The half life for radioactive `""^(14)C` is 5730 years. An archeological artefact containing wood had only 80% of `""^(14)C` activity as found in a living tree. Calculate the age of the artefact.

Explanation

Radioactive decay is a first order reaction.
`k=(2.303)/(t)"log"([A]_(0))/([A]) or t=(2.303)/(t)"log"([A]_(0))/([A])`
For 80% activity , `[A]=0.8[A]_(0)`
Substituting in the above equation, we get
`t=(2.303)/(0.693//t_(0.5))xx"log"([A]_(0))/(0.8[A]_(0))`
or `t=(2.303)/(0.693) xx 5730 xx 0.0969=1845` years
Thus, the artefact in 1845 years old.