Talk:Isotopes of beryllium

⁸Be half life measure

decay width = 5.57 (25) eV per http://www-nds.iaea.org/relnsd/NdsEnsdf/showensdfdata.jsp?NucNo=8&NucID=BE Using standard decay width to time conversion ħ/Γ = half-life in s and applying error range to the same 1.18 (5) x 10^-16s

So I would say that both values are incorrect as listed currently on the Table for this isotope. 24.8.144.79 (talk) 07:05, 28 April 2013 (UTC)[reply]

⁸Be: Shouldn't alpha decay to helium-4 (read: two helium-4 nuclei) be considered fission?

Given that an alpha particle happens to be a helium-4 nucleus in the first place, why is alpha decay ***to*** helium-4 not noted as fission if the daughter nuclei are identical? It would make more sense to note it that way in my opinion (and to that of the average reader) to denote it as such. Thank you. 2602:306:BCA6:AC60:28F0:1D2E:C6A4:796D (talk) 07:23, 24 June 2013 (UTC)[reply]

Positron emission

Given that the isotope EO4Be9 is the isotope that supplies the positrons for the Fermi positron acceleration experiments, why isn't there a description of how this is accomplished? Is that due to the decay of a proton or a neutron? And how can a point source contain a +1 electrostatic charge?WFPM (talk) 21:30, 13 November 2013 (UTC)[reply]

I have no idea what you're talking about and can see no way that Be-9, our normal Be isotope, could supply positrons. Do you have a link? As for how a point source can be charged it's no harder for a positron than an electron. Ultimately it's still a mystery although QED helps explain what happens to space at such high field strengths very near the point charge. Ultimately the vacuum breaks down into virtual particles that prevent infinite field strengths. S B H arris 08:55, 14 November 2013 (UTC)[reply]

It's from the May 1985 National Geographic article about the "Worlds within the Atom", where the element Beryllium is described as the source the "manufactured antiprotons" needed for the proton-antiproton collision experiments. And these "antiprotons would thus be negative unit charged particles with the same mass as the proton. I guess I'm wrong about the production of a positron, because the article says antiproton. But somehow, the interacting positive proton is causing the presumably EO4Be9 atom to emit a negative unit charged particle. And the question is what it would be after it did that? It's a good article, but I don't agree with its drawing portrayal of the nucleus of the EE6c12 Atomic nucleus.WFPM (talk) 03:49, 15 November 2013 (UTC) PS In the Wiki antiproton article, it now says they're using a 29In (Indium) rod as a target. So I guess you have to keep close tabs on these things.WFPM (talk) 02:01, 16 November 2013 (UTC)[reply]

At Fermilab now they use a target of Inconel, an iron/nickel/chromium superalloy steel that can withstand sudden thermal shock stresses from proton beam heating. [1] Antiprotons are then focused with a lithium lens, basically a cylinder of lithium with a current running down it. Lithium because it is the least dense electric current conductor, thus minimizing scattering and annihiliation loss as the antiprotons travel through it. It's a fascinating problem in engineering physics, as is anything to do with accelerators. But I see no role for beryllium except its usual one to keep other metals from oxidizing while itself serving as a low density window for various radiations. S B H arris 02:46, 16 November 2013 (UTC)[reply]

Okay but that's what the article said. And I'm studying "The hunting of the Quark" By Michael Riordan, which details the history of SLAC and trying to understand how both point particles and quark containing composite particles can wind up with a net unit electric charge. Since it would have to be for different reasons, it sounds like quite a coincidence.WFPM (talk) 00:07, 17 November 2013 (UTC)[reply]

Beryllium-17?

Could you tell us where you found that there is a possible ¹⁷Be? There is no site where it is mentionned. — Preceding unsigned comment added by 92.92.224.178 (talk) 20:04, 2 February 2015 (UTC)[reply]

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Beryllium-6

Beryllium-6 is an intermediate in the proton proton chain. 2 He-3 > Be-6 > He-4 + 2 H-1 . 2603:6000:8740:54B1:C941:7ADA:D88:9366 (talk) 17:17, 6 May 2023 (UTC)[reply]

Beryllium-8 natural occurrence

Shouldn't Beryllium-8 be listed as extinct, rather than synthetic? It's in helium fusing stars. 174.103.211.189 (talk) 17:56, 21 November 2023 (UTC)[reply]

It's somewhat tricky to define. I don't believe ⁸Be can be considered extinct because it is constantly being created and destroyed in stellar cores (existing in secular equilibrium), yet due to its extremely short half-life, it does not exist on Earth outside of laboratories and has never been primordial (as is the case with extinct nuclides) anywhere. The infobox in beryllium-8 just states "0 natural abundance" with a footnote; for {{Infobox beryllium isotopes}}, it might be best to leave it as "synthetic" but with a similar footnote. ^Complex/_Rational 19:13, 21 November 2023 (UTC)[reply]

