As we know muon being more massive than electron species is more likely to fuse , but muon decays quickly , does that have any implications in nuclear fusion ?
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$\begingroup$ en.wikipedia.org/wiki/Muon-catalyzed_fusion $\endgroup$– VirgoJun 11, 2017 at 11:18
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1$\begingroup$ Read the link that @Virgo has provided: muons don't fuse; hydrogen atoms with the electron replaced by a muon fuse more readily than normal ones, but that is a different matter. $\endgroup$– dmckee --- ex-moderator kittenJun 11, 2017 at 15:59
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1 Answer
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In the muon-catalyzed fusion process, the muons basically replace the electrons in the hydrogen-isotopes Deuterium and Tritium. This makes the muonic atoms very small, which increases the probability for interaction processes.
Starting with a muon being captured by a Deuterium, it can, for example, be transferred from the Deuterium to a Tritium,
$$\mathrm{Dµ} + \mathrm{T} \rightarrow \mathrm{Tµ} + \mathrm{D},$$
which happens on a time-scale of roughly $10^{-9}\mathrm{s}$. A molecule can then be formed,
$$\mathrm{Tµ} + \mathrm{D} \rightarrow \mathrm{DµT},$$
which happens on an even faster time-scale. Note that I have omitted the released energies on the right-hand side of both reactions!
With Deuterium and Tritium being very close to each other, they fuse (and releases the muon and some energy), this takes roughly $10^{-12}\mathrm{s}$.
So, in summary, all these processes happen on time-scales much shorter than the mean lifetime of a muon (which is $2.2\,\mathrm{µs}$), therefore the answer is no.
Just a note: the muon-catalyzed fusion process cannot be used as a potential source of "free" energy, as it requires a tremendous amount of energy to create the muons and the catalyze-process is far from perfect as the muons have a certain probability to stick to the alpha-particles created in the D-T fusion process, which means they are lost for further fusion-processes.
