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I’m aware that elements like $\ce{^14C}$ have a known half-life, which means that over a span of roughly $5730$ years, half of the $\ce{^14C}$ atoms decay into $\ce{^14N}$.

Are there any substances known to speed up this process, that applying it to $\ce{^14C}$ will catalyze a chemical reaction resulting in $\ce{^14N}$, or sticking them in a particle accelerator and slamming them together will result in $\ce{^14N}$, or otherwise transmute elements?

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    $\begingroup$ No sure what you're asking, but it is impossible to transmute an isotope via any chemical reaction. You of course can transmute isotopes in a particle accelerator - that is what a particle accelerator is designed to do. $\endgroup$
    – MaxW
    Nov 7, 2017 at 16:37
  • $\begingroup$ @MaxW What about in extreme conditions (for instance, abnormally high pressure or heat)? $\endgroup$
    – DonielF
    Nov 7, 2017 at 16:38
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    $\begingroup$ The sun "burns" hydrogen into helium via high pressure and heat. But that isn't considered to be a chemical reaction. $\endgroup$
    – MaxW
    Nov 7, 2017 at 16:40
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    $\begingroup$ How is a really opened ended question that would span a number of books. In general you're asking about astrophysics. en.wikipedia.org/wiki/Astrophysics There is the big bang, how a sun works, and man's effort to harness energy via fusion reactors instead of conventional nuclear reactors that work via neutron emission. $\endgroup$
    – MaxW
    Nov 7, 2017 at 17:18
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    $\begingroup$ This is more a nuclear physics question. Nuclear reaction and cross section data can be found at the National Nuclear Data Center, hosted at Brookhaven National Laboratory (nndc.bnl.gov). $\endgroup$
    – Jon Custer
    Nov 7, 2017 at 18:09

2 Answers 2

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It is possible to modify nuclear decay rates using chemistry, though it is rare and the effect is usually very small. Here I summarize the information available in this link. You may want to see the references within.

There is a type of nuclear decay called electron capture, where a nuclide directly captures an electron from the innermost electron shells and transforms a proton into a neutron. Therefore, there is coupling between the nucleus and the wavefunctions of the innermost electrons in this form of radioactive decay.

Usually the core electrons are very weakly disturbed by changes in chemical environment, but they are changed slightly. For especially light atoms, where the core electrons are very close to the valence shell, such as $\ce{^7_4 Be}$, this change results in a measurable difference in decay rate, varying by 0.1-1% compared to the isolated atom.

In a more dramatic case, there is a subset of beta decays called bound-state beta decay, where the electron released by the decaying neutron is immediately captured by the nucleus after decay. Apparently, if the parent atom is stripped completely bare of its electrons, and if the energy involved in the nuclear decay is comparatively low, once again there is a meaningful coupling between nuclear and electronic states. For the case of rhenium-187, the neutral atom $\ce{^187_75 Re}$ has a half-life of 42 billion years, but upon full ionization to $\ce{^187_75 Re^75+}$, the half-life is reduced to 32.9 years! For dysprosium-163, the neutral atom $\ce{^163_66 Dy}$ has half-life so high it has not been measured to decay, but when ionized to $\ce{^163_66 Dy^{66+}}$ its half-life is reduced to 47 days!

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    $\begingroup$ I should know better than to say "impossible." Great information! $\endgroup$
    – MaxW
    Nov 7, 2017 at 22:59
  • $\begingroup$ Perhaps Randall was incorrect after all? $\endgroup$ Nov 8, 2017 at 1:11
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    $\begingroup$ +1, but now we're left wondering what kind of chemical process can completely strip rhenium or dysprosium of electrons! $\endgroup$
    – ruakh
    Nov 8, 2017 at 6:19
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    $\begingroup$ @ruakh Indeed I am stretching the domain of chemistry by including such ions, but it's interesting enough that I had to do it. $\endgroup$ Nov 8, 2017 at 8:01
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    $\begingroup$ @jheindel I had a feeling I might get asked this when I saw this question bumped yesterday. I am not using these terms in a strict technical sense. All I mean to say is that there is some connection between what happens inside and outside of the nucleus. Apologies for the confusion. That said, it does appear to be possible to accurately speak of electron wavefunctions in this scenario, if I am reading this article correctly. This is probably because it depends on the wavefunction of the hydrogen-like atom after decay. $\endgroup$ Mar 19, 2019 at 22:43
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Nuclear decay is accelerated (artificially produced) all the time. Per wikipedia,

U-238 is not usable directly as nuclear fuel, though it can produce energy via "fast" fission. In this process, a neutron that has a kinetic energy in excess of 1 MeV can cause the nucleus of 238U to split in two.

So a neutron with 1 MeV kinetic energy can split the nucleus. 1 MeV = 1,000,000 eV.

Under normal conditions, we breathe around an ambient available thermal energy of $k_B T \approx 2.5$ kJ/mol. The neutron above has $96485307$ kJ/mol of energy.

A lot more. It takes a lot of energy to produce artificial radioactive decay; but we can and regularly do it.

I think I recall the US government used $\ce{UF6}$ as the primary energy-producer for nuclear development during/after WWII. Different forms (salts) of Uranium seem to make artificial decay easier (I don't know by how much, though).

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    $\begingroup$ +1 . But are you sure about that last sentence? What evidence is there that decay rates of U are different in its different salts? $\endgroup$
    – Curt F.
    Nov 7, 2017 at 18:48
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    $\begingroup$ That whole last paragraph is nonsense. $\endgroup$
    – Jon Custer
    Nov 7, 2017 at 19:11
  • $\begingroup$ I think it was easier to put energy into said salt -- not fundamentally faster decay. i.e. easier in lab. $\endgroup$
    – khaverim
    Nov 7, 2017 at 19:31
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    $\begingroup$ Turns out $\ce{UF6}$ is used for enrichment due to its high vapor pressure (processed as a gas at low temperature) -- i.e. isotope separation of U-235 from U-238. No artificial decay in that process; it's just the method by which the desired isotope is obtained in greater quantities $\endgroup$
    – khaverim
    Nov 7, 2017 at 21:13

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