Decay to two products - Sr 87 obs = 0.7 Sr 86 obs Rb 87...

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Decay to two products Things become more complex if we have more than one decay product. Thus for example with the the Potassium decay where we have two pathways, K 40 can go to Ar 40 after electron capture, or two Ca 40 after beta decay: K 40 + e --> Ar 40 (decay constant L e ) K 40 (beta decay) --> Ca 40 (decay constant L b ) D = P 0 - P 0 e -(Le+Lb)t so D = P n (e (Le+Lb)t -1) so t = [1/(L e +L b )] ln [1+D/P n ] D = D e + D b = (1 + D b /D e )D e so D e = D/[1+D b /D e ] = D/[1+L b /L e ] = DL e /(L e +L b ) so D = D e (L e +L b )/L e and t = [1/(L e +L b )] ln[1 + (L e +L b )D a /L a P n ] Now L e = ln 2 /T e 1/2 Typically we might get amount of Ar 40 per gm = 13.6 10 15 atoms and amount of K 40 = 77 10 15 atoms. This will give a t of around 1.86 10 9 . Rubidium-Strontium Method Sr 87
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Unformatted text preview: Sr 87 obs = 0.7 Sr 86 obs + Rb 87 obs (e Lt-1) (SR 87 /Sr 86 ) obs = 0.7 + (Rb 87 /Sr 86 ) obs (e Lt-1) Rubidium substitutes for Potassium, K, with the same valency 1 and similar radii, 0.13 - 0.15 nm. The method used for this dating technique is to take a rock sample, a solid source, then vaporise it to pass it through a mass spectrometer to obtain the ratios of the elements. Plotting the ratios should give an interrupt of 0.7 as a check on the answer, while the slope gives t. The Potassium and Rubidium methods can be used together and often are as checks on one another....
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