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Author | Topic: Does radio-carbon dating disprove evolution? | |||||||||||||||||||||||||||
ChrisS Junior Member (Idle past 5631 days) Posts: 5 From: Melbourne, Victoria, Australia Joined: |
There has been some early (1980's) theoretical and field work done on the production of (i) neutrons and (ii) Carbon 14 in "soils" which may be of some help. These papers are probably not available in the 'net but should be available in a good university library.
Chris
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PurpleYouko Member Posts: 714 From: Columbia Missouri Joined: |
Beryllium is purified from minerals within the earth so in this purified state its said to beable to liberate 30 neutrons for every million hits from an alpha particle
Same numbers that I came up with. You've been doing your research haven't you
It forms various molecular compounds that due to the coloumb barrier no reason or evidence in the natural its releasing neutrons. Is there any evidence that beryllium molecular chemical compounds (not beryllium) that are found within the earth are violating the coloumb barrier?
I really can't say if there is any evidence of it happening with naturally occurring beryllium or not at the moment. That is a very good question. If refined elemental Beryllium exhibits a different magnitude coulomb barrier than it would were it bonded to another element then this would indeed pose a possible problem for the hypothesis.I will see what i can dig up on the subject. I strongly suspect though , that Be compound will not exhibit any differences due to the usually ionic nature of bonds formed by alkali earth metals such as Be or Li. While searching for research on the subject I came across this reference.
quote:It would appear that Gamma rays can also induce neutrons when they impact Be atoms. I will attempt to pursue this line of investigation a little later.
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PurpleYouko Member Posts: 714 From: Columbia Missouri Joined: |
Thanks ChrisS. I will look into your references and see if I can get them from my local University library
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PurpleYouko Member Posts: 714 From: Columbia Missouri Joined: |
From the first reference given by ChrisS i was able to pull up the abstract pretty easily.
quote:Unfortunately I cannot access an electronic version of the full paper. (I work 3 miles off campus so actually going to the library is difficult) From one of the other papers, I got this.
quote:This would appear to bear out the previously asserted point that C14 concentration in uranium rich soils is proportional to the Uranium concentration. |
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PurpleYouko Member Posts: 714 From: Columbia Missouri Joined: |
I think I hit the jackpot this time.
This is by far the most informative information that I have come across so far. In This Article we see tables of low Z elements that will release a neutron via collision with alpha particles. There are actually a lot more of them than I had previously thought. Sorry for leaving this as a bare link. The darn thing is a pdf and doesn't easily lend itself to conversion to text.(Try loooking at the html version for a laugh)Here are a couple of google's html translations that i felt particularly amusing. The alpha decay process leads to the emission of gamma rays tlom unstable &ughters (see Chapter 1). Also, the alpha @rticles can ”prpduce neutrons through (%n) reactions with certain elements. This source of neuons ean be comparable in intensity to spontaneous fission if isotopes W@ high alpha decay rates such as 233U,2MU,238Fu,or 241Amare pnt: This section describes thep@wtion of neutrons by (n) reactions and provides wrne guidelines for&kxdating@,expeeted neutron yield. Folloiving km two extiples ofn) Aactioni that occur in many nuclear fiel cycle And especially this one.
When the alpha piirticle’dives at ahoter nucleus, the probabili@ of a reaction
So now we know. It's all down to "the diffiwin in bindingergh% between the two initial nuclei".depends on the C@lue,e threshold enew, and the height of the Coulomb barrier. The Qdue is the ,diffiwim in bindingergh% between the two initial nuclei and the two final reaktion ”products! A!tisitive Bet you are all really relieved to hear that. As to the issue of there being a difference between the chance of a Be atom accepting an Alpha particle, depending on its chemical form... Consider the following. From table 11.4, the coulomb barrier of Be9 is 2.9 MeVFrom table 11.3, the average energy of an alpha emitted by the decay of U238 is 4.19 MeV. From this we can see that there is far more energy than is needed to overcome the coulomb barrier. If we compare MeV to the typical bond strengths of covalent bonds, we find that the bond strength is utterly insignificant in comparrison. For example, the strongest covalent bond known is the Nitrogen Nitrogen triple bond. It's strength is 945 Kj/mol.However we are looking at the strength of 1 bond, not a mol of them. 945/6.022 E23 (Avogadro's number) comes to 1.66 E-22 Joules of energy required to break a single (triple)N-N bond. 1 megaelectron volt (MeV) = 1.60217646 E-13 joules That means that one of our alpha particles (4MeV) has about 4,000,000,000 times the amount of energy needed to cleeve that covalent bond. Berylium bonds are a whole lot weaker than N-N so I think you get the picture.
