I'm having some trouble in understanding how the released energy in a nuclear reaction is actually calculated. I think I understand the general approach if we are dealing with a reaction like $$\rm ^{235}U+n\rightarrow{}^{93}Rb+{}^{140}Cs+3n,$$ where I would just find $Q=\Delta m$. This $\Delta m$ would be a function of the binding energies (in the example above we would get $\Delta m = B_{Rb}+B_{Cs}-B_U$) (and maybe the masses of proton/neutron/electron if we talk about a general case).
But how does one exactly find the released energy if we have something like this $$4 p \quad \rightarrow \quad^{4} \mathrm{He}+2 e^{+}+2 \nu_{e}.$$ In my lecture notes it is written "that the positrons created in the fusion reaction will annihilate with two electrons, releasing extra energy" and that we can neglect the rest mass of the neutrinos. Neglecting the rest mass of the neutrinos is equivalent to ignoring them in the whole process, right? If this is the case, why is this assumption justified? Can't Neutrinos carry a significant amount of the released energy in a reaction? The lecture notes also state that the released energy equals to $$Q=4 M_{p}-\left(2 M_{p}+2 M_{n}+2 M_{e}\right)+B_{H e}+4 M_{e}.$$ I'm really confused from where all these electron masses come form. The $2(M_p+M_n+M_e)$ looks like he was counting the protons, neutrons and electrons present in $^4\mathrm{He}$, which would give him the mass of $^4\mathrm{He}$ if he subtracted the binding energy, but he adds it. Why? And how does the $e^+e^-$ annihilation produce $4M_e$ (I just assume that this term comes from there...).
