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Isotopes

# 512638 147sm144ndchur t 0 ma 01967 we can now

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Unformatted text preview: Notation ( ( ) ) Ⱥ 143 Nd / 144Nd Ⱥ rock ,t = 0 Ma εNd = Ⱥ − 1Ⱥ × 10 4 (6) Ⱥ 143 Nd / 144Nd CHUR ,t =0 Ma Ⱥ Ⱥ Ⱥ where: 143Nd/144Ndrock, t=0Ma = measured ratio of rock at present day 143Nd/144Nd CHUR, t = 0 Ma = 0.512638 = the value of the Chondrite Uniform Reservoir at 0 Ma This is valid for t = 0 Ma but with older rocks we need to recalculate the 143Nd/144Nd ratios of both the rock and CHUR for the time in question by rearranging equation 1, inputting the age of the rock (t) and solving for (143Nd/144Nd)0 such that: (143Nd/144Nd)rock,t = (143Nd/144Nd)rock, t = 0 Ma – (147Sm/144Nd)rock, t = 0 Ma*(e t ­1) (7) and (143Nd/144Nd)CHUR, t = (143Nd/144Nd)CHUR,t = 0 Ma – (147Sm/144Nd)CHUR, t = 0 Ma*(e t ­1) (8) Since we know: (143Nd/144Nd)rock, t = 0 Ma, (147Sm/144Nd)rock, t = 0 Ma = these are known by measurement and (143Nd/144Nd)CHUR,t = 0 Ma = 0.512638 (147Sm/144Nd)CHUR, t = 0 Ma = 0.1967 We can now “back ­calculate” to find the initial 143Nd/144Nd rocks of the rock and CHUR. We then take the results from (7...
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