Question

exponential decay app (y = ae^(kt))
the half - life of radium - 226 is 1590 years. if a sample contains 300 mg, how many mg will remain after 2000 years?
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Explanation:

Step1: Find the decay constant \(k\)

The formula for exponential decay is \(y = ae^{kt}\). At the half - life \(t = 1590\), \(y=\frac{a}{2}\). Substitute into the formula: \(\frac{a}{2}=ae^{k\times1590}\). Divide both sides by \(a\) (since \(a
eq0\)) to get \(\frac{1}{2}=e^{1590k}\). Take the natural logarithm of both sides: \(\ln(\frac{1}{2})=\ln(e^{1590k})\). Using the property \(\ln(e^{x}) = x\), we have \(\ln(\frac{1}{2})=1590k\). Since \(\ln(\frac{1}{2})=-\ln(2)\approx - 0.693\), then \(k=\frac{\ln(\frac{1}{2})}{1590}\approx\frac{- 0.693}{1590}\approx - 0.000436\).

Step2: Find the remaining amount after \(t = 2000\) years

We know \(a = 300\) (initial amount), \(k\approx - 0.000436\), and \(t = 2000\). Substitute into the formula \(y=ae^{kt}\), so \(y = 300e^{-0.000436\times2000}\). First, calculate \(-0.000436\times2000=-0.872\). Then \(y = 300e^{- 0.872}\). Since \(e^{-x}=\frac{1}{e^{x}}\) and \(e^{0.872}\approx2.393\), \(y = 300\times\frac{1}{2.393}\approx125.4\).

Answer:

\(125.4\) mg