Talk:Californium/Archive 2
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Initial price of californium?
The article states that Californium was sold by the Atomic Energy Commission starting in the 1970s for $10 a microgram. Several other sources, however, given different figures. An October 1968 newspaper article (The Spokane Daily Chronicle) gives a figure of $100 per tenth of a microgram - there's also an amusing quote from Seaborg in the article, he called californium "a bargain at almost any price". A November 1968 article in the Leader Post (see here) cites the same figure of $100 per tenth of a microgram. This price was repeated in a number of newspapers at around the same time - with the equivalent per pound price arousing considerable interest. On balance, these sources seem to make a much more credible case that Californium first went on sale in 1968 for $100 per tenth of a microgram - rather than the 1970s for $10 a microgram figure currently included. What's not clear, though, is whether anyone actually bought it at this price. It would be great if someone could help clarify. Sir Nils (talk) 19:24, 1 July 2011 (UTC)
- I did some digging but could not verify with a more reliable source than a newspaper that Cf was sold prior to the 1970s. So I'm not comfortable mentioning that. But that is enough for me to copyedit the sentence to get rid of the impression that it was sold starting in the 1970s. I'm not concerned by the price being much higher in the late 1960s; Cf became easier to synthesize as time went by and the AEC set up the Cf-252 sale program specifically to encourage study of the element. --mav (reviews needed) 03:04, 4 July 2011 (UTC)
Stability
From the article:
The most stable of californium's twenty known isotopes is californium-251, which has a half-life of 898 years; this short half-life means the element is not found in the Earth's crust.[note 1] Californium-252, with a half-life of 2.645 years, is the most common isotope used and is produced at the Oak Ridge National Laboratory in the United States and the Research Institute of Atomic Reactors in Russia.
This may be a silly question, but based on the half-life as stated above, isn't Ca-252 more stable than Ca-251? It's half-life is almost 3 times as long. Kleuske (talk) 07:23, 21 July 2011 (UTC)
- 2.645 means a bit more than two and half years. Materialscientist (talk) 07:58, 21 July 2011 (UTC)
- I suspect Kleuske has fallen victim here to differences between how decimal and thousand separators are handled in the MOS and the way they are handled in his/her native culture. As a reminder, here are the relevant quotations from the MOS: "Use commas to break the sequence every three places: 2,900,000" and "use a period character between the integral and the fractional parts of a decimal number, not a comma or a raised dot (6.57, not 6,57 or 6·57)." --Khajidha (talk) 14:42, 21 July 2011 (UTC)
- And I am a native user of that convention, and still read it the wrong way. I find it awkward to read decimal parts of years. It might be clearer to give the half-life of Cf-252 in days (although "966 days" may cause confusion of a different sort). -- Donald Albury 15:37, 22 July 2011 (UTC)
- Just cut a decimal place or two. then the number will be shorter either way and won't look like a Euro style thousands separator. Plus it's un-needed precision for prose.TCO (reviews needed) 01:49, 23 July 2011 (UTC)
- Yep, shortening to 2.64 years does the trick. SBHarris 02:30, 23 July 2011 (UTC)
- Just cut a decimal place or two. then the number will be shorter either way and won't look like a Euro style thousands separator. Plus it's un-needed precision for prose.TCO (reviews needed) 01:49, 23 July 2011 (UTC)
- And I am a native user of that convention, and still read it the wrong way. I find it awkward to read decimal parts of years. It might be clearer to give the half-life of Cf-252 in days (although "966 days" may cause confusion of a different sort). -- Donald Albury 15:37, 22 July 2011 (UTC)
- I suspect Kleuske has fallen victim here to differences between how decimal and thousand separators are handled in the MOS and the way they are handled in his/her native culture. As a reminder, here are the relevant quotations from the MOS: "Use commas to break the sequence every three places: 2,900,000" and "use a period character between the integral and the fractional parts of a decimal number, not a comma or a raised dot (6.57, not 6,57 or 6·57)." --Khajidha (talk) 14:42, 21 July 2011 (UTC)
Introduction: absence from Earth's crust
I have removed the phrase "this short half-life means the element is not found in the Earth's crust." It is already stated that the element does not occur naturally, and that it was artificially produced. The half life is not a definitive reason for absence: absence of that which does not naturally occur scarcely needs explanation. Custard is not found in the Earth's crust either. Kevin McE (talk) 06:28, 21 July 2011 (UTC)
- It's not the same thing. Every one of the 288 nuclides that has a half life long enough to have existed from the solar system's primordial cloud (with nucleosynthesis in the supernovae that seeded it with elements heavier than carbon) exists primordially on Earth. If Cf had an isotope with a half life longer than Pu-244, it would exist as a primordial nuclide also. The fact that it doesn't means it must exist as a non-primordial or not at all.
