The radioactive nuclide plutonium 199 has a half life of 43.0min.A sample is prepared that has an initial activity of 7.56 x 10^11Bq i.How many plutonium 199 nuclei are initially present in the sample ii. What is the activity at this time
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718. We know that the half-life of plutonium 199 is 43.0 min, which means that the decay constant is: λ = ln(2) / t1/2 = ln(2) / 43.0 min ≈ 0.0161 min^-1 Substituting the given values into the formula, we get: N = N0 e^(-λt) N0 = N / e^(-λt) N0 = 7.56 x 10^11 Show more…
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