I read some weird equation on wikipedia about the beta decay of radiocarbon:

${^{14}_{6}C} \rightarrow {^{14}_{7}N} + e^{-} + \overline{\nu_{e}}$

The problem with this equation that it does not conserve charge. In other terms it is like:

$8n + 6p^{+} + 6e^{-} \rightarrow 7n + 7p^{+} + 8e^{-} + \overline{\nu_{e}}$

so the neutron should decay to

$n \rightarrow p^{+} + 2e^{-} + \overline{\nu_{e}}$

which I think is not the case.

I read some interesting explanation here: https://chemistry.stackexchange.com/a/10455/24591 According to this the wikipedia equation is wrong and the proper equation is

${^{14}_{6}C} \rightarrow {^{14}_{7}N^{+}} + e^{-} + \overline{\nu_{e}} \rightarrow {^{14}_{7}N} + \overline{\nu_{e}}$

This appears to be wrong too at least there is an interesting comment by this answer:

If the emitted electron is "captured by the newly formed nitrogen", what does a Geiger counter detect?

So what is the truth here, which is the real equation of radiocarbon's beta decay?


First of all the $\beta $ particle emitting from nucleus is too energetic to be captured in atomic orbit to form atom.So it is definitely not the case. Infact thta's why we are sure that it is coming from nucleus not from atomic orbital. More over while writing for radio active decay we are looking for change in the nucleus as it is basically a nuclear reaction not chemical reaction where atomic changes are considered.So appropriate equation to describe $\beta$ decay should be $$n\rightarrow p^++e^-(\beta)+\bar{\nu_e} $$

If you are too interested about the atom itself I think it captures electron from environment which is suitable to be bound to an atom.

  • $\begingroup$ Okay, thanks! So the N+ is the good equation. I wonder why wikipedia cannot write good equations. :S $\endgroup$ – inf3rno Jan 12 '16 at 6:26

Probably the best way to think of this, and the way I suspect it works in the minds of people who write exactly this sort of equation in physics textbooks, is that it's an equation describing a nuclear process, rather than a chemical process. The electrons initially bound to the nucleus are irrelevant spectators. A less ambiguous way to describe the reaction might be $$ \rm (6e^- + {}^{14}_{~~6}C) \to (6e^- + {}^{14}_{~~7}N) + e^- + \bar\nu_e $$ The ambiguous overlap between the chemical notation for an atom of an isotope and the nuclear notation for a bare nuclide is unfortunate, but that's the price we pay for compressed notation.

Note that the energy required to ionize a neutral nitrogen is about 14 eV, while the beta-decay endpoint energy for carbon-14 is 150 keV. The two-body decay $$ \rm (6e^- + {}^{14}_{~~6}C) \to (7e^- + {}^{14}_{~~7}N) + \bar\nu_e $$ is allowed, but only when the antineutrino carries away 99.99% of the decay energy --- quite rare.

  • $\begingroup$ Thanks rob! I realized that the energy is huge compared to the ionization energy and the energy of chemical bonds. Are there measurements about the energy distribution between the 3 particles? $\endgroup$ – inf3rno Jan 14 '16 at 6:58
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    $\begingroup$ The three particles share the momentum more or less equally, since before the decay there is a frame where the unstable nucleus is at rest. Since energy is more-or-less $p^2/2m$, the energy is shared more or less equally between the two leptons; the exact division depends on the angle between them, which is a random variable. Predicting and measuring beta-decay energy spectra is a big industry. An interesting book on the subject is Allan Franklin's Are there really neutrinos? $\endgroup$ – rob Jan 14 '16 at 16:07

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