Basic Nuclear Fusion question on Binding energy

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SUMMARY

This discussion focuses on the concept of binding energy in nuclear fusion, specifically the fusion of deuterium nuclei into helium-4 (He-4). It establishes that nuclei with higher binding energy per nucleon exhibit lower atomic weight per nucleon. The confusion arises from the negative nature of binding energy, which indicates that a higher absolute value of binding energy corresponds to a more stable, lower-energy state, allowing energy release during fusion. The key takeaway is that the fusion process results in a product with a higher binding energy, leading to a decrease in mass relative to the reactants.

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  • Understanding of nuclear fusion principles
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  • Knowledge of atomic structure and nucleons
  • Basic grasp of mass-energy equivalence (E=mc²)
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JonDawe
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Hey guys, I'm getting a bit confused about nuclear fusion. I understand the basics of binding energy being the energy required to split the nucleons apart etc. and the fact that in nuclear fusion you are combining two particles e.g. (2 deuterium nuclei) with a lowish binding energy to form a particle (he-4 or something) with a higher binding energy. I'm just getting a bit confused at why a higher binding energy means that the mass of the particle relative to the masses of the 2 fusion reactants is lower? Can someone here help explain it?

I know nuclei with higher binding energy per nucleon have a lower*atomic weight*per nucleon, but I'm kind of confused at why?
 
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Binding energy is always negative (otherwise the state is not bound), and "high" is usually meant as absolute value. So helium has a "high negative" binding energy, it is actually a low-energetic state. This means that fusion can release energy while forming this state.
 
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