The question is incomplete, here is the complete question:
The decay constant for 14-C is [tex]0.00012yr^{-1}[/tex] In 1947, the famous cave paintings in Lascaux, France were discovered and testing revealed that charcoal in the cave contained 20% of the 14-C found in living trees. Write a formula for the age of the charcoal (hence of the associated paintings). Show your work to find this formula.
Answer: The formula for the age of the charcoal is [tex]t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}[/tex]
Explanation:
Carbon-14 isotope is a radioisotope and its decay process follows first order kinetics.
Rate law expression for first order kinetics is given by the equation:
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = [tex]0.00012yr^{-1}=1.2\times 10^{-4}yr^{-1}[/tex]
t = time taken for decay process = ? yr
[tex][A_o][/tex] = initial amount of the sample = 100 grams
[A] = amount left after decay process = (100 - 20) = 80 grams
Putting values in above equation, we get:
[tex]1.2\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{20}\\\\t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}[/tex]
Hence, the formula for the age of the charcoal is [tex]t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}[/tex]
