When a particle at rest decays, the momentum of the fragments has to add up to zero, because momentum is a constant when there isn’t any external force. In a two-body decay this means the two fragments have equal and opposite momenta. In a three-body decay, the magnitudes of the different momenta take on different values depending on the angles between them. Computing the details of the spectrum is hard, but the hand-waving approximation is that each fragment carries about the same amount of momentum.
This means that nearly all of the energy in the decay is carried away by the low-mass electron and the ultra-relativistic neutrino: the poor nucleus only gets to carry kinetic energy $\sim p^2/2M$, while the electron gets to carry $\sim p^2/2m_e$.
The reason that we can separate nuclear physics from atomic physics is that the energy scales involved in the interactions are very different. In order to separate an electron from a hydrogen atom, you have to supply it with a minimum of 13 electron-volts (eV) of energy. But the typical energy in a nuclear decay is $10^6$ eV. So in the vast majority of decays, the electron and the nucleus go in different directions, with too much energy for the electromagnetic force to bind them.
However, there is a very small corner of the parameter space where nearly all of the energy is carried away by the neutrino, leaving the daughter nucleus and the decay electron nearly at rest. This is called a “two-body beta decay” or a “bound beta decay.” For the free neutron, whose beta-decay energy is around 0.8 MeV, the bound decay $$\require{mhchem} \ce{n \to H + \nu}$$ is predicted to occur a few times out of every million decays.
This 2014 paper outlines a proposed attempt to measure it, but the experiment is tricky and I wouldn’t be surprised if there were no result yet —— they hadn’t even picked a site for the experiment. The goal would be not just to detect the rare decay mode, but to measure the total spins of the produced hydrogen atoms, which tell you in a direct way about the spins of the invisible neutrinos.
You could in principle apply the same logic to heavier beta emitters. One candidate might be bound tritium decay, $$\ce{^3H \to {}^3He + \nu},$$ where the beta decay energy is much smaller (around 15 keV) and the ionization energy well is deeper: you can imagine the odds of the neutrino carrying away “all” of the energy might be many per million decays, instead of a few per million decays. But [experimentalist rabbit hole deleted] it’s not clear to me that a higher branching ratio would immediately make for a better experiment.
You would never expect to find a decay like
$$\ce{
^{14}C \not\to {}^13C + {}^1H + \nu
}$$
because it takes at least 10 MeV to knock a proton or neutron out of a stable nucleus, and beta decays are typically not that energetic.
tl;dr summary: such decays are predicted, rare, not yet observed, but not really in doubt.