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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. |
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Answer» Solution :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. |
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