Asked by felpin
Naturally occuring boron is composed of only two isoptopes 10/5B and 11/B with isotopic masses of 10.0129 and 11.00931 respectively. To account for the atomic mass of 10.811 what must be the percentage abundance of each isotope?
Answers
Answered by
DrBob222
Let 10B abundance = x
Then 11B abundance = 1-x
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x(10.0129) + (1-x)(11.00931) = 10.811.
Solve for x and 1-x, then convert to percent.
Then 11B abundance = 1-x
-------------------------
x(10.0129) + (1-x)(11.00931) = 10.811.
Solve for x and 1-x, then convert to percent.
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