Answer
a). $(vbar)_{e}+H^{1}_{1}=n^{1}_{0}+e^{0}_{+1}$
By charge conservation, the charged particle product must be a positive electron or a positron.
b). $Q=-1.8MeV$, so this reaction is endothermic since $Q<0$
c). $n^{1}_{0}+H^{1}_{1}=H^{2}_{1}+gamma$ and
$e^{0}_{+1}+e^{0}_{-1}=gamma+gamma$
d). For the neutron-proton process,
$E_{gamma}=2.224MeV$ and for beta-positron annihilation process, $E_{gamma}=1.022MeV
Work Step by Step
a). $(vbar)_{e}+H^{1}_{1}=n^{1}_{0}+e^{0}_{+1}$
By charge conservation, the charged particle product must be a positive electron or a positron.
b). $Q=-1.8MeV$, so this reaction is endothermic since $Q<0$
c). $n^{1}_{0}+H^{1}_{1}=H^{2}_{1}+gamma$ and
$e^{0}_{+1}+e^{0}_{-1}=gamma+gamma$
d). For the neutron-proton process,
$E_{gamma}=2.224MeV$ and for beta-positron annihilation process, $E_{gamma}=1.022MeV$
