Timeline for What happens to a radioactive carbon dioxide molecule when its carbon-14 atom decays?
Current License: CC BY-SA 3.0
19 events
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| Apr 23, 2020 at 20:43 | vote | accept | etc | ||
| Apr 23, 2020 at 20:43 | vote | accept | etc | ||
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| Jan 11, 2017 at 19:49 | vote | accept | etc | ||
| Apr 23, 2020 at 20:43 | |||||
| Jan 8, 2017 at 21:24 | vote | accept | etc | ||
| Jan 8, 2017 at 21:24 | |||||
| Jan 8, 2017 at 4:00 | answer | added | user6999 | timeline score: 15 | |
| Jan 8, 2017 at 2:50 | comment | added | user6999 | @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. | |
| Jan 8, 2017 at 1:29 | comment | added | user6999 | And as Loong points out in his answer, the new electron (beta particle) leaves at relativistic speeds, so it's unlikely to break those bonds either. This doesn't seem like a valid logical step to me. It seems perfectly reasonable to imagine that the electron would interact electromagnetically with the other electrons on its way out of the molecule. In fact, doesn't your argument claim to prove something that contradicts the experimental results described in the Snell paper? The time scale for the electron to exit is short, but its EM radiation is intense due to the rapid time variation. | |
| Jan 7, 2017 at 14:04 | comment | added | MSalters | @BenCrowell: "breaking the bond" requires a physical process, such as one of the atoms recoiling at high enough speeds. The bonds themselves are electrons in orbit, which aren't directly impacted by an event in the nucleus. And as Loong points out in his answer, the new electron (beta particle) leaves at relativistic speeds, so it's unlikely to break those bonds either. | |
| Jan 7, 2017 at 3:13 | comment | added | user6999 | @MSalters: Your argument doesn't work, because your estimate of 1 eV is only a bound on the energy tied up in the center-of-mass motion of the recoiling nitrogen nucleus. It isn't a bound on the amount of energy that can go into other forms of energy, such as breaking the bond. | |
| Jan 6, 2017 at 20:21 | comment | added | Ivan Neretin | @MSalters: it's thousands, not millions (the electron is faster all right, but also lighter). | |
| Jan 6, 2017 at 17:49 | comment | added | MSalters | @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. | |
| Jan 6, 2017 at 15:22 | history | edited | etc | CC BY-SA 3.0 |
Incorrect assumption
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| Jan 6, 2017 at 14:48 | answer | added | user7951 | timeline score: 26 | |
| Jan 6, 2017 at 14:25 | answer | added | Linear Christmas | timeline score: 39 | |
| Jan 6, 2017 at 12:35 | comment | added | Ben Norris | 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. | |
| Jan 6, 2017 at 12:28 | history | tweeted | twitter.com/StackChemistry/status/817347142545186816 | ||
| Jan 6, 2017 at 10:42 | comment | added | Ivan Neretin | 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. | |
| Jan 6, 2017 at 10:33 | history | edited | Martin - マーチン♦ | CC BY-SA 3.0 |
deleted 19 characters in body
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| Jan 6, 2017 at 10:20 | history | asked | etc | CC BY-SA 3.0 |