The PPI Chain is composed of 3 steps: Does only the 2nd step release energy? The PPI chain is
\begin{align}
\rm\ ^1H + {}^1H &\rm\rightarrow {}^2D + e^+ + \nu_e \tag 1
\\
\rm\ ^2D + {}^1H &\rm\rightarrow {}^3He + \gamma \tag 2
\\
\rm\ ^3He + {}^3He &\rm\rightarrow {}^4He + {}^1H + {}^1H \tag 3
\end{align}
I know that, as an overall, the PP chain should release ~26 MeV. I can see clearly that (2) release energy in form of a photon but I can not say the same about (1) and (3). Does it mean that the ~26 MeV comes solely from (2)?
Thank you for your help!
 A: No, all three steps release energy.  The easy way to see this is to define the mass excess for a particle, generally a neutral atom at rest in its ground state, as $\Delta  = M - A\rm\,amu$. This mass difference can totally be measured in energy units, $\Delta_E = \Delta _m c^2$. From a good reference table, we have
$$
\begin{array}{c|c}
\text{particle} & \Delta\text{ (MeV)}
\\\hline
\gamma & 0.0 \\
\nu & \lesssim10^{-6} \\
\rm e^\pm & 0.511 \\
\rm n & 8.0713 \\
^1\rm H & 7.2889 \\
^2\rm H & 13.1357 \\
^3\rm He & 14.9312 \\
^4\rm He & 2.4249
\end{array}
$$
The total mass excesses before and after your reactions are
\begin{align}
2\times 7.2889 &\to 13.1357 - 0.511^* + 0.511 + 10^{-6}
& \text{change} &= -0.42\rm\,MeV
\\
13.1357 + 7.2889 &\to 14.9312 + 0
& \text{change} &= -5.49\rm\,MeV
\\
2\times 14.9312 &\to 2.4249 + 2\times7.2889
& \text{change} &= -11.29\rm\,MeV
\end{align}
In all cases the energy difference is carried away as the kinetic energies of the various decay products, subject to the constraint that, in the rest frame of the reaction, the total momentum is zero.  For the massive particles, where the kinetic energy is approximately $K_i=p^2/2m_i$, that means the lighter products carry more of the energy than the heavy products.  For the reaction with the photon, where momentum conservation makes the energies obey 
$$K_\gamma/c = \sqrt{2m_\mathrm{He}K_\mathrm{He}}$$
the trend of having the massless particle carry away most of the energy continues.

$^*$ Note: since the mass excesses are tabulated for neutral atoms, while the participants in proton-proton fusion are completely ionized, it might be better to write
\begin{align}
\rm \Delta\left( {}^1H^+ + {}^1H^+ \right)
&= \rm 2\times(7.2889 - 0.511)\, MeV
\\
\rm \Delta\left(
{}^2H^+ + e^+ + \bar\nu_e 
\right)
&= \left(
[13.1357 - 0.511] + 0.511 + 10^{-6} \right)\rm\,MeV
\end{align}
The result is the same, though.
