# 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*):

$$\ce{^14_6C -> ^14_7N + e- + \overline{v_e}}$$

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.

• The molecule will probably split apart, together with a good many other molecules that will happen nearby. Nuclear energies are way above anything in chemistry. – Ivan Neretin Jan 6 '17 at 10:42
• The electrons are ejected with maximum kinetic energy of $156\ \mathrm{keV}$, which is on the order of $\mathrm{15\ MJ/mol}$. However, its only $\mathrm{2.5\times 10^{-20}\ J}$ per particle, which is not a lot but enough to get that electron traveling a maximum distance of 22 cm in air. You more likely have $\ce{NO2+}$ before it starts falling apart. – Ben Norris Jan 6 '17 at 12:35
• @IvanNeretin: The problem is how much of that energy is carried away by the ejected electron. You have simultaneous conservation of energy and impulse, in what's now basically a two-body system. Impulse conservation is easiest: since both bodies have opposite relative impulse, the electron since it's several thousand times lighter, must move several thousand times faster. But since kinetic energy goes up with the square of speed, the electron carries millions times more energy. There's less than 1eV left for NO2+, insufficient to split it. Any other molecule hit by the electron is fair game. – MSalters Jan 6 '17 at 17:49
• @MSalters: it's thousands, not millions (the electron is faster all right, but also lighter). – Ivan Neretin Jan 6 '17 at 20:21
• @MSalters: But since kinetic energy goes up with the square of speed, the electron carries millions times more energy. Actually the betas are normally ultrarelativistic, and the neutrinos are always ultrarelativistic. Therefore the proportionality is closer to $KE\propto v$ for them, but $\propto v^2$ for the nucleus. The 14C example is unusual because of the very low decay energy, which makes the electron (but not the antineutrino) approximately nonrelativistic. – Ben Crowell Jan 8 '17 at 2:50

There seem to be three processes to consider here:

1. As the high-energy electron exits, it creates a rapidly varying electrical current, which produces intense electromagnetic fields for a short time. This may act on the other charged particles that are present, possibly creating further ionization and/or breaking the chemical bond.
2. The nitrogen nucleus recoils, and this recoil could break the bond and/or create further ionization.
3. The charge of the nucleus changes, so even if effects 1 and 2 didn't exist, the state of the electrons is no longer the ground state. It is now a linear combination of different states of the new element, which means there is some probability of excitation or ionization.

A paper by Oksyuk and Gerasimenko says that effect 2 is normally the important one.

From Linear Christmas's answer, we know that in the case of $\ce{^14C}$ decay, which has an unusually low energy, there is a fairly small but still appreciable probability of breaking the bond, and this is presumably caused by one or more of the above processes.

For process 1, it certainly isn't safe to assume that it's negligible because the electron leaves the molecule so quickly. If this were true, then beta particles wouldn't produce ionization when they came along from the outside and hit atoms. Although the time-scale for the electron to exit is short, its electromagnetic fields are intense. By the way, the motion of this electron is not insanely fast compared to the velocities of the other electrons. It has a typical velocity in this decay of about $0.5c$, which could be compared to an estimate of $Zc/137\approx 0.04c$ for an inner-shell electron in carbon.

To estimate process 1, let's use the mean rate of energy loss for beta particles in a solid. For $\pu{0.1 MeV}$ betas in a solid made of fairly light elements, this is about $(dE/dx)/\rho\approx 0.3$ MeV.m2/kg. Taking $\rho$ to be the density of water, and $\Delta x=0.2$ nm, we find $\Delta E=0.06$ eV, which seems to be a couple of orders of magnitude too low to break a bond. However, energy loss of betas is a process that has a lot of random variation about the mean, so it doesn't seem unreasonable to me to imagine that there is something like a 1% chance that it deposits 100 times this energy in the parent atom on its way out. This puts us in the right ballpark for this mechanism to contribute significantly to the observed probability of breaking a bond in NO.

