0.714 0.712 0.710 0.703 0.706 0.704 0.702 0.700 0.10 0.15 8TRbSr FIGURE 20.19 Re

Question

0.714 0.712 0.710 0.703 0.706 0.704 0.702 0.700 0.10 0.15 8TRbSr FIGURE 20.19 Relative abundance determinations for a sample obtained in the lunar highlands. (Data from D. A. Papanastassiou and G J. Wasserburg, Proc. Seventh Lunar Sei Conf Pergamon Press, New York, 1976.) Example 20.4.1. Data for one sample obtained in the lunar highlands, based on the beta decay 14 of rubidium-87to strontium-87.3gRb ISr + e-+5, are shown in Fig. 20.19 From Eq. (20.1) and Fig. 20.19 where = 0.0146 x 10-9 yr-1 for #Rh Solving for t, we find that the age of the sample is 4.39 x 10 yr It is important to point out that this procedure assumes that the initial ratio Sr/Sr is a constant throughout the sample, whereas the initial ratio Rbmay vary somewhat (ie.. the sample is not perfectly homogeneous). This is because Sr and Sr are chemically identical, allowing them to be bound up in minerals in the same proportions, whereas the proportion of RbSr need not be constant throughout. 3838 TABLE 20.2 Results from the Analysis of Basalt 10072, Returned from the Sea of Tranquility by the Apollo 11 Astronauts in 1969. (Data from D. A. Papanastassiou, D. J. DePaolo, and G J Wasserburg. "Rb-Sr and Sm-Nd Chronology and Genealogy of Mare Basalts from the Sea of Tranquility, Proceedings of the Eighth Lunar Science Conference, Pergamon Press, New York, 0.1847 0.1963 0.1980 0.51172118 0.51 1998 ± 16 0.512035 ± 21 0.512238 ± 17 0.5 1 378815 0.514154 17 0.2715 0.2879

Explanation / Answer

for beta decay of 87 Rb

let initial number of atoms of 87Rb be No

then

87Rb/86Sr = No/86Sr

now, number of 87Sr atoms = N

N = No[e^(lambda*t) - 1]

hence

87Sr/86Sr = N/86Sr

hence

slope of the graph = 87Sr/87Rb = N/No = e^(lambda*t) - 1

now, e^(lambda*t) - 1 = 0.0062 ( slope of the curve)

t = 4.39*10^9 years

20.2

now, from given data

having the 147Sm/144Nd on y axis and 143Nd/144Nd on x axis

slope m = 42.2

hence

e^(lambda*t) - 1 = 42.2

also, lambda = ln(2)/1.06*10^11 years

hence

t = 5.7589369*10^11 years

comparing this with lumar highland data

this is a 100 times older than the high lands

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