Talk:Triple-alpha process

message: The decay time of 8Be is about 10-16 seconds and not 10-6 seconds as far as I know. It says so here and here. 193.171.121.30 6 July 2005 15:33 (UTC)
:It is, in fact, 2.6 x 10^-16 s. Ref.: Wong, Samuel S.M. Introductory Nuclear Physics, p. 16. Betaneptune (talk) 03:52, 25 December 2012 (UTC)
replies: text: How about 8.7 x 10^-17 seconds in isotopes of beryllium  article.WFPM (talk) 17:40, 7 February 2013 (UTC)
replies: text: Whoever argues about such things as in the comments above do not know how difficult such measurements are to make, and that the digits like "8.7" and "2.6" are just estimates. The numbers listed above round off to 1.0 x 10^-16 for all practical purposes.98.67.97.54 (talk) 02:41, 6 September 2013 (UTC)
thread_title:  <sup>8</sup>Be decay 

message: I'm surprised by the statement that the helium flash lasts minutes and burns a large percentage of the helium. My recollection is that it lasts seconds and does not burn a significant percentage of the helium. Can you give a citation? 128.165.87.144 23:30, 12 March 2007 (UTC)
thread_title:  Helium flash 

message: The text doesn't like a γ in 4He + 4He ↔ 8Be, but the diagram does. Should this be changed?
:On that note, I made some consitancy changes to the page and noticed that it says "Be + He <-> C + γ + e", instead of "Be + He -> C + γ + e" (notice the difference in arrows). I'm assuming this is an error, but I am not a physicist, so I won't change it without confirmation. SkyLined (talk) 23:24, 2 March 2008 (UTC)
replies: text: Another thing I noticed: "The net energy release of the process is 7.275 MeV." whereas the equations imply that "7.367 MeV" is produced... where does the difference come from? SkyLined (talk) 23:31, 2 March 2008 (UTC)
replies: text: Presumably this is the net energy, i.e. add the energy produced by the two equations given. 141.211.184.143 (talk) 15:38, 9 March 2009 (UTC)
replies: text: There's still an inconsistency between the text and diagram as to whether the first stage emits a gamma ray or not. Naively I imagine that it does not, but may be wrong. Either way, it should be consistent. Modest Genius talk 21:53, 4 July 2009 (UTC)
replies: text: All you have to do is to apply the Law of Conservation of Mass-Energy, and you can figure it out about the gamma rays.98.67.97.54 (talk) 02:46, 6 September 2013 (UTC)
thread_title:  4He + 4He ↔ 8Be + γ? 

message: The Fusion process comparative reaction rate is reported as being proportional to an exponent of the very high ambient temperature which doesn't sound reasonable, and probable should be instead related to the ratio of the 2 considered temperatures.WFPM (talk) 18:09, 7 February 2013 (UTC)
thread_title: Reaction Rate

message: "until the central temperature rises to 10^8 K (8.6 keV). "
This sounds like a mistake to me. eV is not a temperature unit. 
Asgrrr (talk) 19:10, 5 May 2016 (UTC)
replies: text: Maybe the author meant average energy per helium nucleus in some model. But if nobody can find a source saying 8.6 keV, it should be removed. Rolf H Nelson (talk) 02:52, 6 May 2016 (UTC)

text: The electron volt can be treated as a unit of temperature.  The conversion factor works out at 11,605 K/eV.  So 8.6 keV comes out at 99.8 million Kelvin, which seems correct to me.  Lithopsian (talk) 15:59, 12 May 2016 (UTC)

text: I don't see what you mean. Your two numbers are 99 800 000 and 100 000 000 -- the same within round-off error. Eric Kvaalen (talk) 11:42, 7 March 2017 (UTC)
thread_title:  Temperature as keV 

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thread_title:  External links modified (January 2018) 

message: Since there is a mixture of carbon and oxygen, could these elements ignite in a conventional fire?  — Preceding unsigned comment added by 142.161.218.253 (talk) 20:04, 15 December 2018 (UTC)
replies: text: That would require elements and molecules.  In the centre of a star, we're talking about bare nuclei, electrons off wandering on their own, so no conventional chemistry.  Plus carbon dioxide, as an example, disassociates above  so molecules don't exist even in the atmospheres of most stars.  Lithopsian (talk) 21:05, 15 December 2018 (UTC)

text: A white dwarf (a lot) cooler than 4,200 K?  Not really a white dwarf any more, eh?  The core is out since it is degenerate, and the atmosphere of even a CO white dwarf is mostly hydrogen and helium, usually in two distinct layers.  A very small number of white dwarfs have atmospheres with a lot of carbon, but usually only trace amounts.  Gravity is very high, so white dwarf atmospheres rapidly stratify into separate elements.  If you really want to look at molecular chemistry around stars, cool ones obviously, you'll find a slightly strange dichotomy between those that are dominated by oxygen and those that are dominated by carbon.  Oxygen-rich stars are the norm, but in some cases sufficient carbon is convected to the surface that it becomes more abundant than the oxygen, and the chemistry changes completely.  Read S-type star and carbon star for some more on this.  Lithopsian (talk) 15:31, 16 December 2018 (UTC)

text: Back to the original question - regardless of everything above, "fire" isn't really on the cards.  There are situations where carbon and oxygen (or oxygen and other elements) can be combined into molecules, but this only happens once and then they're done.  Fire requires free oxygen to be produced, and the only situations where that can really happen is life.  Lithopsian (talk) 15:33, 16 December 2018 (UTC)
thread_title:  Chemical consequences of carbon-oxygen 

message: replaced what has been a mistake with another confusing stuff. A post-hydrogen star practically is in hydrostatic equilibrium. Gradual changes in its structure occur not because the core can’t withstand pressure of upper levels, but because in that age is loses more heat than produces, and consequently cools. Incnis Mrsi (talk) 09:26, 8 July 2019 (UTC)
replies: text: Do you want to change this?  It is accurate, but perhaps you think it misleads in some way?  Or just confuses?  It isn't really critical to this article.  I don't like what was there before, "heat-transfer equilibrium", because that isn't "a thing".  Looking at the current text, I also don't like that it could be read as saying that the production of carbon stops the core collapsing, rather than the production of energy.  It also ignores the fact that, in the vast majority of stars, the onset of core helium fusion occurs in a degenerate core that was already in (a different kind of) equilibrium and not collapsing.  Might be best to ignore the whole issue and just state that helium fusion begins when (and if) the core temperature gets hot enough.  Lithopsian (talk) 12:52, 8 July 2019 (UTC)
thread_title:  Equilibrium 

message: Task1 triple Alpha process: what is the result  of the formation of atoms? 58.69.115.20 (talk) 14:43, 24 August 2022 (UTC)
replies: text: Homework?  Lithopsian (talk) 14:49, 24 August 2022 (UTC)
thread_title:  Science