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In fusion , I have understood so far that two light nuclei fuse to form a heavy nucleus. The nucleons in the light nuclei experience lesser binding energy as compared to the nucleons in heavy nucleus which experience higher binding energy. The mass of the reactants is more than the mass of the product hence energy is released. Since the nucleons experience a change in binding energy, they lose mass. This mass turns into the extra binding energy required to bind the nucleons into place. But where does the mass defect originate from?

The helium nucleus (for instance) has 2 protons and no neutrons-the mass remains the same, so why did the mass defect occur? Also, if the mass converts to binding energy then why is energy released in the fusion. It should remain the same or should in fact, absorb energy. Please correct my understanding if wrong or, please answer the two questions above. I have been thinking about this for a month now.... :(

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    $\begingroup$ A helium nucleus has a neutron or two, unless you're talking about the very short-lived diproton that occasionally forms when protons collide, only to then undergo positron decay, making deuterium. $\endgroup$
    – J.G.
    May 11, 2022 at 11:43

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The helium nucleus (for instance) has 2 protons and no neutrons-the mass remains the same, so why did the mass defect occur?

Helium nucleus contains two protons and one or two neutrons (depending on whether we deal with Helium-3 or Helium-4). The mass of the nucleus is not the same as the sum of the masses of free particles: rather it is the sum of these masses plus the binding energy: $$ Mc^2=\sum_{i=1}^Nm_ic^2 + E_b $$ The binding energy is negative in this case (or we can define it as positive, but change the sign in the equation above). This means that some energy was released when the protons and neutrons merged in the nucleus, and the released energy is the difference between the masses of the separate particles and the mass of the nucleus, i.e., the binding energy taken with the negative sign (plus their kinetic energy before they merged).

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  • $\begingroup$ But to look at , why is the energy released in the first place when the nucleons bind together as a nucleus? What is the need of it ? What is the exact cause for it , specifically? $\endgroup$
    – Hardik
    May 11, 2022 at 14:39
  • $\begingroup$ @Hardik bound state has lower energy than the unbound one - otherwise, there would be no reason for bound states to be stable. Same applies to electrons and nuclei binding to form atoms (electromagnetic forces) or planets/comets/etc. binding to stars (gravitational forces). Except that in cases of electromagnetic and gravitational forces, it corresponds to a negligible change of mass, so it is usually not considered this way. $\endgroup$
    – Roger Vadim
    May 11, 2022 at 14:43
  • $\begingroup$ So @Roger Vadim , do you mean that the reactant nuclei 'experience' low strong force as compared to the product nucleus which 'experiences' higher strong force. Due to this , the product nucleus is stable and this explains for the low energy system in the nucleus? Do you mean this? Also , if this is the correct theory then what causes for the extra binding/strong force experienced by the nucleons of the product nucleus ? And if the energy system is low in the product nucleus , the extra energy is released in the form of nuclear energy? Is this the correct theory ? $\endgroup$
    – Hardik
    May 12, 2022 at 4:46

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