Homework Set #4 Astronomy/EPS 12 de Pater/Jeanloz Spring 2005

Due: Noon, Friday, Feb 25 in Cambell Hall Some reminders and notifications  Star parties or being held every Thursday with clear skies at Cambell Hall, 8:00pm.  Be sure to always show your work and put units on any numbers. They are meaningless without them!!  Be sure to include your section number or time on your homework.  Sign up on blackboard if you have not already.  You should already have started your moon project! Make sure to get your T.A. to check your observations.

Dating Rocks

The objective of this exercise is to familiarize you with the methods used in dating rocks by means of radioactive isotopes. We consider one example, based on the decay of 87Rb to 87Sr, to determine the age of a rock using actual chemical analyses made a few years ago.

Rubidium (Rb) and Strontium (Sr) are present in essentially all materials, although usually in minute quantities, less than a tenth or a hundredth of one percent. Therefore, it is not easy to measure the amounts of Rb and Sr present in rocks. Chemists use a technique called mass spectrometry, which makes use of the different masses of different isotopes to determine the amounts of the different isotopes of Rb and Sr that are present in materials.

The isotope 87Rb of Rubidium is radioactive and decays to the stable (non-radioactive) isotope of Strontium, 87Sr. The half-life for this decay process is 48.8 Gyr (Giga-years, or billions of years), meaning that half the amount of 87Rb originally present in a material decays to 87Sr in 48.8 x 109 years. This is about ten times longer than the age of our Solar System, which is believed to have an age of 4.6 Ga. Of course, we can’t measure the time it takes to lose half the original 87Rb because the age of the Solar System is less than 5 Gyr (and the Universe is estimated to be less than 20 Gyr old). However, out of one gram of 87Rb which would contain more than 1021 atoms, about 10 billion atoms would decay in one year, and this rate can be measured precisely enough to accurately determine the half-life of the radioactive Rubidium.

The rate at which 87Sr is produced due to the decay of 87Rb is given by the amount of 87Rb present times the quantity et – 1, where t is time and  is called the decay constant, which is the rate at which 87Rb decays. ( Note that the half-life can be calculated from : half-life = ln(2)/ = 0.693/, as explained in words below. Also, don’t worry too much if you’ve never come across “e” before. You can use a “scientific” calculator or table 2 below to get values for ex you need). For 87Rb,  = 0.0142 Ga-1, so that we can use the value of 0.0142 for  if we use units of Ga for the time t. This is summarized in the following equation:

87 87 t Srproduced by the decay = Rb (e – 1) [Equation 1]

We have not derived or proven Equation (1) for you, but if you know how to calculate et (or if you use Table 2 below), you can see that the equation works: by inserting the half-life value of 48.8 Ga for t into Equation (1), you will find that the amount of 87Sr produced in one half life is equal (to within 3 or 4 decimal places) to the amount of 87Rb remaining. That is, half the original 87Rb is converted to an equivalent amount of 87Sr during the time period of 48.8 Ga. This shows that the value of  contains the same information as the half-life, as noted above.

1 Now, The total amount of 87Sr present in a rock today is equal to the amount that was initially present when 87 the rock was formed ( Sr initial) plus the amount that has accumulated over time due to the decay of Rb. We can write this as an equation:

87 87 87 Srnow = Srinitial + Srproduced by decay [Equation 2]

Or, using information from Equation (1), we can write:

87 87 87 t Srnow = Srinitial + Rb (e – 1) [Equation 3]

In practice, it is easier to measure the relative amounts of isotopes in a material than to measure the actual amounts, so what is really determined experimentally are the ratios 87Rb/86Sr and 87Sr/86Sr. The reason that 86Sr is used is that this particular isotope is unaffected by radioactive decay, thus its amount does not change with time. To express the information in equation 3 in terms of these measurable ratios, we divide both sides by the amount of 86Sr present (which doesn’t change with time):

87 87 87 Sr now Sr initial Rb ------= ------+ ------(et – 1) [Equation 4] 86Sr 86Sr 86Sr

The only remaining obstacle to obtaining an age from Equation (4) is that we don’t know how much 87Sr was initially present in the rock before 87Rb began to decay. That is, we know the value of  and we can measure the 87 86 87 86 87 87 86 amounts of Sr now / Sr and Rb/ Sr, but we don’t know Srinitial. If only we knew the value of Sr initial / Sr, we could determine the age, t, of the rock (i.e., how long 87Rb has been decaying to 87Sr in the rock). There is a clever way to deal with this problem: by measuring the ratios 87Rb/86Sr and 87Sr /86Sr in different minerals of the same rock. This approach is best illustrated with a specific example that you will work through as part of the homework assignment.

