Astronomy Exercise

Rubidium-Strontium Ages of Cosmic Materials.

Introduction

In this exercise, we will use the spreadsheet Excel to help with the calculation of radioactive age determinations for several cosmic materials of astronomical interest. Planetary and earth scientists use a number of methods to date rocks and soil samples. Perhaps the most straightforward of these uses the decay of $\rm ^{87}Rb$ to $\rm ^{87}Sr$. The half-life of this decay is known to a few percent, so our determinations will be uncertain to at least that much. We shall consider the uncertainty in the age a little later in this write up, but for the present, we state that age uncertainties less than 10 percent are entirely feasable.

The theory for this determination has been outlined in the web text. If a given amount of $\rm ^{87}Rb$ is present in a rock at time zero then after a time t the amount will be decreased by an exponential factor. We now use the common notation in isotope studies of letting the symbol for the nuclide also represent the relative of the species present. Ultimately, we will be taking ratios of numbers of different species, for example, $\rm ^{87}Rb$ to $\rm ^{86}Sr$. If this is the case, it doesn't matter whether we mean by $\rm ^{87}Rb/ ^{86}Sr$, the ratio of the species per cm3, or the ratio of the total numbers in a given crystal.

Theory: This little section uses calculus. Skim it for the notation if you haven't had calculus.

Now the amount of $\rm ^{87}Rb$ that is present in a given crystal is given by the formula for radioactive decay, which states that the change of $\rm ^{87}Rb$ in a time increment dt is directly proportional to the number of $\rm ^{87}Rb$ present, say at time t = 0. The constant giving this proportionality is called the decay constant, and it is usually assigned the Greek letter $\lambda$. Then

\begin{displaymath} d (^{87}{\rm Rb}) = -\lambda (^{87}{\rm Rb})dt\end{displaymath} (1)

Assume the radioactive decay began at time t=0, with an amount $\rm (^{87}Rb)_0$ of the radioactive isotope present. Then at time t, the amount remaining is given by the solution to Equation(1):

\begin{displaymath} ^{87}{\rm Rb} = (^{87}{\rm Rb})_0 \exp(-\lambda t).\end{displaymath} (2)

Equation (2) is the equation of radioactive decay. For $\rm ^{87}Rb$, the decay constant is $1.42 \times 10^{-11}$years-1. Note the units, which mean ``per year''. One can readily see that the half-life is proportional to the decay constant $\lambda$.If $\rm ^{87}Rb/(^{87}Rb)_0) = 0.5$, then exactly one half-life has passed since t = 0. From Equation(2), we find for the half-life, $\tau_{1/2}$

\begin{displaymath} \ln(0.5) = -\lambda \tau_{1/2}.\end{displaymath} (3)

From Equation(3), we find

\begin{displaymath} \tau_{1/2} = 0.6931/\lambda\end{displaymath} (4)

For $\rm ^{87}Rb$, then, the half-life for radioactive decay is $4.881 \times 10^{10}$ years.

The Sample Age

Let us assume for simplicity that a rock freezes at time t = 0. At this point, all ions are locked up in the individual crystals by amounts that depend on their charge and ion size. Strontium ions have the same charge, and similar ionic radii to Calcium, so for a rock containing diopside ($\rm MgCaSi_2O_6$), or anorthite ($\rm CaAl_2Si_2O_8$) one would expect a relatively large amount of both strontium isotopes in these two minerals. Rubidium has a relatively large ion size. Of the simple minerals for which we have learned formulae, it would substitute readily only in K-feldspar. However, there will be a small amount of substitution of all ions in all minerals, and with modern laboratory technologies, these amounts can be measured. Complex minerals may be able to accomodate large ``foreign'' ions more readily than simple ones.

The amount of $\rm ^{87}Sr$ present in any mineral at time t will be the sum of what was there originally, and the amount of $\rm ^{87}Rb$that radioactively decayed to it. We can write

\begin{displaymath} (^{87}{\rm Sr})_t = (^{87}{\rm Sr})_0 + (^{87}{\rm Rb})_0 - (^{87}{\rm Rb})_t.\end{displaymath} (5)

We have no way of knowing the amount of $\rm ^{87}Rb$ that was present when the rock froze, but we can measure the amount present now. That amount is given by Equation(2). With its help, we can write

\begin{displaymath} (^{87}{\rm Rb})_0 - (^{87}{\rm Rb})_t = (^{87}{\rm Rb})_t (\exp(\lambda t) -1).\end{displaymath} (6)

If we put (5) into (4), and divide each term of by $\rm ^{86}Sr$, we obtain the equation for dating rocks that is plotted in the web text, and discussed in various links:

\begin{displaymath} (^{87}{\rm Sr}/^{86}{\rm Sr})_t = (^{87}{\rm Sr}/^{86}{\rm Sr})_0 + (^{87}{\rm Rb}/^{86}{\rm Sr})_t[\exp(\lambda t) -1].\end{displaymath} (7)

It might be helpful to think of this as a simple linear equation: y = a + bx, as in the plots on the web text. The unknowns, are the intercept, $a = (^{87}{\rm Sr}/^{86}{\rm Sr})_0$, and the slope, $b = \exp(\lambda t) -1$. The other quantities, $y = (^{87}{\rm Sr}/^{86}{\rm Sr})_t$, and $x = (^{87}{\rm Rb}/^{86}{\rm Sr})_t$ are measured in the laboratory.

