# How is the energy output released in a proton-proton fusion reaction?

$$\rm p + p \to {}^2H + e^+ + v_e$$

The net energy output for this reaction, not including the electron-positron annihilation that likely would occur, is 0.423 MeV - according to my math and sources I've found online. But these all don't account for that neutrino at the end. Does this mean that the energy is released in the form of that neutrino? Does the neutrino not matter, and the energy is released some other way?

The neutrino mass is less than one eV --- that is, less than $$0.000\,001\rm\,MeV$$. So leaving its mass out of the calculation isn't a big deal.

The energy is carried away by all three reaction products. Consider a case where the initial momentum of the system is zero, and the geometry of the reaction products means that they all three have the same momentum magnitude $$p$$. (Perhaps they all make 120 degree angles with each other's paths.) Then the kinetic energy carried by the (ultra-relativistic) neutrino is $$K_\nu = pc$$, the kinetic energy carried by the non-relativistic deuterium nucleus is $$K_{\rm d} = p^2/2M_{\rm d}$$, and the positron's kinetic energy

$$K_{\rm e}= \frac{p^2}{(\gamma+1)m_{\rm e}} =(\gamma-1)m_{\rm e}c^2 ,$$

with the usual relativistic factor $$\gamma=(1-v^2/c^2)^{-1/2}$$, doesn't usually fit into either of those nice approximation regions. But you can see that the deuteron, with the most mass, carries off the smallest amount of kinetic energy in that (and actually all) geometries. How the energy is shared among the light particles as the geometry of the interaction changes is a fun problem to play with.

• "How the energy is shared among the light particles as the geometry of the interaction changes is a fun problem to play with" with quantum mechanical probability , given by the wavefunctions of the interaction adsabs.harvard.edu/full/1969ApJ...155..501B Sep 2, 2019 at 3:06
• @annav In neutron beta decay (a closely related process), the angular correlations turn out to be a place to hunt for weak interaction terms that aren't the usual $V-A$. But that's a little too complex for this answer.
– rob
Sep 2, 2019 at 5:42