In nuclear fission the total rest mass of the products is less than the mass of the original nucleus. But on the other hand, the binding energy of the products is higher. Doesn't the decrease of the rest mass of the products contribute to the increase of their binding energy? So why then there is enormous energy released?
[Physics] Binding energy in fission
nuclear-physics
Related Solutions
Your basic nuclear reaction conserves the number of nucleons present.1
That is important, because at a bit less than 1 GeV each the mass of the nucleons dominates the total energy of all these states.
So the only place available to get or lose energy in a reaction is by
Changing the flavor of nucleons. Every neutron converted to a proton gets you a neutrino and some gammas (once the positron has captured and annihilated).
By changing the total binding energy. Notice that getting one nucleon more bound doesn't help if another one gets less bound by a large amount. The decision to express this in terms of the average binding energy is completely arbitrary because for $N$ total nucleons (which doesn't change, remember?) $E_\text{total} = E_\text{AVG} * N$.
Another questions addresses what is so special about iron that it has the highest binding energy per nucleon.
1 This is more or less required by Baryon number conservation in the Standard Model, and we will ignore the need for Baryon non-conservation in most beyond the Standard Model candidate theories.
In general if the the fragments nuclei and alpha particles they are moving at speeds much less than the speed of light so relativistic effects do not manifest themselves.
Some beta particles can be moving close to the speed of light and the Kaufmann–Bucherer–Neumann experiments were the first to investigate the increase in apparent mass of the very fast moving beta particles.
In fact they measured the charge to mass ratio and found that the specific charge of beta particles decreased as the speed of the beta particles increased as shown in the graph below.
The amount of energy associated with this change in the beta particle is much smaller than the binding energy released in the decay.
Best Answer
There's a closely related question: binding energy of a nucleus is positive?
Imagine taking the nucleus and pulling it apart into individual protons and neutrons. To do this you have to put in energy, and the amount of energy you put in is equal to the binding energy - call this $E_0$. Now let the individual protons and neutrons come together again to form the daughter nuclei (and free neutrons). As the question I linked explains, to form the daughter nuclei you have to take energy out, and the amount of energy you take out is the sum of the binding energies of the two daughter nuclei - call these $E_1$ and $E_2$.
When you say the binding energy of the products is higher what this means is:
$$ E_0 < E_1 + E_2 $$
So in our notional process of pulling the nucleus apart and letting it reform, we end up getting energy out i.e. the fission process produces energy (typically as kinetic energy of the reaction products).
As for the masses, when we separated the original nucleus we had to add an energy $E_0$, so the mass of the separated nucleons is more than the original nucleus by an amount $E_0/c^2$. When we let the nucleons recombine into the daughter nuclei we get energy out so the mass of the products is less than the separated nucleons by an amount $(E_1 + E_2)/c^2$. Because $E_1 + E_2 > E_0$ overall the mass has decreased giving us the mass deficit. The mass deficit is just the amount of energy released in fission divided by $c^2$.