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The C-14 content of an ancient piece of wood was found to have three tenths of that in living trees. How old is that piece of wood? (log 3= 0.4771, log 7 = 0.8540 , Half-life of C-14 = 5730 years ) |
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Answer» k = \(\frac{0.693}{t^\frac{1}{2}}\) k = \(\frac{0.693}{5730}\) years-1 t = \(\frac{2.303}{k}\) log \(\frac{Co}{Ct}\) let Co = 1 Ct = \(\frac{3}{10}\) so \(\frac{Co}{Ct}\)= \(\frac{1}{(3/10)}\) = \(\frac{10}{3}\) t = \(\frac{2.303}{0.693}\) x 5730 log \(\frac{10}{3}\) t = 19042 x (1-0.4771)= 9957 years |
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