What happens to a radioactive carbon dioxide molecule when its carbon-14 atom decays? When carbon-14 decays, the decay products are nitrogen-14 and an electron (and an electron antineutrino, but that's chemically irrelevant*):

$$\ce{^14_6C -&gt; ^14_7N + e- + \overline{v_e}}$$

Let's assume that the carbon atom in question is part of a carbon dioxide molecule in the atmosphere. What would happen to the molecule when the atom decays into nitrogen? Will it be converted into a $\ce{NO2}$ molecule, or will it split apart? Will the electron created in the decay have sufficient energy to escape the molecule and form a positive ion?

Here's a somewhat related question dealing with the formation of radioactive carbon dioxide.

* Of course, not all the energy from the defect will be transferred to the beta particle's kinetic energy, so this is in fact relevant for the rate. See Loong's answer for details.
An article by Snell and Pleasanton, 'The Atomic and Molecular Consequenses of Radioactive Decay', (J. Phys. Chem., 62 (11), pp 1377–1382, $1958$) supports Ben Norris's comment.


  It is clear ... that $\ce{^{14}CO2}$ remains predominantly bound as $\ce{NO2+}$, a result that is perhaps not surprising. [This occurs in] $81$% of the decays. In $\ce{^{14}CO2 -&gt; NO2^+}$ dissociation yielding $\ce{NO+}$, $\ce{O+}$ and $\ce{N+}$ follows [in], respectively, $8.4$, $5.9$, and $3.6$% of the decays.


A table summarising the results is given.

$$\begin{array}{|c|c|}
\hline
\mathbf{Ion} &amp; \mathbf{\%\  abundance} \\
\hline
\ce{NO2+} &amp; 81.4(16) \\
\ce{NO+} &amp; 8.4(4) \\
\ce{O+} &amp; 5.9(6)  \\
\ce{N+} &amp; 3.6(4) \\
\ce{NO2^{2+}} &amp; 0.40(06)\\
\hline
\end{array}$$