Question

The decay constant for 14C is .00012 In 1947, the famous cave paintings in Lascaux, France were discovered and testing revealed that charcoal in the cave contained 20% of the 14C 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 1

The question is incomplete, here is the complete question:

The decay constant for 14-C is $0.00012yr^{-1}$ 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 $t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}$

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:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant  = $0.00012yr^{-1}=1.2\times 10^{-4}yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = (100 - 20) = 80 grams

Putting values in above equation, we get:

$1.2\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{20}\\\\t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}$

Hence, the formula for the age of the charcoal is $t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}$