How to find abundance of 206, 207 and 208 Pb

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Discussion Overview

The discussion revolves around the methods for determining the abundance of lead isotopes 206Pb, 207Pb, and 208Pb using given isotopic ratios and total lead content. The focus is on the mathematical relationships between the isotopes and their molar fractions.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant suggests that the molar fractions of 206Pb, 207Pb, and 208Pb sum to 1 and that the ratios 207Pb/206Pb and 208Pb/206Pb can be used to set up simultaneous equations for solving the abundances.
  • Another participant presents a specific scenario with a total lead content of 468ppm and the ratios 207Pb/206Pb = 0.2515 and 208Pb/206Pb = 0.0581, proposing to use algebra to find the molar fractions.
  • The same participant reiterates the approach of defining 206Pb as x, 207Pb as 0.2515x, and 208Pb as 0.0581x, indicating a method to solve for x and subsequently the other isotopes.
  • There is a mention of using the definition that 206Pb + 207Pb + 208Pb = 1 as part of the solution process.

Areas of Agreement / Disagreement

Participants appear to agree on the method of using molar fractions and ratios to find the abundances, but there is no consensus on the specific calculations or any potential complications that may arise.

Contextual Notes

The discussion does not clarify any assumptions regarding the isotopic data or the implications of the total lead content on the calculations.

subopolois
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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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