Talk:Spontaneous fission

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I modified the spontaneous fission rates reported, some of which weren't quite right, based on the data on half-lives and branching ratios provided by Brookhaven National Laboratory in their handy-dandy chart of the nuclides (http://www.nndc.bnl.gov/chart/). I also modified the statement that spontaneous fission can't provide the flux needed for a chain reaction, since, if a critical mass is present, it is possible for an INITIAL spontaneous fission to provide the neutrons to initiate a chain reaction.

The question about a code error below seems already to have been resolved before I got here; I fixed the other question below (explaining what Z and A are, and pointing out that it is possible to have spontaneous fission if Z^2/A>45 and not just if it is approximately equal). ---

Is there a code error? The following appears: \hbox{Z}^2/\hbox{A}\approx45. when a simple number or item or something would be expected. I'm not sure what exactly should be here, or how it should be coded, though.

Mathematical criterion

It would be nice to explain what Z and what A means in the equation (I guess A is atomic mass and Z is number of protons but I'm not sure). Additionaly it seems to me that if Z^2/A is greater than 45 then spontaneous fission can occur (but again it is my guess).

Mathematical criterion

It is interesting to note the formula Z²/A ≥ 45, as for Uranium, this doesn't work. Take U-238, which undergoes spontaneous fission. 92² = 8464. Divided by 238, this gives 35.6, which doesn't stack up.

This has been flagged with a citation needed tag. Piperh (talk) 18:17, 9 July 2009 (UTC)[reply]

I suspect the formula should be ≤ instead of ≥. The larger the atomic mass, the smaller the term Z²/A, which would mean that for any given element, there would be a maximum atomic mass that could undergo spontaneous fission, which doesn't sound right to me. In fact, for most elements, there are no isotopes that would fit that formula! So, there must be something wrong with that formula XinaNicole (talk) 11:29, 19 April 2011 (UTC)[reply]

Atomic fission is a failure rate characteristic of atoms with relation to their size and structure. The 2 competing failure modes are alpha particle emission and fission. The fission possibility occurs in the structural area that is more stable than alpha particle emission prone, but less than the neutron values required for a stable element. This implies the existence of a structural condition in heavier isotopes that permits their being fractured by impacting neutrons. At lesser energy levels, the more probable alpha particle emission occurs such as to get rid of the excess energy content.WFPM (talk) 14:58, 30 April 2011 (UTC)[reply]

Rates of spontaneous fission

I have put a citation needed tag in this section because the number for U-235 disagrees with what I have read elsewhere (althought I don't have all that much faith in other sources). It would be nice to have a reliable source. Man with two legs 09:54, 16 October 2006 (UTC)[reply]

Spontaneous fission occurrences

Spontaneous fission is a random failure rate characteristic of atomic nuclides. Evidently nature more or less indiscriminately produced a lot of them and then, over time, the more stable ones stayed in existence longer than the rest. A measure of the relative stability of each isotope is its halflifetime period, when half of the remaining ones will have decayed. Rather than a mishmash of time period values, the value of the halflifetime period is better understood as the base 10 logsecond halflifetime value, which can only vary from negative values (for less than 1 second) to say a value of 18 or more (the life of the universe). The occurrence of spontaneous fission is noted to occur to nuclides that are close to structural conditions that are stable, such that they don't have individual nucleon condition problems or alpha emission tendency problems, but are nevertheless not structurally enough stable to be able to survive a random variation of conditions that can result in their splitting into fragments.WFPM (talk) 19:06, 8 April 2011 (UTC)[reply]

Neutrons per g.s.

What's that then? — Preceding unsigned comment added by 95.146.50.110 (talk) 10:51, 25 July 2011 (UTC)[reply]

I am moderately knowlegeable w.r.t. physics. I have no clue WTF "g.s" is supposed to mean. I wonder why it needs to be said to explain jargon which is OBVIOUSLY not in common usage in an article explaining a technical term to a lay audience. Lame. And oh, gee, you think someone could actually write down specific decay processes as examples? Or would that be too informative?71.31.152.220 (talk) 22:29, 10 August 2012 (UTC)[reply]

Is 92 mass unit?

Why 92 mass unit? (92Zr), not 58Ni? or 116Sn??? — Preceding unsigned comment added by 59.126.202.81 (talk) 15:15, 15 July 2012 (UTC)[reply]

Nb-93

(Z^2)/A>45, but "Nb-93": (41^2)/93=18.0752688<45 — Preceding unsigned comment added by 59.126.202.81 (talk) 15:17, 15 July 2012 (UTC)[reply]

Stub

I am not sure why this article is NOT classified as a stub. I got here from cluster decay. In that article, it was stated that the decay fragments of S.F. are normally distributed. Yet here, NO mention is made of what the products are. Yet an almost completely USELESS sentence is included that cluster decay is super asymmetric spontaneous fission. Anybody really think giving something a name informs? The claimed distinction between S.F. and C.D. is that C.D. results in the same daughter nuclei every time while S.F. results in a distribution of isotopes. I am (obviously) not qualified to decide the accuracy of these statements. But I can say that the best articles answer the Who, What, Why, When, How, and Where questions; both by inclusion AND by exclusion. After reading this, I still don't know whether the process results in 2 fragments, 3 fragments, a variable number of fragments, a variable number of neutrons, or what. As a minimum it should DEFINE THE PROCESS(es) IT IS in a complete way.173.189.75.206 (talk) 23:07, 2 March 2013 (UTC)[reply]