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Strontium-90 (90Sr) is a radioactive isotope with a half-life of 28.8 years. What percentage of this...

Strontium-90 (90Sr) is a radioactive isotope with a half-life of 28.8 years. What percentage of this isotope remains after 11.4 years?

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Answer #1

Given:
Half life = 28.8 yr
use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k
= 0.693/(half life)
= 0.693/(28.8)
= 2.406*10^-2 yr-1
we have:
[90Sr]o = 100 (let initial concentration be 100)
t = 11.4 yr
k = 2.406*10^-2 yr-1


use integrated rate law for 1st order reaction
ln[90Sr] = ln[90Sr]o - k*t
ln[90Sr] = ln(100) - 2.406*10^-2*11.4
ln[90Sr] = 4.6052 - 2.406*10^-2*11.4
ln[90Sr] = 4.3309
[90Sr] = 76.0

Since initial concentration was 100 and 76.0 is remaining
This is same as 76.0 % remaining
Answer: 76.0 %

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