In this case, it is apparently not at all. Absense of a long-lived nuclide means no primordial element, but not necessarily no natural occurence. There exist an additional 50 or so short lived radionuclides that exist naturally on Earth due to cosmogenesis (C-14), radiogenesis (Fr, Ra, Rn, etc), and natural nucleogenic processes (some Pu-239). No Cf appears to exist in that fashion, however. Its only route would be neutron absorption nucleogenesis (geonuclear transmutation) in the crust, and there just aren't enough neutrons around these days to do the job. SBHarris 06:45, 21 July 2011 (UTC)
- To sum up, yes, not only lifetime but also the bombarding particle (required for synthesis from lighter elements) is important. The custard comparison is invalid - all chemical elements found in custard occur in the Earth's crust. Materialscientist (talk) 07:12, 21 July 2011 (UTC)
- Yes, obviously the custard example was tongue in cheek. But Californium (not unlike custard) is only known to have existed by human intervention: there is no assertion in the article that it ever has existed naturally. Without evidence of that, the rate of decay is irrelevant to its absence from the earth's crust today. It seems to be a presumed reason for an assumed change in prevalence. Kevin McE (talk) 19:22, 21 July 2011 (UTC)
- SBHarris can answer better, but for what I know, the rate of decay is very relevant simply because it gives a rough estimate whether an isotope could have survived from the time when the Earth wasn't a solid yet and its particles were originating from the outer space. In other words, if the half-life is short, the element can't be primordial, and this is the primary selection factor for occurrence. However, it doesn't necessarily mean the isotope does not occur in the crust, because an unstable isotope can still be formed in detectable amounts by natural irradiation, provided both the precursor and the irradiating particles are abundant enough. (Here again, the half-life - of precursor though - is a major factor, because if it is too short, the precursor decays before conversion.) Materialscientist (talk) 23:41, 21 July 2011 (UTC)
- Yes. All elements heavier than boron are made in type IIa supernovas. There is every reason to think that such supernovas make every nuclide that CAN be made. If none are left of some of these > 4.6 billion years later, it's because they are radioactive with half lives too short for enough to still be detectable now, AND there are no routes to them having been naturally produced in some process, SINCE the supernova. The short-lived transactinides past plutonium illustrate all that very well. The lighter radioactives are usually made naturually from something heavier, such as fission producing Tc and Pm. But most heavy radioactives with short half lives have no way to be made naturally, except a way that occured too long ago (in supernovae) for them to survive to today. Cf is one of these. And BTW, please capitalize Cf, but not californium. SBHarris 02:43, 23 July 2011 (UTC)
- SBHarris can answer better, but for what I know, the rate of decay is very relevant simply because it gives a rough estimate whether an isotope could have survived from the time when the Earth wasn't a solid yet and its particles were originating from the outer space. In other words, if the half-life is short, the element can't be primordial, and this is the primary selection factor for occurrence. However, it doesn't necessarily mean the isotope does not occur in the crust, because an unstable isotope can still be formed in detectable amounts by natural irradiation, provided both the precursor and the irradiating particles are abundant enough. (Here again, the half-life - of precursor though - is a major factor, because if it is too short, the precursor decays before conversion.) Materialscientist (talk) 23:41, 21 July 2011 (UTC)
- Yes, obviously the custard example was tongue in cheek. But Californium (not unlike custard) is only known to have existed by human intervention: there is no assertion in the article that it ever has existed naturally. Without evidence of that, the rate of decay is irrelevant to its absence from the earth's crust today. It seems to be a presumed reason for an assumed change in prevalence. Kevin McE (talk) 19:22, 21 July 2011 (UTC)
- To sum up, yes, not only lifetime but also the bombarding particle (required for synthesis from lighter elements) is important. The custard comparison is invalid - all chemical elements found in custard occur in the Earth's crust. Materialscientist (talk) 07:12, 21 July 2011 (UTC)
So why does it still state "it is the heaviest element to occur naturally on Earth"?
More information
The "Universitium" + "Ofium" + "Californium" + "Berkelium" anecdote (http://elements.vanderkrogt.net/element.php?num=98) should perhaps be mentioned. Double sharp (talk) 12:59, 19 January 2012 (UTC)
Critical mass
Under "Applications", it says, "Californium-251 has a very small critical mass (about 5 kg)...." It should probably be stated that this is a calculation/extrapolation, as enough californium to directly test this has probably never been produced, and probably shouldn't be, because nuclear blasts are devastating and, in this case, would spray highly radioactive californium into the environment. Also, how credible are such estimates of critical mass?--Solomonfromfinland (talk) 08:03, 5 May 2013 (UTC)
- [1] Double sharp (talk) 13:18, 6 May 2013 (UTC)
- In case this wasn't clear: yes, it is a calculation, but a reliable one. I have edited the article to make it clear that it is a calculation. Double sharp (talk) 15:54, 19 October 2013 (UTC)
Use in processing of radioactive wastes ?