So now let's consider process 2. Let $Q$ be the energy released in the decay and $M$ the mass of the recoiling nitrogen nucleus. We'll see that for kinematical reasons, almost all of the energy goes into the electron and antineutrino, not the nucleus. Suppose we want to find the maximum energy of the recoiling nucleus. This is attained in the case where the electron gets almost 100% of the energy, because for a fixed energy, a more massive particle carries more momentum. If the electron and neutrino were to share the energy, then their momentum vectors could also partially cancel, further reducing the recoil.

The decay energy of $\ce{^14C}$ is unusually low, but in most beta decays the beta is much more relativistic. Let's do the ultrarelativistic case first, both because the math is simpler and because it's a better guide to our intuition about what happens in general.

In the approximation that the beta is ultrarelativistic, its momentum (in the case where it carries all the energy in this example) is $p\approx Q/c$, and by conservation of momentum, this is also the momentum of the recoiling nucleus. Since the nucleus is nonrelativistic, its kinetic energy is $K=p^2/2M\approx Q^2/2Mc^2$.

As a typical example, let's take $\ce{^40K}$, which is the strongest source of naturally occurring beta radioactivity in our environment. It has an 89% probability of decaying to $\ce{^40Ca}$ plus an electron and an antineutrino. The energy for this decay mode is $\pu{1.33 MeV}$, which is almost $10$ times that of $\ce{^14C}$. The ultrarelativistic approximation for the electron is not too ridiculous; in the case where it gets almost all the energy (none to the neutrino), its velocity is about $0.96c$. The maximum kinetic energy of the recoiling calcium nucleus in this approximation is about $\pu{24 eV}$, which is clearly plenty of energy to break a chemical bond.

Without the ultrarelativistic approximation, the momentum of the beta, in the maximum-recoil case, isn't $Q/c$ but rather $\sqrt{(x+Q)^2-x^2}/c$, where $x=mc^2$, and $m$ is the mass of the electron. Even in the case of $\ce{^40K}$, it turns out that the ultrarelativistic approximation is not so great. The actual maximum recoil energy for the calcium is $\pu{41.7 eV}$, so although the approximation gives the right order of magnitude, it's off by almost a factor of $2$.

In the example of $\ce{^14C}$, the result for the maximum energy of the recoiling nitrogen is $\pu{7.0 eV}$, in agreement with Loong's answer. This is plenty of energy to break the bond. Another interesting case with a very low energy is $\ce{^3H}$ , in which the recoiling $\ce{^3He}$ has a maximum energy of only about $\pu{3 eV}$. People have used this to try to study the chemistry of helium.

So the take-away here is that in almost all cases, beta decay is very likely to break up the molecule in which it occurs.

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 $\ce{C^{14}O2}$ remains predominantly bound as $\ce{NO2+}$, a result that is perhaps not surprising. [This occurs in] $81$% of the decays. In $\ce{C^{14}O2 -> NO2^+}$ 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.

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

• Nice find! I guess the main point is that the electron goes flying off - the 'kick' to the nucleus is small enough that the (remaining) bonds are OK with it. – Jon Custer Jan 6 '17 at 14:36
• @Jon Custer As Loong shows in their answer, the recoil energy transferred to the nucleus does in fact appear to (potentially) be able to break the bonds. The question then is whether the other ions result from this process or from the instability of $\ce{NO_2^+}$. – Marcel Jan 6 '17 at 15:43
• @Marcel - of course the recoil of the nucleus could break bonds. Whether it has to seems to be a no, which is just nice to see. I would presume there would be a fairly complex and hard to calculate dependence on dissociation and the direction of nuclear recoil with respect to the molecular bonds. – Jon Custer Jan 6 '17 at 16:26
• @Marel $NO_2^+$ is a stable ion. $N_2O_5$ is $NO_2^+ NO_3^-$ in the solid state. Other stable salts of $NO_2^+$ are known, such as $NO_2SbF_6$ – Demi Jan 6 '17 at 19:40
• @Marcel It is stable in the sense of "won't break apart spontaneously". It will nevertheless react with the nearest available reducing agent and/or nucleophile (probably water). – Demi Jan 7 '17 at 0:08

The antineutrino is not completely irrelevant. The decay energy is shared between the beta particle and the neutrino. Therefore, the beta particles appear with an energy distribution that ranges from zero to the maximum beta energy. The maximum beta energy for the decay of $\ce{^14C}$ is $E_\text{max}=0.1565\ \mathrm{MeV}$; the average beta energy is $E_\text{avg}=0.0495\ \mathrm{MeV}$.

