|WikiProject Physics||(Rated C-class, Mid-importance)|
|WikiProject Astronomy||(Rated C-class, High-importance)|
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- 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)
- 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.126.96.36.199 (talk) 02:41, 6 September 2013 (UTC)
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? 188.8.131.52 23:30, 12 March 2007 (UTC)
4He + 4He ↔ 8Be + γ?
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)
- 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)
Diagram is wrong. Does not match text
Modest Genius is right. The diagram shows an incorrect reaction.
The initial reaction is endothermic. It is the inverse of Alpha decay. There are no other particles involved.
- There may well be two channels for the release of energy after Be-8 combines with an alpha. 7 Mev is more than enough energy to create e+,e- pairs. SBHarris 17:33, 7 October 2010 (UTC)
- The gamma ray on the first reaction is likely correct. Remember that in order to fuse, the two alphas have to overcome the coulomb barrier to get close enough. I'd expect the resulting Be nucleus to be in an excited state and have to emit a gamma to relax to the ground state.
- That said, it's quite possible that my expectation is wrong about this. Does anyone have an astrophysics text handy? That should sort out the question very quickly. --Christopher Thomas (talk) 18:37, 7 October 2010 (UTC)
- I read through one of the references: "Nuclear Reactions in Stars Without Hydrogen". It looks like your (Christopher Thomas) expectation is correct. Even though there is a overall energy loss in the first reaction, it requires 95 kev of input energy. Since the mass of Be-8 is 8.005305, and the mass of two He-4 is 8.005206, there is a net mass gain of 0.00099. Since 1 u = 1.49 x 10-10 J = 931 MeV, the extra mass equals 92.1789 kev of energy that stays with the atom as increased mass and leaves 2.8 kev to be released as a gamma ray. Contrary to my first assumption, it looks like the diagram is right, and the text is wrong.--Bartosik (talk) 11:40, 9 October 2010 (UTC)
- Bartosik, Gamma Rays have energy greater than 100 keV. A 2.8 keV photon would be a soft X-ray. If the excited Be-8 state releases a 2.8 keV photon its NOT a gamma ray, and the diagram should show an emitted X-Ray instead. Or, as Christopher says, 2.8 keV is a modest kinetic recoil for a Be-8 nucleus, and no photon need be released at all? Zirconscot (talk) 14:59, 30 June 2012 (UTC)
SI Units Please
Can we have energy in Joules please ? Surely this is not too much to ask for now we are in the 21st century ? Also can someone please show us how these energies are calculated ? — Preceding unsigned comment added by 184.108.40.206 (talk) 05:15, 17 June 2012 (UTC)
- Physicists use electron volts, not joules, when working with nuclear reactions. Actually, MeV (Mega electron volts) is commonly used. This is much more convenient than joules. Joules are good for ordinary earthbound phenomena involving energy, especially work and heat. (That's "work" in the physicist's sense: when one system does work on another, the work is the amount of energy transferred. For example, you can compress a gas with a compressor. The compressor does work on the gas.) Calories are convenient in other cases.
- SI units are not always the best:
- Astronomy: light-year, parsec, astronomical unit (equal to the radius of the earth's orbit). Each is useful for a different reason in astronomy. And astronomers don't convert years to seconds.
- Pressure: That on the earth is 1 atmosphere. Useful when comparing to the Earth's atmosphere.
- G forces: You wouldn't use newtons, would you?
- The electrical charge of particles, in many cases, is more conveniently measured in multiples of that of a proton, as in the charge on an electron (which would be -1), or quarks, which, depending on which quark you're dealing with, is +/- 1/3 or +/- 2/3 of that amount of charge. Would you instead use coulombs or esu? The charge on a proton is 4.803 X 10^-10 esu, or 1.602 X 10-19 coulombs. I think not.
- There's also a foot (of cable, or patch cord, if you will), which is how far light travels in one nanosecond. This is useful in high-energy physics experiments for estimating signal delays.
- Time: years, months, weeks, and days are not metric units, yet scientists use days and years, at least. The dinosaurs were wiped out 65,000,000 years ago. No one converts that to seconds.
- Bottom line: Use the units that are appropriate for the job. Betaneptune (talk) 03:46, 25 December 2012 (UTC)
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)
The third reference is 404.
G. Audia,§, O. Bersillonb, J. Blachotb and A.H. Wapstrac, http://www.nndc.bnl.gov/amdc/nubase/Nubase2003.pdf/ The NUBASE evaluation of nuclear and decay properties, (2001)
Temperature as keV
- 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)
- 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)
Improbability and fine-tuning
The section "Improbability and fine-tuning" seems out of place and a bit of a stretch for this article. Perhaps is should be moved to Fine-tuned universe with a mention of the idea on this article. Zedshort (talk) 16:43, 24 August 2016 (UTC)