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As far as I understand, when $\ce{^{14}C}$ decays into $\ce{^{14}N}$, a neutron is turned into a electron and a proton, and a electron is emitted. But here is the part I don't understand.

If an electron is emitted (witch I guess means that one electron is sent away?) while we also get a new electron from the neutron, wouldn't that be +1 electron and −1 electron, witch would result in no change from the initial 6 electrons because they cancel each other out? How can it become a $\ce{^{14}N}$ that has 7 electrons when the resulting electron amount is 6?

numbers of particles

I am very new to chemistry and I have searched everywhere to find out how this makes sense but I haven't found anything.

andselisk
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    Also, a duplicate on Physics.SE: When Beta decay of C-14 happens, why doesn't the Nitrogen atom produced have a positive charge? – andselisk Sep 26 '21 at 14:43
  • The $\ce{^{14}C}$ atom emits an energetic electron that leaves a $\ce{^{14}N+}$ cation. Later on, the emitted electron will eject some of its own peripheral electrons and also hit other atoms to ionize them, producing more cations and more electrons. After a while, the $\ce{^{14}N+}$ has become a $\ce{^{14}N^{z+}}$ ion surrounded by a lot of secondary electrons and cations. When these secondary electrons have lost their excess of kinetic energy, $z$ electrons will fall on the $\ce{^{14}N^{z+}}$ ion and discharge it. The other electrons will be redistributed on all the hit atoms around. – Maurice Sep 26 '21 at 15:47

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