About binding energy, mass defect and enegy released in a nuclear fusion I know this question looks very similar to many others but I actually think it is not the same, because I was looking for a possible answer for hours.
The question is about the first step in the pp chain (i.e. the following reaction) 
$$\rm^{1}H+{}^{1}H \to {}^{2}H+e^{+}+\nu_{e}$$
I wanted to know if the following reasoning is correct. In order to point out easier the possible mistakes I separated it in steps:
Context: Imagine the two initial $\rm^{1}H$ has zero kinetic energy (I know this is difficult but I suppose there are in some point of the potential such that this is possible), because I am not really interested in kinetic energy (at least before the fusion takes place).

*

*The initial sistem before the fusion has an energy equal to $2mc^2$ where $m$ is the mass of the $\rm^{1}H$ (when it is "free" or alone).


*After the process takes place the two $\rm^{1}H$ stay together (forming $\rm^{2}H$) and due to this a negative potential energy appears i.e. binding energy  (like when two bodies come closer in a gravitational field).


*As $E_{\rm^{2}H}=m_{\rm^{2}H}c^2$ the previous fall in energy of this part of the products is readed as a reduction in mass (because even taking into account the energies due to the masses of the positron and the neutrino is there a fall) and is called "mass defect".


*To balance energy before and after the fusion takes place an equal amount of POSITIVE released energy has to be created (usually called $Q$). 


*This energy (which is 0.42 Mev) is usually the kinetic energy associated with the products of the reaction (and sometimes in other reactions even with radiation).
I really appreciate an answer, because many books are not absolutely clear about this point.
 A: I would tweak your description a few different ways.
First, it’s fine that you start off with zero kinetic energy.  The reaction is exothermic, so it’s technically allowed at zero temperature. The rate of the reaction rapidly approaches zero as the temperature falls, so spontaneous fusion is not a thing in neutral hydrogen molecules, but for the sake of argument it’s fine.
I would split the formation of deuterium into two steps:

*

*The strong interaction forms the remarkably unstable nucleus $\rm^2He$, with two protons and zero neutrons.


*Instead of immediately disintegrating via the strong interaction, the $^2\rm He$ has some minuscule chance of decaying via the weak interaction to deuterium.
The energy of the “virtual” helium state is not well-defined. But you are quite correct that the total (mass-)energy of the final state is reduced because of an attracting interaction between the nucleons. The comparison with gravitational potential energy is good.
The energy released in the reaction is shared among all three of other particles in the final state. In the center-of-momentum reference frame, the three particles will have roughly equal momenta, depending on the angles they make with each other. If you know that the kinetic energy is $T=pc-mc^2$ for massive particles (which becomes $T≈p^2/2m$ for non-relativistic particles), you should convince yourself that most of the energy leaves the system with the least-massive particles.
A: In completing the answer by rob, in detail to your point 2.



*After the process takes place the two 1H stay together (forming 2H) and due to this a negative potential energy appears i.e. binding energy  (like when two bodies come closer in a gravitational field).


I would add :
The process of fusion is quantum mechanical; in the "stay together" in you 2H there are the two interactions, the strong force which is attractive and the electromagnetic which is repulsive due to the same charge. In addition quantum numbers and the Pauli exclusion principle have to enter to get a complete quantum mechanical description of the process. More than special relativity is involved.
