Is it an ion or a neutral atom that is created through beta decay?

For example: $_6^{14}\mathrm C \to ~ _7^{14}\mathrm N + \mathrm e^- + \mathrm{v{_e}}^{-}$

Isn't $_7^{14}N$ an ion since neutron gives us only a proton and an electron, not a proton and 2 electrons? If so, why isn't it stated in the example above (it's from Wikipedia)?

  • $\begingroup$ @MelanieShebel there's one more "Beta" to decapitalize in the text, and the minus sign on the neutrino needs to be removed. Also thanks for fixing the spelling error in my question earlier! $\endgroup$
    – uhoh
    Apr 9 '17 at 16:39

An ion is created. Usually this is not noted down in the nuclear reaction because while talking of nuclear reactions we only concern ourselves about nuclei, not the entire atom. The reaction is to be read as "A 14-$\ce{C}$ nucleus decays into a 14-$\ce{N}$, an electron, and an antineutrino."

Also, note that the neutrino has no charge.


Further to @ManishEarth's answer, note that the distinction is important in doing mass-energy calculations. In the above example, you can find the energy released by subtracting the tabulated mass of a $N^{14}$ atom from the tabulated mass of the $C^{14}$ atom, and ignore the electron mass. Why? Because the actual Nitrogen atom produced only has 6 orbital electrons, while tabulated values include the mass of 7 orbital electrons. So, in using the tabulated mass, you already include an extra electron...


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