Question

plutonium - 240 decays according to the function $q(t)=q_{0}e^{-kt}$. how long will it take 27 grams of plutonium - 240 to decay to 9 grams? k is the decay constant, 0.00011 2,100 years 9,987 years 1.44 years 18,900 years

Explanation:

Step1: Substitute values into formula

Given \(Q(t) = 9\), \(Q_0=27\), \(k = 0.00011\). Substitute into \(Q(t)=Q_0e^{-kt}\), we get \(9 = 27e^{- 0.00011t}\).

Step2: Simplify the equation

Divide both sides by 27: \(\frac{9}{27}=e^{-0.00011t}\), so \(\frac{1}{3}=e^{-0.00011t}\).

Step3: Take natural logarithm

Take \(\ln\) on both sides: \(\ln(\frac{1}{3})=\ln(e^{-0.00011t})\). Since \(\ln(e^{x}) = x\), we have \(\ln(\frac{1}{3})=- 0.00011t\).

Step4: Solve for \(t\)

We know \(\ln(\frac{1}{3})=-\ln(3)\approx - 1.0986\). Then \(t=\frac{-\ln(\frac{1}{3})}{0.00011}=\frac{\ln(3)}{0.00011}\approx\frac{1.0986}{0.00011}\approx9987\).

Answer:

9,987 years