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. Estimate the age of the sample.

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Solution

Decay constant $k = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{5730 y e a r} = 1.209 \times 10^{- 4} / y e a r s$

The rate of counts is proportional to the number of C-14 atoms in the sample.
$N_{0} = 100, N = 80$

The age of the sample $t = \frac{2.303}{k} l o g (\frac{N_{0}}{N})$

$t = \frac{2.303}{1.209 \times 10^{- 4}} \times l o g (\frac{100}{80}) = 1846 y e a r s$

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The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.

Decay constant k=0.693t1/2=0.6935730year=1.209×10−4/yearsThe rate of counts is proportional to the number of C-14 atoms in the sample.N0=100,N=80The age of the sample t=2.303klog(N0N)t=2.3031.209×10−4×log(10080)=1846years

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. Estimate the age of the sample.