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Author | Topic: Isochron data puzzle | |||||||||||||||||||
allenroyboy Inactive Member |
I have a question about isochron data. Below is some data that puzzles me.
______________________________________________computed Sr86 ppmRb87___Sr87____Rb87/Sr86_error__Sr87/Sr86_error____Rb87/_____Sr87 / ppm____ppm____________________________________Rb87/Sr86_Sr87/Sr86 135.86__595.95__0.6606__0.0134__0.71990__0.00005__205.6615__827.8233161.60__582.60__0.8040__0.0230__0.72240__0.00005__200.9950__806.4784 126.72__612.52__0.5995__0.0106__0.72022__0.00009__211.3761__850.4624 141.78__641.18__0.6413__0.0108__0.72005__0.00006__221.0822__890.4659 140.79__642.96__0.6345__0.0110__0.71984__0.00010__221.8913__893.1985 151.26__570.05__0.7690__0.0129__0.72105__0.00007__196.6970__790.5832 On the left are data that I found in an on-line paper [ SBGEO - Sociedade Brasileira de Geologia ]. (This data is typical of others that I found on the internet.) It gives the ppm of Rb87 and Sr87 for 6 samples. Then it gives the ratios of Rb87/Sr86 and Sr87/Sr86 for the same samples. It occurred to me that one ought to be able to compute the ppm of SR86 for each of the samples. So I divided the Rb87 (ppm) by the ratio of Rb87/Sr86 and obtained what should be Sr86 (ppm) for each sample. Then I divided Sr87 (ppm) by the ratio of Sr87/Sr86 and obtaind what should be Sr86 (ppm) for the same samples. HOWEVER! Sr86 (ppm) computed from Rb87 averages at 209 (ppm), while Sr86 (ppm) when computed from Sr87 averages at 843 (ppm)!!!! Why? How can the same sample have 209 ppm and 843 ppm of Sr86? Is this typical of isotopic measurements?? Have I missed something? The formula for computing Sr87 over time from Rb87 is: SR87 = Sr87i + Rb87 ( e^t*lamda - 1) For Isochrons the formula is modified by dividing by Sr86 (where Sr86 is a stable isotope that must be the same ) SR87 = Sr87i + Rb87(e^t*lamda - 1)------ = -------__----- SR86 = Sr86i__Sr86 Yet the computations above indicate that ther are different amounts for SR86 from the same sample. Anybody know what is going on? Am I missing something? I have Faure, 1986, Principles of Isotope Geology, 2nd ed.; Dalrymple, 1991, The Age of the Earth; and Attendorn, 1997, Radioactive and Stable Isotope Geology. None of them point this out, nor explain why this should be. Allen
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AdminJar Inactive Member |
Thread moved here from the Proposed New Topics forum.
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PaulK Member Posts: 17826 Joined: Member Rating: 2.3 |
It's really simple. In the actual paper the second column is labelled Sr. So it is the total quantity of Strontium - and not just the Sr87 isotope.
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Matt P Member (Idle past 4800 days) Posts: 106 From: Tampa FL Joined: |
I was waiting for this one to come off PNT...
Isotopes are measured as ratios as this is the easiest thing to do with our current instruments. The Rb and Sr columns are not isotope specific, nor does the author claim them to be (take a look at the column headings of table 2 again). They're just the trace abundances of these two elements. This message has been edited by Matt P, 07-22-2005 04:10 PM
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allenroyboy Inactive Member |
Indeed, I did mislabel the first two columns through my misunderstanding. I want to thank those who pointed that out.
Through further study of the books I mentioned above I find that the 87Sr/86Sr ratio is measured directly by mass spectrometer. However, the 87Rb/86Sr ratio is computed from the Rb/Sr ratio using this formula 87Rb_____Rb_____Ab(87Rb) x W(Sr)------__=__---__X__-------------------- 86Sr _____Sr_____Ab(86Sr) x W(Rb) (Ab = abundance W = atomic weight)This is as of 1997 when the latest book was written. I don't know if they have found a way to directly measure the 87Rb/86Sr ratio yet. I have some concerns about this formua because the atomic weight of an element is the sum of the abundances of the isotopes times their mass. Yet in the formula they use the abundance of an isotope and atomic weight of that element while ignoring that the atomic weight is the sum of isotope weights. Because the atomic weight is the sum of isotope weights, then the abundance is not the correct ratio by which to compute the mass of a specific isotope from the atomic weight. Lets take an example using clorine which has two isotopes:________Mass___X___Abundance______isotope amu 35Cl____34.968_______0.7577____=____26.495 37Cl____36.965_______0.2423____=_____8.957 ___________________________________---------- ___________________________________35.452 total amu But, what they are doing in the forumla above is this 0.7577 / 35.452 = 0.0214 which is the inverse of a computed 35Cl mass of 46.789 -- which is way to large (should be 34.968). The correct ratio between the mass of 35Cl and the total amu is 34.968 / 35.453 or 0.9863 not 0.7577. So my question is, why do they use the abundance of an isotope and the total amu of an element rather than the amu of just the isotope?? What am I missing here?
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allenroyboy Inactive Member |
Come on guy's! Where are the experts here?
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roxrkool Member (Idle past 1014 days) Posts: 1497 From: Nevada Joined: |
Why don't you contact the author?
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PurpleYouko Member Posts: 714 From: Columbia Missouri Joined: |
I'm not an expert at Isochron dating but I do have a considerable amount of experience with Mass Spectrometers, including multi-collector systems designed for measuring ratios.
You cannot directly measure the ratio of 87Rb/86Sr. The problem is that There is always a certain amount of 87Sr present (abundence = about 7% of all Strontium). Mass specs don't have the necessary resolution to tell the difference between 87SR and 87Rb. The count rate for 87Rb has to first be corrected by subtracting the 87Sr contribution, This is normally done by measuring the Sr88 isotope, which doesn't decay and is known to make up82.6% of the total Strontium. The exact correction is a little more complex because of mass bias (mentioned below) but that is the gist of it. I am pretty sure that the equation you have there is actually one designed to quantify the exact correction factor for the mass bias that results in greater sensitivity for higher masses due to a number of factors in the physics of a mass spectrometer. Having said that It doesn't look quite right still and doesn't return anything like the right numbers. For one thing Rb/Sr actually represents the measured count rate of the two isotopes so it is likely to differ slightly from the predicted ratio due to instrumental error. I can give a better example of this with Lead ratios since there are no initial corrections to be applied before plugging the numbers into the formula. (The examples are based on ICP-MS rather than TIMS)
quote: In the case of Sr you could spike the sample with another element of similar mass to use as an internal standard (as the examples uses Tl. Zr would work well as isotopes 90 and 91 are clear of isobaric interferences (other isotopes at the same mass)You would first find the applicable mass bias using the Internal Standard. This should be a constant in correction/AMU so can be applied to 87Sr based on measured 88Sr. After subtracting 87Sr from the total 87 counts you would be left with a direct measurement of 87Rb/86Sr.Apply the mass bias correction as outlined above and you will have your corrected ratio. Maybe this helps or maybe it confuses the issue even worse. Sorry if it did.Without direct access to your book I can't help much more. It's been a long time since I did this on a regular basis.
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allenroyboy Inactive Member |
I've read your note and think I understand. I will study it more as I continue to study this out...
Nice to find someone who actually understands what I'm talking about. Allen
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paleolutheran Inactive Member |
Usually when the isotopic ratios are inverted they are usually applying the natural logarithm of the ratio in order to make the curve linear. Frankly I'm surprised they think they had luck on granite that had foliations from what seems like minor metamorphism as that usually resets the "clock," even in zircons.
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