You could argue that beryllium-8 is the most important isotope of beryllium (along with beryllium-6 in the proton-proton chain). Who changed it from "extinct " to "synthetic " ? Besides, doesn't it occur naturally in the sun (stars above around 0.5 solar masses fuse helium) ? 174.103.211.189 (talk) 02:49, 27 November 2023 (UTC)[reply]

Shouldn't there be "intermediate " for nuclei like this. 174.103.211.189 (talk) 03:49, 27 November 2023 (UTC)[reply]

Another classification could potentially be beneficial, but "natural abundance" generally refers to abundance on Earth, now. Otherwise, we'd have to consistently list every nuclide found in stars or the interstellar medium that doesn't occur on Earth (e.g., ¹⁷F in the CNO II cycle; ⁵²Fe in the alpha ladder; various isotopes potentially observed in Przybylski's Star), for which inclusion criteria are not quite as clear. ^Complex/_Rational 23:25, 27 November 2023 (UTC)[reply]

By that logic, wouldn't we have to call many more nuclides extinct, considering the r-process path? Double sharp (talk) 05:20, 3 December 2023 (UTC)[reply]

Yes, but beryllium-8 is an intermediate in a major chain (triple alpha) in the Sun's lifetime. Sure, secular equilibrium is far beyond current technology (anyone who achieves this would be a surefire Nobel Prize winner) but it is in secular equilibrium as part of the triple alpha process. Was it created in the Big Bang or no? 174.103.211.175 (talk) 16:29, 10 May 2024 (UTC)[reply]

Almost certainly yes during the odd triple-alpha event, but the abundance of the CNO produced would be too low to detect (and it would also be swamped by stellar CNO). Last I remembered, it was something like one ¹²C atom produced for every 10¹⁴ atoms of ¹H. Double sharp (talk) 14:17, 22 May 2024 (UTC)[reply]

If it was created in the Big Bang, that means it was primordial, right? 174.103.211.175 (talk) 18:12, 28 July 2024 (UTC)[reply]

"Primordial" generally carries the implication that it hasn't gone extinct since. I wouldn't call ⁷Be primordial either, though it was created in Big Bang nucleosynthesis. Double sharp (talk) 06:19, 30 July 2024 (UTC)[reply]

If it was created in the Big Bang, it had been primordial, right? 2600:1008:B138:A08D:39E7:F942:305C:F0FE (talk) 17:01, 5 August 2024 (UTC)[reply]

Is the 2p decay of ⁶Be a ²He-emission?

The decay ⁶Be → ²He + ⁴He is energetially possible. ⁵He, ⁵Li, ⁶Be and ⁸Be are the only light nuclides that undergo alpha decay. 129.104.241.214 (talk) 02:44, 3 March 2024 (UTC)[reply]

"⁵He, ⁵Li, ⁶Be" someone else is thinking outside the box! 24.115.255.37 (talk) 23:11, 20 April 2024 (UTC)[reply]

Yeah, it's probably better to consider ⁸Be as being like these light nuclides than like the more normal heavy alpha emitters, because in these cases the alpha itself is no longer small compared to the nucleus. Double sharp (talk) 14:21, 22 May 2024 (UTC)[reply]

what

there are 6 isomers shown in the table but the first sentence says there are 3 isomers can someone fix 24.115.255.37 (talk) 23:10, 20 April 2024 (UTC)[reply]

Decay from ⁸Li or ⁸B

According to this link, ⁸Li and ⁸B can decay to ⁸Be as an intermediate product (rather than decay directly into ⁴He). 14.52.231.91 (talk) 00:50, 16 August 2024 (UTC)[reply]

⁸Be is quite stable in terms of binding energy

Firstly, ⁸Be has quite high binding energy among nuclides with A ≤ 11, only second to ⁴He.

Secondly, ⁹Be has quite low S(n) value of 1664.54 keV (compared to the high S(p) and S(n) values of ⁸Be), so its first excited level readily emits a neutron to become ⁸Be. ⁹B is even unbound. Imagine that ⁴He were sufficiently destabilized, then ⁵He and ⁵Li would become bound, but ⁹B wouldn't since it is unstable with respect to ⁸Be.

Lastly, note that the Q_α-value of ¹²C is only -7366.59 keV, while ¹³C, ¹⁴N, and ¹⁵N all have Q_α-values lower than -10 MeV (see here). If we agree that ¹²C is more stable than ¹³C, ¹⁴N, and ¹⁵N, then it is quite obvious that ⁸Be is more "stable" than ⁹Be, ¹⁰B, and ¹¹B. 129.104.241.231 (talk) 10:51, 1 October 2024 (UTC)[reply]

The high binding energy per nucleon of ⁸Be, and how it fails to translate into high stability, is picturesquely noted here. Double sharp (talk) 13:33, 1 October 2024 (UTC)[reply]

8Be half life measure

8Be: Shouldn't alpha decay to helium-4 (read: two helium-4 nuclei) be considered fission?