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johnfolton  Suspended Member (Idle past 5591 days) Posts: 2024 Joined: |
The earth has sediments and are they not shielding the beryllium from the alpha particle? In the air the alpha particle can travel a couple of centimenters before its energy has expired, yet within the earth is not the problem that "the alpha particle is immediately absorbed by the sediments"?
The molecular compounds, nitrogen oxides (n14) would not produce neutrons and if beryllium did by chance get hit directly by an alpha particle. Would it not take 1,000,000 direct hits before 30 neutrons could be generated? Would not those direct alpha hits on the beryllium in an beryllium compound need to be not shielded by the surrounding sediments so the alpha particles energy would not be depleted in anyway below the coloumb barriers strength? Why would not the sediments that are not beryllium not immediately absorb the alpha particles energy depleting its ability to overcome the coloumb barrier? The sediments that are shielding the beryllium within the earth could explain why an alpha particle would not overcome the coloumb barrier? Edited by johnfolton, : No reason given.
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NosyNed Member Posts: 8996 From: Canada Joined: |
You keep bringing up reasons (that you more or less make up) why there can be no very low level of C14 produced by natural radioactivity.
If it can't happen then why does the amount of residual C14 correlate with the radioactivity of the surroundings? In what way does this explain the high degree of correlations between C14 and other methods when the dates are in a range that C14 is appropriate for? Why is there only a problem when we reach the dates that we expect there to be very, very little C14? Do you think the earth is made up of solid lumps of one kind of mineral and then another? Do you see a big lump of uranium there and then a lump of berylium? The soil is a mixture of very small amount of all sorts of things. This way the low atomic weight elements are, in some cases, very close to the sources of alpha particles. And so is the C12 close to the neutron sources. There are very, very large numbers of atoms in a very, very small amount of soil and we only need to produce very low numbers of C14 atoms to give a residual reading.
The sediments that are shielding the beryllium within the earth could explain why an alpha particle would not overcome the coloumb barrier? This shows that you have, at best, a most tenuous grasp of what is going on. The shielding by other elements in the soil would stop the alpha particle from reaching an appropriate target. The coloumb barrier is only an issue when the alpha particle actually reaches a target. It has already been shown the the coulomb barrier is then no problem. Do not ignore what has already been answered.
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johnfolton  Suspended Member (Idle past 5591 days) Posts: 2024 Joined: |
There are very, very large numbers of atoms in a very, very small amount of soil correct
The coloumb barrier is only an issue when the alpha particle actually reaches a target. incorrect
Penetration and Shielding Alpha particles interact with matter primarily through coulomb forces between their positive charge and the negative charge of the atomic electrons within the absorber. Edited by Admin, : Shorten URL.
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Percy Member Posts: 22391 From: New Hampshire Joined: Member Rating: 5.2 |
johnfolton writes: The earth has sediments and are they not shielding the beryllium from the alpha particle? In the air the alpha particle can travel a couple of centimenters before its energy has expired, yet within the earth is not the problem that "the alpha particle is immediately absorbed by the sediments"? The various elements and compounds in soil and sedimentary layers are all mixed together. The alpha particle that hits a beryllium atom is not one that has completed a miraculous journey of several centimeters, but one from a very nearby atom of a radioactive element.
The molecular compounds, nitrogen oxides (n14) would not produce neutrons and if beryllium did by chance get hit directly by an alpha particle. Would it not take 1,000,000 direct hits before 30 neutrons could be generated? I'll let PurpleYouko handle this one, but it seems to me that a direct hit is a direct hit, and that as long as the energy of the alpha particle is sufficient it should always result in a neutron.
Would not those direct alpha hits on the beryllium in an beryllium compound need to be not shielded by the surrounding sediments so the alpha particles energy would not be depleted in anyway below the coloumb barriers strength? As explained just above, the alpha particle would come directly without any intermediate collisions from a very nearby radioactive element. --Percy
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JonF Member (Idle past 168 days) Posts: 6174 Joined: |
The coloumb barrier is only an issue when the alpha particle actually reaches a target. incorrect
Penetration and Shielding Alpha particles interact with matter primarily through coulomb forces between their positive charge and the negative charge of the atomic electrons within the absorber. Wrong. The Coulomb force is not the Coulomb barrier. Try again.