A recent article in Nature Chemistry, [2], reports a major advance in the chemistry of Californium that is a "game-changer" for the processing of radioactive wastes. The Nature article costs $32, and the abstract does not make the story clear to a non-chemist [like me.] Can anyone illuminate this claim? Wwheaton (talk) 22:52, 24 March 2014 (UTC)
Comments
The comprehensiveness of this article is quite disappointing for an FA. The surrounding actinides (Ac, Pa–Fm) are much better in this regard. I'd work on this, but Th is a much higher priority.
This doesn't mean that the material here is not good – it is. But I think there is a lot more that could be covered. The entry barrier is kept low, but I see no reason to excise so much content.
Lede
- "Californium is one of the few transuranium elements that have practical applications." – really? All transuraniums from Np to Es (except Bk) have been used. Beyond that they haven't been produced in large enough amounts, so they can't be used yet. I think this sentence misses the point, that almost all the transuraniums that can be produced in quantity are useful.
- Something could be said about Cf's lanthanide-like chemical behaviour, not like the earlier actinides.
Physical properties
- No mention is made of Cf's location in the PT (under Dy, between Bk and Es). This info is useful to showcase the trend in the actinides. Most of the surrounding actinide articles have this info. See Np and Bk for examples.
- Volatility isn't mentioned, and isn't linked well enough to the other properties of Cf. This property makes Cf harder to study than Am, Cm, and Bk.
- It's not mention that Cf will form "mirrors" on a quartz surface if brought to 300 °C in those vacuum-containing quartz containers. Also, it's not mentioned that it will react with the quartz.
- Valency of the metal isn't mentioned. This is a cool point, because it makes the Cf metallic structure complicated. Mention the how Cf is on the cusp between being a divalent metal (like Yb or Es) and being a trivalent metal (Am, Cm, Bk). It could be a case of film vs. solid, like Sm. Also mention and explain the link to enthalpy of sublimation. Finally it's never mentioned explicitly what the reason for this is (5f electrons require high energy to be promoted). Mention also that Cf (g) is divalent.
- High-pressure delocalization of the 5f electrons is cool, but it's never mentioned that they are not delocalized at STP.
- Alpha-beta transition temperature isn't mentioned (600–800 °C). The transition pressure is also missing (16 GPa). Also, it's not mentioned that the high-pressure structure is the same as alpha-U. This is important because they're both actinides, and I think comparison between actinides is important.
- Bulk modulus is introduced weirdly. I would just link it, or put the definition in parentheses – it's not directly related to Cf. Also I'm not sure why it is compared only to the Ln and Al values – I get that it is for comparison, but wouldn't it make more sense to compare it to other actinides? But don't cut the Al value, this is important for the reader.
Chemical properties and compounds
- Why the big list of compounds with no context? Firstly compounds should be its own large section – I doubt that enough Cf compounds are known that it should be split, like for F and Zn.
- The relative reactivity should be mentioned. I understand that it is a quite reactive actinide metal, about as reactive as the first few lanthanides, though less reactive than Eu, the most ignoble lanthanide. Eu's reactivity is mostly because it is a divalent metal. So what does this say about Cf?
- Colour change on reaction isn't mentioned (it becomes golden or bluish).
- "Californium is only water soluble as the californium(III) cation" – this is wrong, evidence for Cf(II) and Cf(IV) in solution has been reported, although it's not completely proven. Cf(V) has also been claimed, which would have a stable half-filled 5f subshell. Mention the usual conditions for the stabilization of high oxidation states like CfO2+ or CfO+
2. - Cf complexes are totally not mentioned. At least give a brief mention. Even a statement that they exist would be fine. Already there is not much chemistry and we should be thinking about what to add, rather than what to subtract.
Isotopes
- It should be mentioned here that Cf-252 is the major isotope produced.
- Mention the difference between Cf-249 and Cf-252, and which isotope is suitable for what work.
Stopping here for now. Double sharp (talk) 07:24, 20 July 2014 (UTC)
History
Looks good.
Occurrence
Looks good.
