Question

4) An igneous Martian meteorite contains four different minerals with the following isotopic compositions: Mineral Covidite Trumpite Insleeite Faucite 87Rb/86Sr 10 7.00 5.00 2.00 87St/86Sr 1.12 0.101 0.89 0.79 86Sr and 878r are stable isotopes but 87Rb is radioactive and decays to 875r with a decay constant (time to decay by e-?) of 7 x 1010 years. When the minerals formed by crystallization from a liquid they had identical 875r/86Sr ratios. Plot the data on a graph of 87Sr/86Sr (y axis) versus 87Rb/86Sr (x axis) and compute the age of the rock. What was the initial Sr isotopic composition when the rock solidified? If all four minerals have the same abundance in the rock and the present rock was melted again how much higher [a ratio] would its 87Sr/86Sr be compared to its initial composition ?

Answer 1

ANSWER:

Plot of 87Sr/86Sr (y-axis) vs 87Rb/86Sr (x-axis):

1.2 1 y = 0.0424x +0.6984 R = 0.9889 0.8 875r/ssr 0.6 0.4 0.2 . 0 0 2 4 6 00 10 12 87Rb/65r

• Note: This plot must give a straight line. So, there is an error in the 87Sr/86Sr value of trumpite. The correct value must be 1.01, so I'm doing calculations withs value instead of 0.101.

This plot follows the isochron equation:

$\left ( \frac{^{87}Sr}{^{86}Sr} \right )_{t}=\left ( \frac{^{87}Sr}{^{86}Sr} \right )_{0}+\left ( \frac{^{87}Rb}{^{86}Sr} \right )_{t}\left ( e^{\lambda t}-1 \right )$

where

(87Sr/86Sr)t and (87Rb/86Sr)t = present ratios of 87Sr/86Sr and 87Rb/86Sr respectively

(87Sr/86Sr)0 = y-intercept = inital ratio of 87Sr/86Sr

eλt - 1 = slope

1/λ = inverse of constant decay of 87Rb = 7x1010 years

t = age of rock

Then for the plot we have the following equation:

$y=0.0424x+0.6984$

or

$\left ( \frac{^{87}Sr}{^{86}Sr} \right )_{t}=0.6984+0.0424\left ( \frac{^{87}Rb}{^{86}Sr} \right )_{t}$

This means that

• Age of rock

$slope= e^{\lambda t}-1 =0.0424$

$e^{\lambda t} =0.0424+1=1.0424$

$\lambda t =ln\left (1.0424 \right )$

$t =\frac{1}{\lambda}\times ln\left (1.0424 \right )=7x10^{10}\, year \times ln\left (1.0424 \right )$

$\mathbf{t =2.91x10^{9}\, years}$

• Initial Sr isotopic composition

​​​​​​​$y-intercept=\mathbf{\left ( \frac{^{87}Sr}{^{86}Sr} \right )_{0}=0.6984}$

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If the present rocks is melted again, all mineral will be mixed and we get a new 87Sr/86Sr value. As all minerals have the same abundance in the rock, the new 87Sr/86Sr value could be calculated as an average of 87Sr/86Sr values of all minerals:

$\left ( \frac{^{87}Sr}{^{86}Sr} \right )_{0'}=\frac{\left ( \frac{^{87}Sr}{^{86}Sr} \right )_{covidite}+\left ( \frac{^{87}Sr}{^{86}Sr} \right )_{trumpite}+\left ( \frac{^{87}Sr}{^{86}Sr} \right )_{insleeite}+\left ( \frac{^{87}Sr}{^{86}Sr} \right )_{faucite}}{4}$

$\left ( \frac{^{87}Sr}{^{86}Sr} \right )_{0'}=\frac{1.12+1.01+0.89+0.79}{4}=\mathbf{0.9525}$

Then, compared with the initial composition, the 87Sr/86Sr in melted rock is higher than the initial value (0.6984) by

$\frac{\left ( \frac{^{87}Sr}{^{86}Sr} \right )_{0'}}{\left ( \frac{^{87}Sr}{^{86}Sr} \right )_{0}}=\frac{0.9525}{0.6984}=1.36$

The melted present rock will hava a 87Sr/86Sr that is 1.36 times higher than its initial composition