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I am talking about the nuclear reaction

$$ ^6\text{Li} + n \rightarrow\ ^4\text{He} +\ ^3\text{H} + 4.78\text{MeV} $$

A neutron hits a lithium-6 nucleus and together they form an alpha and triton particle. Is it valid to say that the lithium nucleus "decays" when hit by a neutron? Is there any other verb which better describes the change of the lithium nucleus?

I am interested in the correct terminology.

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  • $\begingroup$ You already have the right verb: "reacts". $\endgroup$ Jan 14, 2018 at 23:58

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A ground state $^7\mathrm{Li}$ nucleus is stable, so this reaction is either direct or involves a unstable, intermediate, excited state of the lithium-7 nucleus.

If you are studying that excited state1 then you consider this reaction as $$ ^6\mathrm{Li} + n \longrightarrow \, ^7\mathrm{Li}^* \longrightarrow \, ^4\mathrm{He} + ^3\!\mathrm{H} + \text{4.78 MeV} \,, $$ and would definitely refer to the intermediate state's eventual breakup as "fission" or "decay".

That said, the reference I used to look up the properties of lithium-7 doesn't explicitly list an excited state for the system and it does list long-lived meta-stable states of other isotopes. From this I conclude that this is probably a very short-lived state which means that for most purposes you can treat the reaction as direct in which case it would not be appropriate to talk of "decay".


1 This is the kind of thing that old school nuclear physicists dd a lot. Much of the information on $^{13}\mathrm{C}(\alpha,n)^{16}\mathrm{O}$ is coached in terms of the properties of the $^{17}\mathrm{O}$ intermediate state.

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  • $\begingroup$ <<it would not be appropriate to talk of "decay".>> Do you know of a word better suited in this example? One which fits the formulation "When hit by a neutron, the lithium-6 nucleus <verb>s into an alpha and a triton particle"? Or do you think that this line of thinking is misleading in this case? $\endgroup$
    – M.Herzkamp
    May 12, 2015 at 11:24
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    $\begingroup$ Interesting. Table 7.4 is for $^6\mathrm{Li}(n,\alpha)^3\mathrm{H}$ (see page 14, item 10) and seem to suggest a lifetime for the excited state in the neighborhood of $10^{-21}\,\mathrm{s}$. (Response to an early version of @uhoh's comment below.) $\endgroup$ Jan 15, 2018 at 0:58
  • $\begingroup$ Fig. 9 Table 7.2 tunl.duke.edu/nucldata/ourpubs/07_2002.pdf there are several states above ${}^6$Li+n, narrow and broad. Certainly seems plausible that formation of a particle-unstable state followed by decay is a good way to describe it. If a plot of the neutron cross-section vs energy shows resonant structure, that would be another indicator, here, Fig. 1, Fig. 3 $\endgroup$
    – uhoh
    Jan 15, 2018 at 0:59
  • $\begingroup$ There's a nice peak but also a continuum in the plots, so probably the reaction is some superposition of several channels, some long lived (I'm calling 10E-21 long) and some better termed direct. $\endgroup$
    – uhoh
    Jan 15, 2018 at 1:02
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Decays happen to individual nuclei ( particles). When more than one nucleus(particle) are involved it is called an "interaction". In this case neutron Li scattering

Neutron capture by a nucleus is a possibility, in this case there is an intermediate nucleus formed , which can then decay.

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  • $\begingroup$ Are you saying that Neutron capture is a possibility for lithium-6, forming an excited state of lithium-7? Or is that part unrelated to my example? $\endgroup$
    – M.Herzkamp
    May 11, 2015 at 13:15
  • $\begingroup$ I have to look up if there exists neutron capture for lithium6. Both 6 and 7 are stable en.wikipedia.org/wiki/Lithium , and the capture crossection is small. I am trying to differentiate the vocabulary, a decay comes from one particle, unstable nuclides created with capture and fragmenting into lower A nuclides could be described as decaying, because they will have a temporary existence as one particle. $\endgroup$
    – anna v
    May 11, 2015 at 17:09
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The process by which the lithium becomes fissile due to neutron capture is called neutron activation. The subsequent decay is simply a fission reaction.

There seems to be a precedent on various sites for such a process to be called a 'neutron capture induced fission reaction', although most of the Google results for the term refer to the more usual fission of uranium and other heavy nuclei.

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  • $\begingroup$ Since one of the reaction products is a helium nucleus, could one also say that the subsequent decay is an alpha decay? $\endgroup$
    – M.Herzkamp
    May 11, 2015 at 13:12
  • $\begingroup$ P.S. Since one of the fission product in this case is an alpha particle $(^4\mathrm{He}$ nucleus$)$, 'neutron-induced alpha decay' would be equally valid as a name for the process. $\endgroup$
    – tok3rat0r
    May 11, 2015 at 13:14
  • $\begingroup$ So we both had the same idea at the same time :D $\endgroup$
    – M.Herzkamp
    May 11, 2015 at 13:15
  • $\begingroup$ Looks like we did! $\endgroup$
    – tok3rat0r
    May 11, 2015 at 13:15
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    $\begingroup$ It is a prompt reaction, Li6(n,T)He4 (and the major portion of the total neutron cross section). I'm not sure 'decay' or 'neutron activation' is entirely appropriate, since those indicate some uncertainty in exactly when it will occur. Here the neutron and the Li6 collide, and the T and He4 come out with no delay. $\endgroup$
    – Jon Custer
    May 11, 2015 at 14:05
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What happens depends on the energy of the neutron. At lower energy, it can simply merge with the $^6$L to form $^7$L. At a higher energy it can cause the $^6$L to fission, and at even higher energy it can break the $^6$L into pieces (spallation). This last mechanism is being studied for neutron induced fission, where the neutrons produced by high energy particles (protons, neutrons, deuterons) blast neutrons out of the lithium to stimulated fission events in Uranium, etc.

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  • $\begingroup$ Hi Royce, welcome to PSE! This answer is nice, but it would drastically improve if you were to include a reference where the interested reader can find more details. Thanks! $\endgroup$ Jan 14, 2018 at 21:38
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    $\begingroup$ This answer contains several errors. There is no neutron energy where the reaction $\rm n+{}^6Li\to {}^7Li+\gamma$ occurs. (The explanation is some argument about isospin which I don't recall at the moment, but it's the same reason that $\rm n+{}^3He \to{}^4He + \gamma$ does not occur.) The absence of a photon-emitting reaction makes lithium-6 an excellent neutron shielding material. Also, spallation generally refers to the emission of a large number of fragments from a heavy nucleus, which lithium is not. $\endgroup$
    – rob
    Jan 14, 2018 at 21:47

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