Proton and neutron can't collide and form a ground-state deuteron; what happens instead?

Question

One might expect that a proton and neutron could easily collide and form a deuteron. However, since $m_d < m_p + m_n$, the process $p + n \to d$ can't conserve both energy and momentum. For example, if the proton and neutron approached with equal and opposite momenta, conservation of momentum says there is nowhere to "put" the $$K_p + K_n + (m_p + m_n - m_d)c^2$$ worth of kinetic energy one would expect the deuteron to have.

Now of course protons and neutrons CAN in fact collide and produce deuterons + gamma rays (i.e. $p + n \to d + \gamma$ is an allowed process) -- but how does this square with conservation of momenergy? Intuition suggests the proton and neutron could create an excited state of the deuteron with $m_d^* > m_p + m_n$ in such a way that $p + n \to d^*$ conserves both energy & momentum, and then the decay $d^* \to d + \gamma$ occurs soon afterwards. Is this what typically happens?

Answer 1

This old, but interesting experiment performed back in 1932 at a Cavendish Laboratory, shows that some materials (for example liquid hydrogen) when bombarded with neutrons, gives rise to a gamma radiation, which is propagated at a $120°-180°$ angles with respect to incident neutrons. I.e. backwards scattering happens in this case. Assuming that neutron / proton mass ratio is about $1.001$ and that usually this kind of collision just produces recoil protons knocked out from a liquid Hydrogen atom - this strange backwards scattering can be only explained that approximately $1/4$ of impact neutrons combines with protons and forms a Deuterium nucleus, $^2H$. Researchers suggests such reaction:

$$ n+p = d + \gamma ~+ \left[K_n/2+(m_p + m_n - m_d)c^2\right] $$

Where last term is gamma photon energy.