EAS4561/6560 Isotope Geochemistry

EAS4561/6560 Isotope Geochemistry
Problem Set 2
Due Feb. 16, 2017
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 = 196.8 kJ/mol and D0 = -0.00077m2/sec. Assume a = 140 µm and
A = 27. The value of R is 8.314 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).
Tc =
EA
æ ARTc2 D0 ö
R ln ç - 2
÷
è a E At 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 551.7 K (279˚C) for 100 K/Ma and 524.8 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 34.82 Ma to cool to closure, so our dates
would differ by 31.61 Ma.
2. You measure the following K2O and 40Ar on minerals from a small pluton. Calculate the age for
each. What do you think the ages mean? Use the following:
branching ratio is 0.1157, e = 0.58755  10-10 yr-1, total= 5.5492 x 10-10 yr-1. (These are newly
recommended values).
40K/K = 0.0001167, atomic weight of K is 39.03983.
K2O (wt. %)
Radiogenic 40Ar, mole/g
biotite
8.45
6.016 x 10-10
hornblende
0.6078
0.4642 x 10-10
Are the ages the same? If not, speculate on why not?
Our first step is to convert K2O in wt percent to moles/g of 40K. To do so, we
divide K2O by the molecular wt of K2O, 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 0.01167 to get moles of 40K. The age may then
be calculated by solving equation 2.33 for t:
t = ln(40Ar*/40K x  +1)/
These calculations are shown in the spreadsheet below:
1
EAS4561/6560 Isotope Geochemistry
Problem Set 2
Due Feb. 16, 2017
K2O%
biotite
hornblende
8.45
0.6087
mole wt K
mole wt
K2O
40K/K
Lam total
lam e.c.
39.398
moles K
0.00178
0.00013
moles 40K
2.0805E-07
1.4987E-08
40Ar*
6.02E-10
4.64E-11
t, Ma
48.6
52.0
94.796
0.0001167
5.55E-10
5.88E-11
The older age of the hornblende likely reflects its higher closure temperature.
3. Use following data to answer this question:
Rb 1.42  10-11 yr-1; 86Sr/88Sr: 0.11940; 84Sr/88Sr: 0.006756, 85Rb/87Rb = 2.59265, atomic weight of Rb:
85.46776
atomic masses of Sr:
88Sr: 87.9056
87Sr: 86.9088
86Sr: 85.9092
84Sr: 83.9134
Calculate the abundances of the isotopes and atomic weight of Sr given that 87Sr/86Sr = 0.7045.
The abundance of an isotope, for example, 84Sr, is calculated as:
(84Sr/88Sr)/(84Sr/88Sr + 86Sr/88Sr + 87Sr/88Sr + 88Sr/88Sr) where 87Sr/88Sr can
be calculated as 86Sr/88Sr x 87Sr/86Sr and 88Sr/88Sr is, of course, 1. In this way,
we calculate the following abundances:
88
Sr: 82.63%
87
Sr: 6.95%
86
Sr: 9.87%
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
87.617
4. 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.
87Sr/86Sr
Rb,
Sr,
ppm
ppm
Muscovite
Biotite
K-feldspar
238.4
1080.9
121.9
1.80
12.8
75.5
1.4125
1.1400
0.7502
2
EAS4561/6560 Isotope Geochemistry
Problem Set 2
Due Feb. 16, 2017
Note: very important! In order to calculate the 87Rb/86Sr ratio, we need to (1)
calculate the atomic weight of strontium for each sample and (2) calculate the
fractional abundance of 86Sr for each sample. Because of the large variation in
87
Sr/86Sr, we cannot just use a standard value. Doing that was the point of the
previous question.
Rb, ppm
Sr, ppm
87Sr/86Sr
Moles
Rb
moles
87Rb
moles
Sr
moles
86Sr
87Rb/86Sr
87Sr/86Sr
Muscovite
238.4
1.8
1.4125
2.789
0.7764
0.02055
0.0019
409.62
1.4125
Biotite
K-feldspar
1080.9
121.9
12.8
75.5
1.14
0.7502
12.647
1.4263
3.5202
0.3970
0.14614
0.86174
0.0138
0.0846
254.65
4.6908
1.14
0.7502
fraction
87Rb of Rb
0.278346
slope
lambda
87Rb
(Ma^-1)
1.42E-05
intercept
Age=ln(slope+1)/la
mbda
84/88
86/88
87/88
(=87/86*
86/88)
88/88
SUM
0.00163
0.737790
787
114.58
Ma
Muscovit
e
0.006756
Biotite
Kspar
0.006756
0.1194
0.168652
5
0.1194
0.136116
0.00675
6
0.1194
0.08957
388
1
1.2948
1
1
1.2623
1.2157
abundances
0.0054
0.0056
84Sr:
masses
83.9134
0.0052
86Sr:
85.9092
0.0922
0.0946
0.0982
87
86.9088
0.1303
0.1078
0.0737
88Sr
87.9056
0.7723
0.7922
0.8226
87.571
87.588
87.614
Sr:
atomic
wt
3
EAS4561/6560 Isotope Geochemistry
Problem Set 2
Due Feb. 16, 2017
1.5
1.4
87Sr/86Sr
1.3
1.2
1.1
1
0.9
0.8
0.7
0
100
200
300
400
500
87Rb/86Sr
The above was done using simple linear regression in Excel. Using Isoplot gives an
age of 115±48 Ma.
5. The following 40Ar*/39Ar ratios were measured in step heating of lunar Basalt 15555 from Hadley
Rile. The flux monitor had an age of 1.062  109 yrs and its 40Ar*/39Ar ratio after irradiation was 29.33.
The 40K/39K ratio is 0.000125137. Calculate the age for each step and plot the ages versus percent of
release. From this release spectrum, estimate the age of the sample.
40
Cumulative % Ar released
Ar*/39Ar
3
10
27
61
79
100
58.14
61.34
72.77
80.15
83.32
79.80
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
EAS4561/6560 Isotope Geochemistry
Problem Set 2
Due Feb. 16, 2017
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 87Sr/86Sr ratio, and (4) the uncertainty on the initial
ratio? The relative uncertainty on the 87Sr/86Sr is 0.005% and that of the 87Rb/86Sr is 1%. (Hint: this is best
accomplished using the Isoplot.xla Excel add-in written by Ken Ludwig and available at
http://bgc.org/isoplot_etc/isoplot.html).
87Rb/86Sr
87Sr/86Sr
Sample
P1
46.77
0.848455
P2
40.41
0.828490
P3
34.73
0.810753
P4
33.78
0.807993
P5
0.1829
0.706272
P6
0.1373
0.705616
P7
1.742
0.710498
5
EAS4561/6560 Isotope Geochemistry
Problem Set 2
Due Feb. 16, 2017
87Sr/86Sr
0.85
0.80
0.75
0.70
0
20
40
87Rb/86Sr
6
60

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