Reversibility of the binding energy of the nucleus

Question

I have been taught that mass defect arises due to loss of mass as energy during nucleus formation. Neutrons and protons have less mass in the nucleus than their theoretical mass. So my question is, if we supply the nucleus with energy equal to its binding energy will the now free protons and neutrons have their original mass?

Answer 1

Yes, that is correct. It is (in principle!) possible to supply a given nucleus with enough energy to match its mass defect, which will cause it to disintegrate and leave you with a bunch of protons and neutrons flying off (in principle, at zero velocity, unless you supply a little extra energy to get them to move along instead of re-bonding and re-emitting the energy as gamma rays). The resulting protons and neutrons will then have their free rest masses as usual (though the neutrons will decay to protons via beta decay in relatively short order, but that's another story).

It's important to stress, though, that it's incorrect to think of each individual particle having less mass than its free version. Instead, the mass defect is a property of the system as a single unit: that is, the whole has less mass than the sum of its parts, but you shouldn't think of it as that being caused by the parts themselves being lighter.

Answer 2

When you think about the nucleus, you have think about it as an entire physical system. When neutrons and protons bound together to form a nucleus, they release energy. This means that the mass of the nucleus will be less than the sum of the masses of its individual nucleons (protons and neutrons). Yes, with enough energy you can break the nucleus into its constituent pars. The energy you need is that nuclear binding energy.

The important thing here is that the mass of neutrons and the mass of protons always remain the same. What makes the mass of the nucleus to be less that the sum of the masses of its constituent parts is the interactions between the constituents.