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How to find abundance of 206, 207 and 208 Pb

by subopolois
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subopolois
#1
Nov21-08, 03:10 PM
P: 69
how would i find the abundance of Pb 206, 207 and 208 if im given the isotopic data of 207Pb/206Pb and 208Pb/206Pb?
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chemisttree
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Nov21-08, 04:36 PM
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Nov24-08, 05:30 PM
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OK... Assume that the expression 206Pb means the molar fraction of lead-206. The same is true for 207Pb and 208Pb. You know that the molar fractions sum to 1 as well as the molar ratios of 207Pb/206Pb and 208Pb/206Pb. You can also calculate 207Pb/208Pb. From here it is merely solving for the simultaneous equations.

subopolois
#4
Nov24-08, 07:05 PM
P: 69
How to find abundance of 206, 207 and 208 Pb

Quote Quote by chemisttree View Post
OK... Assume that the expression 206Pb means the molar fraction of lead-206. The same is true for 207Pb and 208Pb. You know that the molar fractions sum to 1 as well as the molar ratios of 207Pb/206Pb and 208Pb/206Pb. You can also calculate 207Pb/208Pb. From here it is merely solving for the simultaneous equations.
ok, i think i get it. heres my scenerio,
the Pb content is 468ppm
207Pb/206Pb= 0.2515
208Pb/206Pb= 0.0581
i could just use abgebra to find each right?
206=x
207=0.2515x
208=0.0581x
i just solve for x and put it back into the previous equation?
chemisttree
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Nov25-08, 08:49 AM
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Quote Quote by subopolois View Post
ok, i think i get it. heres my scenerio,
the Pb content is 468ppm
207Pb/206Pb= 0.2515
208Pb/206Pb= 0.0581
i could just use abgebra to find each right?
206=x
207=0.2515x
208=0.0581x
i just solve for x and put it back into the previous equation?
I would use the definition... 206Pb + 207Pb + 208Pb = 1


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