Understanding products of triated water decay

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SUMMARY

The discussion centers on the decay process of tritium in tritiated water (^{3}_{1}H_{2}O), which results in the formation of helium-3 (^{3}_{2}He^{1+}) and an electron (e^{-}), along with hydroxide (^{3}_{1}HO). The decay products are energetic and lead to ionization of surrounding water molecules. The hydroxide ion is positively charged and requires an additional electron to achieve stability. The conversation highlights the complexity and variability of the decay mechanisms involved.

PREREQUISITES
  • Understanding of nuclear decay processes
  • Familiarity with isotopes, specifically tritium and helium-3
  • Basic knowledge of ionization and electron behavior in chemical reactions
  • Awareness of the properties of hydroxide ions
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  • Research the decay pathways of tritium and their implications in nuclear chemistry
  • Study the properties and behavior of hydroxide ions in aqueous solutions
  • Explore the ionization effects of energetic particles in water
  • Learn about the applications of tritiated water in scientific research
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Chemists, nuclear physicists, and students studying nuclear decay processes and their chemical implications.

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Figured this was more a chemistry question than nuclear question, so I put it here.

When an atom of tritium in ^{3}_{1}H_{2}O decays, it becomes ^{3}_{2}He^{1+} + e^{-} + ^{3}_{1}HO

This then quickly becomes ^{3}_{2}He + ^{3}_{1}HO

I'm a little confused about the hydroxide. It needs another electron, so it's...positively charged? Anything else I have wrong?
 
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And I just now realized I misspelled "Tritiated" in the title.

Darn.
 
These are rather difficult things to predict. Note that both products of the fission are quite energetic and they both jump away from the initial point, ionizing water around. In the end it will all come to some kind of rest/equilibrium, but exact mechanism can take many paths.
 

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