EAS4561/6560 Isotope Geochemistry

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1 1. Use Dodson s equation (equation 2.37) to calculate the closure temperatures of biotite for the cases of a slowly cooled intrusion discussed in Section 2.3.1, namely at 10 /Ma and 100 /Ma. Use the data given in Figure 5.1, which corresponds to EA = kj/mol and D0 = m 2 /sec. Assume a = 140 µm and A = 27. The value of R is J/K-mol. If we were to do K-Ar dating on these biotites long after they cooled (say 100 Ma later), how much different would the two ages be assuming the intrusion cooled at these rates from an initial temperate of 600 C? (Hint, you can easily do this in Excel, either using the solver, or iterating manually following an initial guess of the closure temperature you can base that on the discussion in the text. Be careful to use consistent units). T c = E A æ Rln - ART c ç è a 2 E A t 2 D 0 ö ø To answer this, we first need to solve the above equation to obtain the closure temperatures. Problem is that we can t, at least not directly. The problem is very suitable for iterative indirect solution, for example using the solver. Doing so, we get closure temperatures of K (279 C) for 100 K/Ma and K (252 C) for 10 K/Ma. In the former case, it would take 3.21 Ma to cool to closure, in the second case it would take Ma to cool to closure, so our dates would differ by Ma. 2. You measure the following K2O and 40 Ar on minerals from a small pluton. Calculate the age for each. What do you think the ages mean? Use the following: branching ratio is , e = yr -1, total= x yr -1. (These are newly recommended values). 40 K/K = , atomic weight of K is K2O (wt. %) Radiogenic 40 Ar, mole/g biotite x hornblende x Are the ages the same? If not, speculate on why not? Our first step is to convert K 2 O in wt percent to moles/g of 40 K. To do so, we divide K 2 O by the molecular wt of K 2 O, divide by 100 (convert from percent), then multiply by 2 (there are 2 moles of K for every mole of K2O). This gives us moles/g of K. We then multiply by to get moles of 40 K. The age may then be calculated by solving equation 2.33 for t: t = ln( 40 Ar*/ 40 K x +1)/ These calculations are shown in the spreadsheet below: 1

2 K2O% moles K moles 40K 40Ar* t, Ma biotite E E hornblende E E mole wt K mole wt K2O K/K Lam total 5.55E-10 lam e.c. 5.88E-11 The older age of the hornblende likely reflects its higher closure temperature. 3. Use following data to answer this question: Rb yr -1 ; 86 Sr/ 88 Sr: ; 84 Sr/ 88 Sr: , 85 Rb/ Rb = , atomic weight of Rb: atomic masses of Sr: 88 Sr:.9056 Sr: Sr: Sr: Calculate the abundances of the isotopes and atomic weight of Sr given that Sr/ 86 Sr = The abundance of an isotope, for example, 84 Sr, is calculated as: ( 84 Sr/ 88 Sr)/( 84 Sr/ 88 Sr + 86 Sr/ 88 Sr + Sr/ 88 Sr + 88 Sr/ 88 Sr) where Sr/ 88 Sr can be calculated as 86 Sr/ 88 Sr x Sr/ 86 Sr and 88 Sr/ 88 Sr is, of course, 1. In this way, we calculate the following abundances: 88 Sr: 82.63% Sr: 6.95% 86 Sr: 9.% 84 Sr: 0.558% The atomic weight may be calculated by multiplying the mass of each isotope by its fractional abundance and summing the result. Doing so, we obtain a weight of The following data was obtained on 3 minerals from a pegmatite. Calculate the age of the rock using the isochron method (you may use conventional regression for this problem). The data and approach used in question 3 will prove useful. Rb, ppm Sr, ppm Sr/ 86 Sr Muscovite Biotite K-feldspar

3 Note: very important! In order to calculate the Rb/ 86 Sr ratio, we need to (1) calculate the atomic weight of strontium for each sample and (2) calculate the fractional abundance of 86 Sr for each sample. Because of the large variation in Sr/ 86 Sr, we cannot just use a standard value. Doing that was the point of the previous question. Rb, ppm Sr, ppm Sr/ 86 Sr Moles Rb moles Rb moles Sr moles 86Sr Rb/ 86 Sr Sr/ 86 Sr Muscovite Biotite K-feldspar fraction Rb of Rb slope lambda Rb (Ma^-1) 1.42E-05 intercept Age=ln(slope+1)/la mbda Ma Muscovit Biotite Kspar e 84/ / masses /88 (=/86* 86/88) / SUM abundances 84 Sr: Sr: Sr: Sr atomic wt

4 Sr/86Sr EAS4561/6560 Isotope Geochemistry Rb/86Sr The above was done using simple linear regression in Excel. Using Isoplot gives an age of 115±48 Ma. 5. The following 40 Ar * / 39 Ar ratios were measured in step heating of lunar Basalt from Hadley Rile. The flux monitor had an age of yrs and its 40 Ar*/ 39 Ar ratio after irradiation was The 40 K/ 39 K ratio is Calculate the age for each step and plot the ages versus percent of release. From this release spectrum, estimate the age of the sample. Cumulative % Ar released 40 Ar*/ 39 Ar We use equation 2.42 first to solve for C, then to calculate the ages: We see that the last 3 release steps, which cumulatively represent 73% of the Ar, form an approximate plateau. Averaging these three steps and weighting each for the amount of Ar released, we calculate an age of 2.10 Ga: 4

5 6. The following data were measured on phlogopites (P) and phlogopite leaches (LP) from a kimberlite from Rankin Inlet area of the Hudson Bay, Northwest Territories, Canada. What is the (1) the age of the rock (2) the uncertainty on the age, (3) the initial Sr/ 86 Sr ratio, and (4) the uncertainty on the initial ratio? The relative uncertainty on the Sr/ 86 Sr is 0.005% and that of the Rb/ 86 Sr is 1%. (Hint: this is best accomplished using the Isoplot.xla Excel add-in written by Ken Ludwig and available at Sample Rb/ 86 Sr Sr/ 86 Sr P P P P P P P

6 Sr/86Sr EAS4561/6560 Isotope Geochemistry Rb/86Sr 6