Given the masses of various atomic particles $m_p = 1.0072$ u, $m_n = 1.0087$ u, $m_e = 0.000548$ u, $m_{\bar\nu} = 0$, $m_d = 2.0141$ u, where $p \equiv$ proton, $n \equiv$ neutron, $e \equiv$ electron, $\bar\nu \equiv$ antineutrino and $d \equiv$ deuteron. Which of the following processes is allowed by momentum and energy conservation?
Answer: (C) $n + p \rightarrow d + \gamma$
$n + p \rightarrow d + \gamma$: the mass on the left is $1.0072 + 1.0087 = 2.0159$ u, more than $2.0141$ u, so energy is released and carried by the photon and the recoil of $d$. Momentum and energy can both be conserved, so this is allowed.
The others fail:
$n + n \rightarrow$ deuterium atom: charge is not conserved (an electron appears from nothing).
$p \rightarrow n + e^+ + \bar\nu$: the products are heavier than a free proton, so energy cannot be conserved.
$e^+ + e^- \rightarrow \gamma$: in the centre of mass frame the total momentum is zero, but a single photon always carries momentum, so momentum cannot be conserved.
Solution by Sreeraj P, M.Sc Physics