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I'm confused about the fact that in a fission process the masses of the "products" are less than the mass of the "reactants". And in a fusion process the masses of the "products" are less than the masses of the "reactants".

My first question is then: Can I have an atom, fission it, then fusion it, then fission it, etc, etc,. And arrive at no mass and pure energy? I know this is wrong, but I don't know why so.

If you tell me this process is possible then how is it possible to recover a certain isotope of Uranium. I see this as a problem because having less mass in the "products" than in the "reactants" shows that nuclear processes aren't reversible, unless I inject mass in some way.

My second question is: Does the mass difference manifest itself as, e.g., "nucleons weigh less now", or as in "$x$ number of nucleons have disappeared".

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    $\begingroup$ What is pure energy, btw, according to you? (Photons?) Particles decay into their products, and the decay products satisfy numerous conditions. This link lists a bunch of them: profmattstrassler.com/articles-and-posts/… $\endgroup$
    – Hritik Narayan
    May 30, 2015 at 18:11
  • $\begingroup$ Well it could be that you obtain $mc^2$ amount of work where $m$ is the mass of the original mass that undergoes fission. $\endgroup$
    – DLV
    May 30, 2015 at 18:14
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    $\begingroup$ The usable energy would be much much lesser than that! (With fission and fusion, that is. Matter antimatter annihilation would do this, though.) $\endgroup$
    – Hritik Narayan
    May 30, 2015 at 18:15
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    $\begingroup$ You should have a look into the mechanisms of these processes too, that'll let you know why you're wrong. $\endgroup$
    – Hritik Narayan
    May 30, 2015 at 18:17
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    $\begingroup$ All of these processes are subject to the laws of thermodynamics, just as well, which means that one can not extract all of their internal energy as work. Just because something is a nuclear or particle physics process and not a chemical one does not mean that thermodynamics doesn't apply, it might just take getting used to how the thermal baths are defined in these cases. $\endgroup$
    – CuriousOne
    May 30, 2015 at 18:21

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It is not true that in all fusion and fission processes the mass of the products is less than the mass of the reactants. This is only valid for exothermic reactions.

The change of mass is due to the change in the binding energy of the nucleons (note that the change in binding energy is in the order of 1 MeV, while the mass of the nucleons is around 940 MeV). The total mass of the nucleus is the sum of the masses of its nucleons minus the binding energy, so a tighter binding means a smaller mass. The binding energy of a nucleus can be computed with Weizsäcker's semi-empirical mass formula: Binding energy per nucleon

In this plot, you can see the binding energy per nucleon. For isotopes lighter than iron, nuclei can become more stable if they fuse (they move to the right in the plot), and they liberate energy (lower their mass) in the process. In the other hand, fission occurs in heavier nuclei (they go to the left), and also liberates energy. But in both cases the maximum energy liberated is given by the above plot: you can only get in total, no matter what intermediate steps you make, is the 9 MeV/nucleon difference between ${}^{56}$Fe and ${}^1$H, wich corresponds to fuse 26 protons and 30 neutrons together to form an iron nucleus. And obviouslly, iron nucleii do have mass!

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  • $\begingroup$ yes ! nice with the graphic $\endgroup$
    – user46925
    May 30, 2015 at 18:59
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My first question is then: Can I have an atom, fission it, then fusion it, then fission it, etc, etc,. And arrive at no mass and pure energy? I know this is wrong, but I don't know why so.

No , you cannot. In addition to special relativity where invariant masses of complex objects are the "length" of their energy momentum summed four vector, which allows for the gain in energy, the atomic/nuclear level is quantum mechanical, not classical.

In quantum mechanics we have in addition to energy momentum and angular momentum conservation, conservation of quantum numbers, and in this particular proposal, it is conservation of baryon number that will not allow the scenario.

In the best/worst case, an atomic number A nucleus may have A nucleons ( neutrons/protons) free.

In addition, in fission some of the fragments are stable , again due to quantum mechanical rules: how the energy levels are arranged in the effective nuclear potential.

In fusion , again baryon numbers are conserved and again not all states are allowed to be bound due to the nuclear potential.

My second question is: Does the mass difference manifest itself as, e.g., "nucleons weigh less now", or as in "x number of nucleons have disappeared".

The excess energy appears as kinetic energy of the fragments and also might appear as a gamma ray ( photon) if spin quantum numbers allow it. Again, no disappearing nucleons, baryon number is conserved.

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  • $\begingroup$ With regards to the second part of your answer, I wanted to know about the mass difference. In a fission process nucleons might have KE, and I guess that's what you call "excess energy", but they as well have less mass. Is then a case of "nucleons weigh less"? Is baryon number always conserved? Thanks. $\endgroup$
    – DLV
    May 30, 2015 at 19:15
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    $\begingroup$ In a fission process the nucleus breaks up into fragments that contain the baryons (protons and neutrons in this case) in bound states of the nuclear force. Bound means "binding energy" can be assigned per nucleon by taking the difference of the summed-mass-of-free-nucleons to the mass of the nucleus, and dividing this difference by the number of nucleons. en.wikipedia.org/wiki/Nuclear_binding_energy . Baryon number conservation is absolutely conserved within the accuracies of our experiment. No proton decay has been seen. $\endgroup$
    – anna v
    May 31, 2015 at 3:17
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If uranium undergoes fission, its products are too massive to release energy in fusion. Atoms must be smaller than iron to release energy through fusion.

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