The isotope strontium-90 is produced during the testing of nuclear weapons. If 100.0mg of strontium-90 was released in the atmosphere in 1960, how much of the radioisotope remains 85 years later? The half-life of strontium-90 is 29 years. Show all your work.
Solution:
Strontium-90 ⇨ Sr-90
The half-life (t1/2) of strontium-90 is 29 years
From law of radioactive decay:
t1/2 = ln(2) / λ
where λ = decay constant
λ = ln(2) / t1/2
λ = (0.693) / (29 years) = 0.0239 years−1
Radioactive decay follows the kinetics of a first order reaction:
ln[Sr-90]t = −λt + ln[Sr-90]o
λ = 0.0239 years−1
t = 85 years
[Sr-90]o = 100.0 mg
[Sr-90]t = unknown
Therefore,
ln[Sr-90]t = −(0.0239 years−1 × 85 years) + ln(100.0)
ln[Sr-90]t = −2.0315 + 4.60517 = 2.57367
[Sr-90]t = e2.57367 = 13.11
[Sr-90]t = 13.11 mg
Answer: 13.11 mg of the radioisotope (Sr-90) remained
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