= Solution
The first step of the <proton–proton chain> is
$$
\boxed{p+p\longrightarrow d+e^++\nu_e,\qquad Q\simeq0.420\ \mathrm{MeV}.}
$$
Here $d$ is the <deuteron>. A <weak charged current> converts one <proton> into a <neutron>, while the resulting <proton>–<neutron> pair becomes bound. The <Q value> follows from the difference of initial and final rest energies:
$$
Q=(2m_p-m_d-m_e)c^2
=B_d-(m_n-m_p)c^2-m_ec^2
\simeq2.2246-1.2933-0.5110=0.4203\ \mathrm{MeV}.
$$
This includes the emitted <Positron>'s rest energy in the final-state mass and excludes its subsequent <electron-positron annihilation>. The <nuclear binding energy> of the <deuteron> makes this reaction possible although the isolated conversion $p\to n+e^++\nu_e$ is energetically forbidden. The released energy is shared by final-state kinetic energies, including the escaping <neutrino>.
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