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PurpleYouko Member Posts: 714 From: Columbia Missouri Joined: |
The earth has sediments and are they not shielding the beryllium from the alpha particle? In the air the alpha particle can travel a couple of centimenters before its energy has expired, yet within the earth is not the problem that "the alpha particle is immediately absorbed by the sediments"?
Of course the alpha is pretty much immediately absorbed by the sediment.The vast majority of alpha particles are going to hit something else than beryllium. It only takes a few. The molecular compounds, nitrogen oxides (n14) would not produce neutrons and if beryllium did by chance get hit directly by an alpha particle. Would it not take 1,000,000 direct hits before 30 neutrons could be generated?
Sure the conversion factor of the alpha-neutron reaction is only about 30 parts per million. I just don't see whay that is a problem. There are many many billions of radioactive disintegrations in a single kilogram of soil in any given year. It only takes one or two to make a C14 atom in order to raise the background to a continuous level.
Why would not the sediments that are not beryllium not immediately absorb the alpha particles energy depleting its ability to overcome the coloumb barrier?
I'm sure they do in a lot of cases.Remember that an alpha particle from a high mass radioactive isotope has somewhere in the vicinity of 4-5 MeV of energy. Beryllium has a coulomb barrier of about 2.9 MeV. The alpha can interact with a whole bunch of stuff before it loses enough energy to prevent it overcoming beryllium's coulomb barrier. The sediments that are shielding the beryllium within the earth could explain why an alpha particle would not overcome the coloumb barrier?
The sediments will certainly take away some of the energy from the alpha. The further it has to travel, the less energy is left. Again, though, all it takes is the odd successful transmutation to produce a continuum C14 background.
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PurpleYouko Member Posts: 714 From: Columbia Missouri Joined: |
I'll let PurpleYouko handle this one, but it seems to me that a direct hit is a direct hit, and that as long as the energy of the alpha particle is sufficient it should always result in a neutron.
Not quite percy, although it might appear that it should.Almost all nuclear interactions like this are able to proceed along various different pathways. A specific pathway has a specific yield percentage. JohnFolton is correct that for ever million Alpha-Beryllium impacts, only 30 neutrons producing reactions occur. However these neutrons often impact other heavier nuclei resulting in "Induced fission" in some cases. Since fission often produces 3 or more neutrons, it becomes possible that a single Alpha-Be fusion can set off a cascade of 10 or more neutrons.
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NosyNed Member Posts: 8996 From: Canada Joined: |
There are many many billions of radioactive disintegrations in a single kilogram of soil in any given year. This feels very high to me. I'd have guessed at 1,000 to a million times lower. Is it right? Edited by NosyNed, : fix dbcodes
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PurpleYouko Member Posts: 714 From: Columbia Missouri Joined: |
This feels very high to me. I'd have guessed at 1,000 to a million times lower. Is it right?
Dunno really. it was a "off-the-top-of-my-head" estimate due to limited time to do the actual math. Tell you what, i will do it noe and find out for sure. From my earlier post Message 100That means that for every kilogram of soil, there are 10 miligrams of Uranium.
The half life of U238 is 4.47E9 (4.47 billion) years1 mol of Uranium (238g) = 6.02E23 atoms (Avagadros number) Therefore 10 miligrams of soil would contain 2.52E19 atoms of Uranium (sounds a lot doesn't it?)That means that half of the U238 present in soil right now will be gone in 4.47 billion years so a quick estimate (if it were linear rather than exponential) would be just a case of dividing the number of atoms lost in that time by the time itself. 2.52E19 / 4.47E9 / 2 = 2.82E9 That makes an average of 2.82 billion disintigrations per year from Uranium alone. The real number is much higher for right now since the real fit is exponential. Nearer to 10 billion. Now take into account the disintigrations from U235, Thorium, Radium, Radon and all of the other semi and meta-stable isotopes in the Uranium decay chain I'm not going to bother doing all the math for every one of them but I think you must get the picture by now. I think many many billions of disitigrations per year per Kg of natural soil is actually quite a conservative estimate. Scary isn't it?
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