Nevertheless, even beta particles with a relatively low beta energy have a much larger kinetic energy than any chemical bond dissociation energy or ionization energy. Thus, the beta particle cannot be captured in the electron shell of the affected atom. The beta particle leaves the atom with almost relativistic speed before the rest of the molecule understands what went wrong. This process is too fast for immediate chemical reactions. The daughter nuclide remains in the same chemical structure. Thus, the beta decay of a $\ce{^14C}$ atom in $\ce{CO2}$ simply leaves a $\ce{NO2+}$ ion in an excited state. This primary product is unstable and can dissociate slightly later. The resulting secondary products can be radicals or ions, which tend to react with almost any other molecule or atom.

However, the energy released in beta decay is not only distributed between the neutrino and the beta particle. A small amount is also allocated to the recoil of the nucleus. Since the mass of the nucleus is much larger than the mass of the beta particle, the recoil energy is much smaller than the energy of the neutrino and the beta particle and can usually be neglected in a first approximation.

For the beta decay of $\ce{^14C}$, the maximum recoil energy of the $\ce{^14N}$ nucleus can be calculated as $E_\text{recoil, max}=7.08\ \mathrm{eV}$. This value is much smaller than the total decay energy of $0.1565\ \mathrm{MeV}$ and could thus be neglected, but it is larger than the energy of a chemical bond. By way of comparison, the bond dissociation energy in $\ce{NO2}$ is only about $306\ \mathrm{kJ/mol}$ or $3.17\ \mathrm{eV}$. Therefore, the recoil of the $\ce{^14N}$ nucleus can be sufficient to break a bond immediately during the beta decay of $\ce{^14C}$.

• Thanks, great explanation! The graph you provided is interesting - could you explain why it has that shape? Also, how did you reach the calculated value for $E_{recoil}$? Did you apply conservation of momentum and assumed the electron's velocity to be directed orthogonally away from the nucleus? Lastly, how would the situation be if we assume that there is a large amount of $\ce{C^14O_2}$ gas instead of an isolated molecule? Would it be possible for the $\ce{NO_2+}$ ions and the electrons to "pair up", forming stable $\ce{NO_2}$? – Marcel Jan 6 '17 at 15:38
• I suspect my incorrect assumption is the result of too many "assume spherical cow"-type high school physics exercises. – Marcel Jan 6 '17 at 15:47
• @Marcel 1) Analytical equations that can approximately describe beta spectra are usually complicated since you have to take into account the conservation of energy and momentum for three particles, relativistic effects and Coulomb interaction between the nucleus and the beta particle. In this case, the plot is simply based on tabulated data and only serves to illustrate the continuous shape of the energy distribution. 2) Yes, the maximum recoil is calculated using conservation of momentum, taking into account the relativistic momentum of the beta particle. – Loong Jan 6 '17 at 17:20
• @Marcel 3) The beta particle will lose its energy in the surrounding gas until it is slow enough to be captured by another molecule. The maximum range in $\ce{CO2}$ at $p=1\ \mathrm{bar}$ and $T=25\ \mathrm{^\circ C}$ is about $19\ \mathrm{cm}$. If the excited $\ce{NO2+}$ ion is able to deexcite without dissociating, it could attract an electron from somewhere to form $\ce{NO2+}$. However, considering the short lifetime of the $\ce{NO2+}$ ion and the low specific activity of $\ce{^14C}$, it is extremely unlikely that this electron is the beta particle from another $\ce{^14C}$ decay. – Loong Jan 6 '17 at 17:20