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The half life for radioactive decay of `.^(14)C` is 5730 years. An archaeological artifact containing wood had only `80%` of the `.^(14)C` found in a living tree. Estimat the age of the sample. |
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Answer» Correct Answer - 1845 years `(N)/(N_(0))=80% or 0.80` `t_(1//2)=5730 years` `k=(0.693)/(t_(1//2))=(0.693)/(5730years)=1.21xx10^(-4)year^(-1)` `k=(2.303)/(t)log``(N_(0))/(N)` `1.21xx10^(-4)=(2.303)/(t)log``(1)/(0.8)` `T=(2.303)/(1.21XX10^(4))log1.25` `=(2.303)/(1.21xx10^(-4))log1.25` `=(2.303)/(1.21xx10^(-4))xx0.0969=1845years` Alternatively Use direct relation, `(t_(80%))/(t_(1//2))=(log(100)/(80))/(0.3)` `t_(80%)=(5730years xx log0.8)/(0.3)~~1845years.` |
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