Data

We shall determine rubidium-strontium ages for three samples of cosmic materials. The first will be for the lunar rock 15016,46, which is the 46th split from rock 16 of the Apollo 15 mission. We give data for 4 minerals, plagioclase, and ilmenite, which we discussed in class, and two spinels (mostly $\rm MgAl_2O_3$) which we designate s1 and s2. I have tried to avoid additional mineral names, but the data in the table are for real rocks and the names are unavoidable. You need not memorize their chemical formula.


 
Table: Rubidium-Strontium Data
Sample 87Rb/86Sr 87Sr/86Sr
     
3cLunar Rock 150116,46    
Plagioclase 0.0055 0.699 36
s1 0.0162 0.699 94
s2 0.0582 0.701 86
Ilmenite 0.135 0.705 44
3cMartian Meteorite ALHA 77005    
PG 1.10 0.7128
OL1 0.60 0.7118
OL2 0.50 0.7116
AUG 0.35 0.7112
3cBaltimore Gneiss    
Mica 116.4 1.2146
K-Feldspar 3.794 0.7633
Plagioclase 0.2965 0.7461

The second sample whose age we will find is that of the Martian meteorite ALHA 77005. The minerals are Plagioclase (PG), two Olivine fragments (OL1 and OL2), and an augite fragment (AUG). The third sample is from a terrestrial gneiss, a metamorphic rock found near Baltimore, Maryland.

Step-by-Step Procedure

We will describe the method to analyze the first sample. Use the same technique for the other two.

On the PC's go to the start menu, highlight ``programs'' and then Microsoft's Excel. When the Excel spreadsheet appears, type in the data for the first sample. Unless you tell it otherwise, Excel will expect data for the x-coordinate in the first column and for the y-coordinate in the second. Highlight both columns. You should have a block two columns wide, and four rows deep. Go to the main menu, and click ``Chart'' (colored vertical bars, just to the right of the Z-to-A sort. Highlight ``x-y scatter'', and click ``next'' until you get Step 4 of 4 of the Chart Wizard. I recommend you put the graph on a new sheet, so check that button. The default placement is to make a small graph on the same sheet as your data, which can be OK, but can also be messy.

After checking the button for a separate sheet, click on finish, and your data will be displayed on a large graph. You may label the axes in previous steps, but it is not necessary.

Go to the main menu, and pull down the Chart options. Highlight ``Add trend line''. You should get a dialog box with various choices, and two ``tabs'' at the top. The ``linear'' graph, which is what you want, should be highlighted by default. Now click the ``Options'' tab. Click the button ``Display equation on chart'', and then ``OK''. You will get the graph back, with a line throught the data, and its equation. The line is fit by a technique known as the method of least squares, which is widely used in scientific work. It is discussed in advanced Astronomy courses, as well as numerous other places. Ask your GSI about it.

The coefficient of x, is [$\exp(+\lambda t) -1$]. If you put in $\lambda = 1.42 \times 10^{-11}$ years-1, the t that satisfies the relation is the age of the rock. Thus, with b equal to the slope $t = \ln(b+1.00)/\lambda$The y intercept gives $(^{87}{\rm Sr}/^{86}{\rm Sr})_0$, the initial ratio at the time of the freezing. If you highlight the displayed equation, and then right click, you will get a dialog box that will give you options for significant figures, and formats (e.g. scientific) for the constants.

Interpretation of the intercept is less certain, but there are general trends that are well established on the earth, where rocks are frozen and remelted many times. After each re-freezing, we can expect the $(^{87}{\rm Sr}/^{86}{\rm Sr})_0$ to be higher. Thus a high value of this initial ratio could indicate a complicated history of melting and remelting. Explain how this might account for the higher age of the terrestrial rock, the Baltimore Gneiss than ALHA 77005, while the $(^{87}{\rm Sr}/^{86}{\rm Sr})_0$ is actually higher for the older rock. Present your answer to this question at the end of the next section.

Results

1.
List the slopes, intercepts, and ages that you derive for the three samples.
2.
Mica is a mineral that is ``late'' in the sense of the Bowen series. Suggest some reasons that might account for the large amount of $\rm ^{87}Rb$ in the mica of the Baltimore Gneiss-even more than in the K-feldspar.
3.
Suppose decay constant for $\rm {87}Rb$ were in error by 5%. To what values would your calculated ages change if $\lambda$ were 5% bigger or 5% smaller then the value given here. A 5% error is a very liberal estimate of the uncertainty on $\lambda$.
4.
How could you account for a larger value of $(^{87}{\rm Sr}/^{86}{\rm Sr})_0$ in the (older) Baltimore Gneiss than in the ALHA 7705?

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