Assignment

1. Table 1 summarizes measurements of Rb and Sr isotope ratios for different minerals within a particular rock. Plot the measurements on the accompanying piece of graph paper. Your graph should have 87Rb/86Sr 87 86 along the horizontal axis and Sr now / Sr along the vertical axis. 2. Use a ruler to draw a straight line that goes as closely as possible through all the points, and extend your line from the left edge to the right edge of the plot. 87 86 3. Use the line you just drew to estimate Sr / Sr for a hypothetical mineral in the rock that has no rubidium. Since radioactive decay never fully eliminates a radioactive element, such a mineral must never have 87 contained any rubidium. If there never was any rubidium in our hypothetical mineral, no new Sr would 87 86 87 86 have been produced, and the present Sr / Sr ratio would be the same as Sr initial / Sr. 3- It turns out that the initial value of 87Sr/86Sr is given by the intercept of the line (the amount of 87Sr/86Sr at the left end of the line, where 87Rb/86Sr = 0). You can convince yourself that this is so: a (hypothetical) mineral containing no 87Rb would plot at the intercept; no extra 87Sr would be produced by radioactive decay, since there is no 87Rb available to decay, and the amount of 87Sr present now would be the same as 87 86 the initial amount. Based on this reasoning, determine the value of Sr initial / Sr from your plot. 4- Now using equation 4 we can, at any given time find the amount of rubidium that would be present in a mineral in the rock containing a given amount of 87Sr. For times t = 0, 0.5, 2, 3, 4, 4.5, 4.7, 4.9, and 5.1, 87 86 87 86 solve equation 4 to find Sr now / Sr for hypothetical minerals that have Rb/ Sr = 1.2 at the given time. 87 86 87 86 Plot the Sr now / Sr values you calculate on your chart. Now, remember that a mineral with Rb/ Sr = 0 87 86 will have the same value of Sr / Sr at all times: the value you found in part 3. So now draw lines 87 86 connecting the points you just plotted with the point representing Sr / Sr in a mineral containing no rubidium and label each of these lines with the corresponding time. You now have predictions for 87 86 87 86 Sr now / Sr at any of these times for any given Rb/ Sr ratio. Notice that the slope of the lines increases as the rock ages.

2 87 86 4. At t = 0, Equation (4) would plot as a straight horizontal line at the value of Sr initial / Sr, the intercept from Question 3. With increasing time, Equation (4) describes a straight line that has increasing slope extending from the same intercept value. For various times (specifically, t = 0.5, 2, 3, 4, 4.5, 4.7, 4.9, 5.1 Ga), solve 87 86 Equation (4) and plot the values of Sr now / Sr that correspond to each of these times for a value of 87Rb/86Sr = 1.2; clearly indicate on the right hand side of your graph which times correspond to which lines. 87 86 In other words, you know Sr initial / Sr (from the previous question), so you can use Equation (4) to calculate 87Sr/86Sr at different times, t, for an arbitrary value of 87Rb/86Sr =1.2 using values of (et – 1) from Table 2 below; plot these 8 points for the 8 times given above and draw 8 lines extending from the intercept to the point given by your calculated value of 87Sr/86Sr and 87Rb/86Sr = 1.2 (and don’t forget to label each line with the corresponding time, t). Note: the initial value you found in Question (3) is not necessarily the origin of the plot. Again, note that we provide you with values of (et – 1) in Table 2. Check us on this with a calculator, if you can, because sometimes we make mistakes- do you get the same results? 5. Estimate the age of the rock by comparing the slope of the measurements you plotted in Question (1) with the different slopes corresponding to different ages, t, that you plotted in Question (4). How precisely do you feel the age is determined from the data you have plotted, assuming that  is known extremely precisely? Your answer should be in the following form: “The age is _____±_____ billion years” and you fill in the blank spaces.

Table 1 87 86 87 86 Rb/ Sr Sr now / Sr 0.3 0.72 0.53 0.74 0.74 0.75 0.9 0.76 1.4 0.80 2.7 0.87

Table 2 t (Ga) (et – 1) 0.5 0.0071 2 0.029 3 0.044 4 0.058 4.5 0.066 4.7 0.069 4.9 0.072 5.1 0.075

3 Isotope Ratios

1

0.9 r S 6 8 / w o n r S 7 8 0.8

0.7 0 0.5 1 1.5 2 2.5 3 87Rb/86Sr

4