Double sharp (talk) 07:39, 20 July 2014 (UTC)
Natural Occurrence
This article states that californium is the heaviest element to occur naturally on Earth. What is the source for that statement and is it reliable? Every source I have seen states that plutonium is the last known naturally occurring element on Earth/in the Earth's crust. — Preceding unsigned comment added by 72.84.185.135 (talk) 16:31, 27 October 2012 (UTC)
- Emsley, John (2011). Nature's Building Blocks: An A-Z Guide to the Elements (New ed.). New York, NY: Oxford University Press. ISBN 978-0-19-960563-7. Double sharp (talk) 02:35, 28 October 2012 (UTC)
This source is available on Google Books: https://books.google.com/books?id=4BAg769RfKoC&pg=PA109&lpg=PA109&dq=natural+californium&source=bl&ots=k13oE1PMtL&sig=aQ7r-ojSQDF7Dv-kc4rfgdeJOVw&hl=en&sa=X&ei=4ckxVcLqEoXSoATdiYCQAQ&ved=0CEsQ6AEwCDgK#v=onepage&q=natural%20californium&f=false
That is the only source I've ever seen make that claim, and it doesn't cite any outside source nor does it state that any natural californium has ever been detected. It's not implausible that it might occur, but has there ever actually been any detection of it, or any other transplutonium element? If not, should Wikipedia be making this claim? - Bootstoots (talk) 18:01, 6 May 2015 (UTC)
I am extremely doubtful of the claim that californium occurs naturally in the Earth's crust. The book says that 5 isotopes exist naturally, including Cf-253, which has a half-life of only 17.8 days. That seems really far-fetched, even with neutron capture. There is only one source, it is a rather non-technical book, and it doesn't cite any sources for its claims. The claim that californium is natural should be removed from Wikipedia, in my opinion, but I am not sure enough to remove it myself.Fluoborate (talk) 07:32, 15 August 2015 (UTC)
Chemistry
http://www.osti.gov/scitech/servlets/purl/1225199/ Double sharp (talk) 11:59, 19 August 2016 (UTC)
Article discussion
Excuse me, some questions were discussed on Chinese Wikipedia:
- "Californium metal starts to vaporize above 300 °C (570 °F) when exposed to a vacuum."
- Anything will be vaporized when exposed to a vacuum, so what's the condition of this discription?
- "melting point of 900 ± 30 °C (1,650 ± 50 °F) and an estimated boiling point of 1,745 K (1,470 °C; 2,680 °F)."....
- Which kind of isotope? The way of estimate?
- 252Cf. Estimated by extrapolated vapour pressure, probably. Double sharp (talk) 11:59, 19 August 2016 (UTC)
- Which kind of isotope? The way of estimate?
- "Below 51 K (−220 °C) californium metal is either ferromagnetic or ferrimagnetic (it acts like a magnet), between 48 and 66 K it is antiferromagnetic (an intermediate state), and above 160 K (−113 °C; −172 °F)"
- What about 66 K to 160 K? Curie point?
- "A double-hexagonal close-packed form dubbed alpha (α) and a face-centered cubic form designated beta (β).[b] The α form exists below 900 °C (1,650 °F) with a density of 15.10 g/cm3 and the β form exists above 900 °C with a density of 8.74 g/cm3.[15] At 48 GPa of pressure the β form changes into an orthorhombic crystal system due to delocalization of the atom's 5f electrons, which frees them to bond.[16][c]"
- Which kind of isotope? 15.10 possesses four effective digit, what is it's measurement uncertainty? What's the condition of "8.74 g/cm"?---Koala0090 (talk) 09:57, 8 July 2015 (UTC)
- If the quoted statement is true, then β-californium is a crystal that exists only above the melting point. How can that be?
- Solo Owl 04:14, 23 November 2016 (UTC)
- The transition temperature is very poorly established, as is very common for elements like this: after all in 1974 two purported Cf allotropes turned out to really be hexagonal Cf2O2S and fcc CfS. The α-β transition temperature is probably somewhere around 600–800 °C. The melting point is not really known very well either anyway, as it was measured for very small particles, and similarly the crystal structures were determined from microgram quantities of Cf. There is a good reason why knowledge of the detailed chemistry of the actinides is mostly concentrated on the long-lived elements from Th to Pu! (These five and Tc are really the only radioactive elements that have been chemically investigated to a level on par with the stable elements.) Double sharp (talk) 04:30, 23 November 2016 (UTC)
- Thanks for your quick response. I thought it was like that. Solo Owl 13:49, 23 November 2016 (UTC)
- Which kind of isotope? 15.10 possesses four effective digit, what is it's measurement uncertainty? What's the condition of "8.74 g/cm"?---Koala0090 (talk) 09:57, 8 July 2015 (UTC)
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