The isotope of hydrogen, tritium, is indeed radioactive resulting in $\ce{^3He}$ through beta decay.

But what of the radioactively 'stable' isotope deuterium?

Are all deuterium atoms destined to be deuterium for eternity or is there any process other than radioactive decay that could lead to transition such as say $\ce{^1H}$?


2 Answers 2


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 helium-2… which would then decay back to deuterium (if it didn't break apart first).

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    $\begingroup$ Perhaps it is interesting to also point out that while deuterium is stable, it is relatively weakly bound. One consequence is that it undergoes fusion comparatively easily, at temperatures as low as ~$8 \times 10^5 K$. Almost all the deuterium present in the Universe was created during the big bang nucleosynthesis, and since then it has been the first nuclear fuel to ignite in any object with mass above approximately 13 times that of Jupiter. $\endgroup$ Commented Feb 10, 2015 at 3:38
  • $\begingroup$ @NicolauSakerNeto Interesting. I knew about fusion and nucleosynthesis, but I didn't know that it's relatively weakly bound. Is the fact that it's weakly bound unusual for stable isotopes? $\endgroup$
    – HDE 226868
    Commented Feb 10, 2015 at 3:40
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    $\begingroup$ As can be seen here, deuterium is the composite nucleus with the smallest binding energy per nucleon of any naturally found isotope; it would seem there are just not enough nucleons to allow for lots of binding via the strong force. Not only that, its composition is very close to a much more stable nucleus, helium-4. The combination makes deuterium quite special in astrophysics and nuclear physics. Precisely why it should be that way is a rather difficult question to approach quantitatively. $\endgroup$ Commented Feb 10, 2015 at 3:59
  • $\begingroup$ Nice answer, +1. This isn't my area, but is deuterium "completely stable" if it turns out protons are unstable? Some physics theories predict proton decay and the experimental lower bound on proton half-life is "only" 10^34 years. $\endgroup$
    – Curt F.
    Commented Feb 10, 2015 at 5:35
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    $\begingroup$ Well, if protons are unstable, the entire idea of stable vs radioactive nuclei kind of goes out the window - nothing would be completely stable. $\endgroup$
    – Jon Custer
    Commented Feb 10, 2015 at 14:26

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.

  • $\begingroup$ Technically protons would not remain fully stable. The antiproton generated in your reaction (which I recognize as $-1$ rather than $+1$ baryon) would consume a proton unfortunate enough to be at the wrong place and time. $\endgroup$ Commented 2 days ago

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