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I have a follow-up question to this one:

Solar neutrino Spectrum - Why are there discrete energies for Be and pep?

I understand why the lines for $^{7}\text{Be}$ and $pep$ are discrete, but why do we have $\textit{two}$ of them in the case of $^{7}\text{Be}$?

  • Because there are two different reactions leading to two different products. – Jon Custer Nov 20 '22 at 19:12
  • @JonCuster I know that the 860 keV line corresponds to the electron capture. What is the reaction responsible for the lower-energy line? – Igor Valuev Nov 20 '22 at 19:15
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    Check ENSDF for the 7Li energy levels. (Sorry, two energy levels, not totally separate reactions). – Jon Custer Nov 20 '22 at 19:22
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    @JonCuster Ah, I see, so the lower-energy line corresponds to the production of 7Li in its first excited state? – Igor Valuev Nov 20 '22 at 19:28
  • The energy difference seems about right by eyeballing it. – Jon Custer Nov 20 '22 at 19:30
  • @JonCuster Feel free to post it as an answer :) – Igor Valuev Nov 20 '22 at 19:32
  • That might be a while since I’m on mobile-only on travel. Feel free to self-answer! – Jon Custer Nov 20 '22 at 19:40
  • @JonCuster I am not quite clear how atomic spectra Be, Li, etc. can be detected on earth, given that the protosphere of the sun is around 6000 K. Wouldn't the blackbody thermal radiation of the sun overwhelm all atomic spectra lines? How can blackbody radiation noise be filtered clearly from the observation records taken on the earth? – James Nov 21 '22 at 08:46

1 Answers1

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If the final state contains three particles, solar neutrinos exhibit a continuous spectrum; if it's only two products, the solar neutrino spectrum is a discrete line. That's just energy and momentum conservation. The reaction you point out is

$e^-+{}^7\mathrm{Be}\rightarrow {}^7\mathrm{Li}+\nu_e$

But the ${}^7\mathrm{Li}$ has an energy state that is accessible in this decay; from the iaea.org Live Chart of Nuclides:

enter image description here

The lowest 477keV energy state is accessible, so the de-excitation gamma from that may carry those 477keV away. Hence, two discrete neutrino energies are possible with either 0.86MeV or 0.38MeV.

rfl
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