Question

carbon - 14 decays at a constant rate, so it can be used to determine the age of fossils. in particular, if the original amount of carbon - 14 present is ( a_0 ), then ( a(t)=a_0e^{-kt} ) can be used find the amount of amount of carbon - 14 remaining after ( t ) years. given that the half - life of carbon - 14 is 5,730 years, what is the value of the decay constant ( k ) to 5 decimal places? ( a_0 = a_0e^{-k\times5730} ) ( kapprox )

Explanation:

Step1: Substitute values into the formula

Given \(A(t)=\frac{A_0}{2}\) (half - life) and \(t = 5730\) into \(A(t)=A_0e^{-kt}\). So \(\frac{A_0}{2}=A_0e^{-k\times5730}\). Divide both sides by \(A_0\) (since \(A_0
eq0\)), we get \(\frac{1}{2}=e^{- 5730k}\).

Step2: Take the natural logarithm of both sides

\(\ln(\frac{1}{2})=\ln(e^{-5730k})\). Using the property \(\ln(e^{x}) = x\) and \(\ln(\frac{1}{2})=-\ln(2)\), we have \(-\ln(2)=-5730k\).

Step3: Solve for \(k\)

\(k=\frac{\ln(2)}{5730}\). Since \(\ln(2)\approx0.693147\), then \(k=\frac{0.693147}{5730}\approx0.00012\).

Answer:

\(k\approx0.00012\)