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I'm aware that tritium undergoes beta decay forming Helium-3. But in the case of HTO, what does it decay into? I don't suppose it decays into HeHO.

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Very shortly after a radioactive element decays in a molecule, you could consider that for a brief moment, it does indeed form a molecule with the new element in the same place as its originating radioisotope. So yes, after the decay of $HTO$ you could get $HeOH^+$ via the unusual reaction $TOH → HeOH^+ + e^- + \bar\nu_e$. Of course, though, this is a very unstable compound and it would immediately participate in further chemical reactions, very likely releasing the helium atom at some point. Even if the molecule were relatively stable after having one of its atoms transmuted, the molecule might still decompose as the nuclear decay releases a lot of energy, which can tear the comparatively fragile molecule to pieces.

A curious tidbit is that, in fact, the decay of tritiated compounds has been considered as a means to produce legitimate helium compounds, given that directly reacting helium with other substances has been fruitless other than for very trivial species, such as $HeH^+$. The problem is that tritium has a rather long half life of about 11 years, so it would be quite hard to produce reasonable quantities of helium compounds in a reasonable timespan, along with the likelihood of the helium compound being very poorly bound and the emission of vast quantities of nuclear decay energy which could easily decompose the fleeting substance.

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  • $\begingroup$ Thanks for your reply. What would HeOH+ be likely to decay into? $\endgroup$ – darrentnh Nov 3 '13 at 12:12
  • $\begingroup$ That's rather hard to say for sure. An easily seen pathway is for the helium atom to leave as a neutral atom, releasing $OH^+$, which itself would very quickly react with water, possibly giving the $H_3O_2^+$ ion, which would lose a proton to form $H_2O_2$, hydrogen peroxide. There are likely several types of reactions that take place simultaneously, since the initial $HeOH^+$ molecule has such a high amount of energy that even strange reaction pathways can have their activation energy met and will have more stable products than the reactant. $\endgroup$ – Nicolau Saker Neto Nov 3 '13 at 12:46

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