Chemistry - On the stability of deuterium

Solution 1:

No, deuterium is completely stable. I found the answer at Hyperphysics, and it has to do with the mass energies of the products and reactants of this hypothetical reaction.

The decay of deuterium would be $$\ce{D -> P + N + e + \bar{\nu}}_e$$ where $\ce{D}$ is deuterium, $\ce{P}$ is a proton, $\ce{N}$ is a neutron, $\ce{e}$ is an electron and $\ce{\bar{\nu}}_e$ is an electron antineutrino. The combined mass energies total to $1877.05\,\mathrm{MeV}$ (mega electron volts). But the mass of $\ce{D}$ is $1875.6\,\mathrm{MeV}$, so the decay cannot happen because conservation of mass/energy would be broken. There's no way to get around this, not even imbuing the initial deuterium atom with extra energy of some other sort. Deuterium will never decay unless, for some reason, the proton turns out to be unstable.

If, for some reason, deuterium did decay, it could go to $\ce{^2He}$ … which would then decay back to deuterium.

Solution 2:

In the Standard Model of physics (plus some assumptions on gravitons), baryon number can be violated in multiples of three, as long as the difference between baryons and leptons remains constant. The process is called sphaleron transitions. Theoretically we could have $$ D \to \bar p + 2e^+ + \bar \nu_e$$

The problem is that baryon number violation has never been measured (it would have a very, very small probability) and in any case it would not belong to chemistry.

Note that in any case, protons would still be stable in this context.