Answer:
Amount of C-14 taken were 2.5 pounds and 3.5 pounds respectively.
Step-by-step explanation:
Radioactive decay is an exponential process represented by
\(A_{t}=A_{0}e^{-kt}\)
where \(A_{t}\) = Amount of the radioactive element after t years
\(A_{0}\) = Initial amount
k = Decay constant
t = time in years
Half life period of Carbon-14 is 5730 years.
\(\frac{A_{0} }{2}=A_{0}e^{-5730k}\)
\(\frac{1}{2}=e^{-5730k}\)
Now we take ln (Natural log) on both the sides
\(ln(\frac{1}{2})=ln[e^{-5730k}]\)
-ln(2) = -5730kln(e)
0.69315 = 5730k
\(k=\frac{0.69315}{5730}\)
\(k=1.21\times 10^{-4}\)
Now we have to calculate the weight of samples of C-14 taken for the remaining quantities \(\frac{5}{8}\) and \(\frac{7}{8}\) of a pound.
\(\frac{5}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}\)
\(\frac{5}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}\)
\(\frac{5}{8}=A_{0}e^{(-1.3863)}\)
\(A_{0}=\frac{5}{8}\times e^{1.3863}\)
\(A_{0}=\frac{5}{8}\times 4\)
\(A_{0}=\frac{5}{2}\)
\(A_{0}=2.5\) pounds
Similarly for \(\frac{7}{8}\) pounds
\(\frac{7}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}\)
\(\frac{7}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}\)
\(\frac{7}{8}=A_{0}e^{(-1.3863)}\)
\(A_{0}=\frac{7}{8}\times e^{(1.3863)}\)
\(A_{0}=\frac{7}{8}\times 4\)
\(A_{0}=\frac{7}{2}\)
\(A_{0}=3.5\) pounds
