Question

A sample of a wood carving from an archaeological dig gave 20,400 disintegrations per gram of carbon per day. A 1.00g sample of carbon from a modern source gave 22,080 disinegrations per day. If the decay constant of carbon-14 is 1.2x10^(-4) year^(-1) , how old is the wood?

Answer 1

Decay constant of C-14 = 1.2x10^(-4) per year
Thus, half life of the carbon source from the archaeological dig = ln (2) / decay constant (k) = 5775 years

(Half life of a first order reaction)

Now, the fraction of C-14 remaining in the old sample of wood:
20400/22080 = 0.9239

Thus, the number of half-lives that have already elapsed for the old wood sample:
(1/2)^n = 0.9239
Taking log on both sides,
n*log (0.5)= log (0.9239)=
n= 0.1141913875224835

Number of years (age of the wood)= 0.1142*5775 = 659.5 years