AMHERST COLLEGE Department of Geology GEOLOGY 41 - Environmental and Solid Earth Geophysics Lab 2: Geochronology Solution

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1 AMHERST COLLEGE Department of Geology GEOLOGY 41 - Environmental and Solid Earth Geophysics Lab 2: Geochronology Solution U/Pb data aliquot # U (ppm) Pb (ppm) 206 Pb/ 204 Pb 207 Pb/ 204 Pb 208 Pb/ 204 Pb z z z z ) The concentrations of U and Pb are given in mass units (ppm) and need to be converted to the number of atoms of U and Pb in each aliquot. To so this, divide the concentration of U and Pb of each aliquot by the atomic weights of U and Pb. (This method gives you the number of atoms in an arbitrary mass of zircon, but it works because the age equations deal with atom ratios). For this first aliquot this is: U = 228.4/ = Pb = 23.21/ = for all of our aliquots we get: U (atoms) Pb (atoms) z z z z ) For each aliquot determine the number of atoms of each of the U ( 235 U, 238 U) and P ( 208 Pb, 207 Pb, 206 Pb, 204 Pb) isotopes. For U, this is relatively easy as the ratio of U isotopes in all U is constant. For Pb, this is a bit more complicated because 3 of the Pb isotopes are radiogenic and the abundance of each isotope must be calculated from the analytical data each aliquot. (Do you know how to do this?-if not see Peter) U = 238 U U U/ 235 U = 238 U/ 235 U + 1 = 1 + 1/ = U/U = 1/ = U/U = 235 U/U x 238 U/ 235 U = / = U = U (atoms) x 235 U/U = U (atoms) x U = U (atoms) x 238 U/U = U (atoms) x

2 We can get the Pb ratios the same way, but we need to do this for each aliquot. Pb = 204 Pb Pb Pb Pb Pb/ 204 Pb = Pb/ 204 Pb Pb/ 204 Pb Pb/ 204 Pb 204 Pb/Pb =1/ ( Pb/ 204 Pb Pb/ 204 Pb Pb/ 204 Pb) and like U 204 Pb = U (atoms) x 204 Pb/Pb this gives us 204 Pb, we can get the others by: 206 Pb/Pb = 204 Pb/Pb x 206 Pb/ 204 Pb 207 Pb/Pb = 204 Pb/Pb x 207 Pb/ 204 Pb 208 Pb/Pb = 204 Pb/Pb x 208 Pb/ 204 Pb 204 Pb for all of our aliquots this is: 206 Pb 207 Pb 208 Pb z z z z ) Some of the Pb in your zircon was present when it crystallized. You know that this is the case because there is some non-radiogenic 204 Pb in your samples. Use the common Pb isotopic ratios given below to determine the amount of radiogenic 207 Pb & 206 Pb in each aliquot (i.e., determine the amount of 207 Pb* & 206 Pb* in each aliquot). 235 U..OK this is getting fussy, but is really just algebra. 206 Pb* = 206 Pb Pb common = 206 Pb Pb x ( 206 Pb/ 204 Pb) common 206 Pb* = 206 Pb Pb x U for all of our aliquots this is: 206 Pb* 207 Pb* z z z z ) For each aliquot determine the 207Pb*/235U & 206Pb*/238U ratios and using the decay equations for 235U & 238U determine a 207Pb*/235U and 206Pb*/238U age. Are these ages concordant? Are the ages for each aliquot the same? This is just plug and chug using the U-Pb age equations. t 238 = 1 ln ( 206 Pb " + 1 )! U or for aliquot z1:

3 t 238 = 1 ln ( !! x )! x 10 6 years for all of our aliquots this is: 206 Pb* /238 U 207 Pb* /235 U 206 Pb /238 U 207 Pb /235 U age (Ma) age (Ma) z z z z The ages are not concordant and vary by 100 Ma from aliquot to aliquot

4 5) Plot the 207 Pb*/ 235 U & 206 Pb*/ 238 U ratios for each aliquot on a concordia diagram. Can you now make sense out of your zircon ages? P b / U P b / U Aliquots fall along a discordia array with an ~600 Ma upper intercept indicating a 600 Ma age for the igneous? Protolith of the gneiss. 6) What observations can you make about the tectonic evolution of the Pelham Hills based on your U/Pb zircon and 40 Ar/ ages? T closure (zircon) > 600 C 600 Ma T closure (hornblende) = 550 C 360 Ma T closure (muscovite) = 400 C 335 Ma Plutonism at ca. 600 Ma, followed by cooling to low? temperature. Metamorphism (and tectonism) at or before ca. 360 Ma followed by slow cooling (6 deg.ma) through 335 Ma (and younger times?). Isostatically driven unroofing? 40 Ar/ data T( C) of heating % / 40 Ar released 36 Ar/ 40 Ar

5 step this step 56.7 Ma biotite standard fusion hornblende muscovite fusion fusion ) In order to determine the 40 Ar/ ages of your samples, you must know the ratio 40 Ar*/, the ratio of radiogenic 40 Ar to. To do this, you must correct your measured ratios of 40 Ar/ for any Ar the was introduced during sample preparation. You know that some Ar must have been introduced into each aliquot that was analyzed because there is some 36 Ar in each aliquot. Assume that the source of 36 Ar is modern atmosphere ( 40 Ar/ 36 Ar = 295.5) and correct each of the measured values of 40 Ar/ for 40 Ar that was introduced during sample preparation and determine 40 Ar*/ for each aliquot. To correct for atmospheric 40 Ar we need to subtract it from the total 40 Ar. (a) 40 Ar! = 40 Ar " 40 Ar atm = 40 Ar " x 36 Ar We only have Ar isotope ratios, but that is OK, because we need 40 Ar*/, so lets divide (a) by to get: 40 Ar! = 40 Ar " x 36 Ar = 40 Ar " x 40 Ar 36 Ar 40 Ar

6 ...for this biotite standard this is: 40 Ar! = " x x = For all of our aliquots we get: standard fusion hornblende muscovite fusion total gas fusion total gas ) Determine the neutron flux parameter J for your analyses using the Ar isotopic ratios collected from the 56.7 Ma biotite standard. This comes directly from the equation for J, the age of the standard, and the 40Ar*/39Ar ratio calculated above. 40 Ar! = 1 J ( e " t # 1 ) J = 1 / 40 Ar! or using the standard data we get J = ( e x x 10! 4! 1 ) = ( e " t # 1 ) 3) Determine an 40 Ar/ age for each of the heating steps for the two samples (muscovite and hornblende). Also determine a total gas 40 Ar/ age for both samples. We can solve our equation for t and calculate ages for each aliqot. 40 Ar! = 1 J ( e " t 1 # 1 ) t =! ln ( J x 40 Ar " + 1 ) or for the 600 deg hornblende aliquot this is:

7 t = x 10! 1 0 ln ( x ) = x10 6 years

8 For all of our aliquots we get: standard fusion hornblende fusion total gas muscovite fusion total gas ) Using the ages determined above, plot an Ar age spectrum (apparent age vs. cumulative percent of Ar) for each of the 2 samples.

9 Age (Ma) hbl musc cumulative Ar

10 5) Assuming a ±3 Ma error in the age of each release step, do your ages represent a plateau? YES