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