Question

archeologists use radiocarbon dating to determine the age of ancient tools. they measure levels of the isotope carbon - 14, which has a half - life of 5,730 years. this means that after 5,730 years, only half of the original amount of carbon - 14 will remain. which function models the decay of carbon - 14?
○ ( a(t)=a_{0}e^{0.000121t} )
○ ( a(t)=a_{0}e^{-0.000121t} )
○ ( a(t)=a_{0}e^{0.0000121} )
○ ( a(t)=a_{0}e^{-0.0000121t} )

Explanation:

Step1: Recall the formula for exponential decay

The general formula for exponential decay is \(A(t)=A_{0}e^{kt}\), where \(A_{0}\) is the initial amount, \(t\) is the time, and \(k\) is the decay constant. Since it is a decay process, \(k<0\).

Step2: Calculate the decay constant \(k\)

We know that the half - life \(T = 5730\) years. When \(t = T\), \(A(T)=\frac{A_{0}}{2}\). Substitute into the formula \(A(t)=A_{0}e^{kt}\):
\(\frac{A_{0}}{2}=A_{0}e^{k\times5730}\). Divide both sides by \(A_{0}\) (since \(A_{0}
eq0\)), we get \(\frac{1}{2}=e^{5730k}\).
Take the natural logarithm of both sides: \(\ln(\frac{1}{2})=\ln(e^{5730k})\).
Using the property \(\ln(e^{x}) = x\) and \(\ln(\frac{1}{2})=-\ln(2)\approx - 0.693\), we have \(-0.693 = 5730k\).
Solve for \(k\): \(k=\frac{-0.693}{5730}\approx - 0.000121\).

Answer:

\(A(t)=A_{0}e^{- 0.000121t}\) (the second option)