When a C-14 atom undergoes Beta decay, the Nitrogen atom produced has 7 protons but only 6 electrons. Why is it not positively charged? I'm baffled.


The nitrogen does have a positive charge i.e. only 6 electrons. Eventually it captures an electron from its surroundings to form a neutral atom.

The reaction is often written as $^{14}_6C \rightarrow {}^{14}_7N + \beta^- + \bar \nu_{\rm e}$ which has a net negative charge on the right-hand side, whereas it should be written as $^{14}_6C \rightarrow {}^{14}_7N^+ + \beta^- + \bar \nu_{\rm e}$ to balance charges.

When doing energy calculations when the relative atomic masses are given the mass of $^{14}_7N^+ + \beta^-$ is taken to be the mass of a neutral nitrogen atom as the binding energy of an election in an atom (~eV) is much less that the of the energies involved in the nuclear decay process (~MeV).

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  • $\begingroup$ So, the products of Beta decay are often written with no charge? That is quite misleading! $\endgroup$ – Mathematician Oct 23 '16 at 9:59
  • $\begingroup$ Charges are also not usually shown for alpha decay but they are implied. It is a matter of convenience. $\endgroup$ – Farcher Oct 23 '16 at 11:57
  • $\begingroup$ @Mathematician Usually the nuclear decay equations, like those in this answer, can be thought of as describing only the nucleus, despite very similar notation used in chemistry used to describe the entire atom. You could also add "six bound spectator electrons" to both sides of Farcher's first expression to clarify. $\endgroup$ – rob Oct 23 '16 at 15:00
  • $\begingroup$ Wow, it's amazing how dumb people teaching elementary physics think physics students are. Such gaps need to be pointed out for the sake of clarity! I feel more concerned about the charges on in an alpha decay equation now. How would we deal with that? Does the starting element have to lose two electrons first? That may explain the conversion of some mass to energy. [Note: I will ask a separate question about the mass conversion to energy during beta decay.] $\endgroup$ – Mathematician Mar 11 '17 at 12:54
  • $\begingroup$ Your comment is incomplete. May you please elaborate? Thank you very much :) $\endgroup$ – Mathematician Mar 11 '17 at 12:59

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