Strontium-90 is radioactive and has a half life of 28.8 years. Calculate the activity of a 4.9mg sample of strontium-90. Give your answer in becquerels and in curies. Be sure your answer has the correct number of significant digits.BqCi
Question
Strontium-90 is radioactive and has a half life of 28.8 years. Calculate the activity of a 4.9mg sample of strontium-90. Give your answer in becquerels and in curies. Be sure your answer has the correct number of significant digits.BqCi
Solution
To calculate the activity of a radioactive sample, we need to know the number of atoms in the sample and the decay constant of the isotope.
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First, we need to calculate the number of atoms in the sample. The atomic mass of strontium-90 is approximately 90 g/mol. Therefore, a 4.9 mg sample is 4.9 x 10^-3 g.
Number of moles = mass (g) / molar mass (g/mol) = 4.9 x 10^-3 g / 90 g/mol = 5.44 x 10^-5 mol
Since 1 mol contains Avogadro's number (6.022 x 10^23) of atoms, the number of atoms in the sample is:
Number of atoms = moles x Avogadro's number = 5.44 x 10^-5 mol x 6.022 x 10^23 atoms/mol = 3.27 x 10^19 atoms
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Next, we need to calculate the decay constant. The decay constant (λ) is related to the half-life (t1/2) by the equation:
λ = ln(2) / t1/2 = 0.693 / 28.8 years = 0.024 yr^-1
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Finally, we can calculate the activity (A), which is the product of the decay constant and the number of atoms:
A = λN = 0.024 yr^-1 x 3.27 x 10^19 atoms = 7.85 x 10^18 decays/yr
Since 1 Bq is defined as one decay per second, and there are approximately 3.15 x 10^7 seconds in a year, the activity in becquerels is:
A = 7.85 x 10^18 decays/yr x 1 yr / 3.15 x 10^7 s = 2.49 x 10^11 Bq
And since 1 Ci is defined as 3.7 x 10^10 decays per second, the activity in curies is:
A = 2.49 x 10^11 Bq x 1 Ci / 3.7 x 10^10 Bq = 6.73 Ci
So, the activity of a 4.9 mg sample of strontium-90 is approximately 2.49 x 10^11 Bq or 6.73 Ci.
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