A radioactive source contains two isotopes of phosphorus: P-32 with half-life T1 and P-33 with half-life T2 (T2 > T1). Initially 10% of all decays come from P-33. How long must one wait until 90% of the decays come from P-33?

t = 4*ln(3) / (ln(2) * (1/T1 - 1/T2))
t = 4*ln(3) / (ln(2) * (1/T1 + 1/T2))
t = 2*ln(3) / (ln(2) * (1/T1 - 1/T2))
None of these

Correct answer: t = 4*ln(3) / (ln(2) * (1/T1 - 1/T2))

Solution

Let lambda1 = ln2/T1, lambda2 = ln2/T2. Initially A33/A32 = 1/9. At time t: A33(t)/A32(t) = (A33₀/A32₀)*exp(-(lambda2 - lambda1)*t) = (1/9)*exp((lambda1 - lambda2)*t) = 9. So exp((lambda1-lambda2)*t) = 81, giving t = ln(81)/(lambda1-lambda2) = 4*ln3 / ((ln2/T1 - ln2/T2)) = 4*ln3/(ln2*(1/T1 - 1/T2)).

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