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When protons and neutrons interact attractively and coalesce to form an atomic nucleus, their energy in this state must be less than what it was when they are separated, so they lose mass which is then converted into energy by $E=mc^2$. Now, when electrons interact attractively with an atomic nucleus and orbit it under act of electrostatic forces, their energy in this state must be also less than what it was when they were free. So what this energy difference between the two electron states could come from? They will lose mass also?

And if they lose mass, how they could when they are, unlike protons and neutrons, are not composed from intraparticles, so they cannot convert their intraparticles total potential energy into binding energy.

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Your confusion arises from a misunderstanding of how the mass of composite systems works in special relativity.

When you have a composite system where objects interact, the energy of that interaction counts towards the total energy of the system at rest (in other words, the mass). This is a totally separate, standalone contribution to the mass of the system. It does not modify the mass of the constituents of the system. None of the constituents "lose mass".

So, let's look at your example. If we consider a system that consists of a proton and an electron very far from each other, the energy of their interaction is basically zero. This system has a mass of $m_p+m_e$. Now, if we have a system where the electron and proton form a hydrogen atom's ground state, the energy of that interaction is $-13.6$ eV. This negative interaction energy is part of the system's total mass, which is $m_p+m_e-(13.6\text{ eV})$. Neither the proton nor the electron lose mass. There's simply another thing that's part of the system's total mass.

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  • $\begingroup$ so how protons and neutrons lose mass? $\endgroup$ – Elsayed Jul 5 '20 at 18:35
  • $\begingroup$ @Elsayed they don't, that's the point of this answer. $\endgroup$ – fqq Jul 5 '20 at 18:38
  • $\begingroup$ @Elsayed As fqq said, they don't. Their interaction energy is negative, and the interaction energy is part of the total mass of the nucleus. $\endgroup$ – probably_someone Jul 5 '20 at 18:48
  • $\begingroup$ Yes i got you now, thank you very much, you corrected for me one belief i had for a really long time!! $\endgroup$ – Elsayed Jul 5 '20 at 18:54

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