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...).


1 Answer 1


Suppose we convert the 4 protons into two protons and two neutrons and two positrons and two neutrinos Before anything else happens we have gained $$ 4M_p -(2M_p +2Mn +2M_e). $$ This is a negative number because neutrons are heavier than protons and we have also had to make the positrons. Now the neutrons and protons bind to form the helium nucleus. This releases energy $B_{He}$ --- probably in the form of gamma rays. Now the two positons find two electrons and annihilate. This releases 4 gamma rays each with 512 keV energy, ie $4M_e$. The net energy gain is therefore

$$ 4M_p -(2M_p +2Mn +2M_e)+B_{He}+ 4M_e. $$ Some this energy went into the gamma rays already mentioned (from binding and $e^++e^-$ annihilation), but part of it goes into the kinetic energy of the neutrinos. Thus the total energy balance is $$ 4M_p -(2M_p +2Mn +2M_e)+B_{He}+ 4M_e= E_{gamma\phantom. rays}+E_{neutrinos} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.