How to find abundance of 206, 207 and 208 Pb

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To find the abundance of lead isotopes 206Pb, 207Pb, and 208Pb using given isotopic data, one can set up equations based on the molar fractions, which must sum to 1. The ratios 207Pb/206Pb and 208Pb/206Pb provide the necessary relationships to express 207Pb and 208Pb in terms of 206Pb. By defining 206Pb as x, the other isotopes can be expressed as 207Pb = 0.2515x and 208Pb = 0.0581x. Solving these equations allows for the determination of the molar fractions of each isotope. This algebraic approach effectively calculates the isotopic abundances from the provided data.
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how would i find the abundance of Pb 206, 207 and 208 if I am given the isotopic data of 207Pb/206Pb and 208Pb/206Pb?
 
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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.
 
chemisttree said:
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. here's 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?
 
subopolois said:
ok, i think i get it. here's 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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