not released.
• Option 4: In nuclear fusion, when small nuclei combine to form a heavier nucleus, the binding energy of the heavier nucleus is greater than the binding energy of the lighter nuclei. The difference in binding energy (related to mass - energy equivalence, since mass defect $\Delta m$ is related to binding energy by $E_{binding}=\Delta mc^{2}$) is released as energy. When the binding energy of the product (heavier nucleus) is greater than the sum of the binding energies of the reactants (lighter nuclei), the mass defect (mass of reactants - mass of products) is positive, and energy $E = \Delta mc^{2}$ is released.

Answer:

D. For small nuclei, the binding energy of the lighter nuclei is less than the binding energy of the heavier nucleus.