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# A wooden artifact from an ancient tomb contains 65% of the carbon-14 that is present in living trees. How long ago was the artifact made?

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A wooden artifact from an ancient tomb contains 65% of the carbon-14 that is present in living trees. How long ago was the artifact made? (The half-life of carbon-14 is 5730 years.

Guest Jul 9, 2014

#1
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The radioactive decay equation is N = N0e-ln(2)*t/τ, where τ is the half-life, N0 is the number of atoms at time t = 0 and N is the number at time t.

Rewrite this as (N/N0) = e-ln(2)*t/τ, where, here, N/N0 is 0.65  (i.e. 65% expressed as a fraction). So:

0.65 =  e-ln(2)*t/5730.

Take logs of both sides:

ln(0.65) = -ln(2)*t/5730.

Rearrange:

t = -5730*ln(0.65)/ln(2) years

$${\mathtt{t}} = {\mathtt{\,-\,}}{\frac{{\mathtt{5\,730}}{\mathtt{\,\times\,}}{ln}{\left({\mathtt{0.65}}\right)}}{{ln}{\left({\mathtt{2}}\right)}}} \Rightarrow {\mathtt{t}} = {\mathtt{3\,561.128\: \!398\: \!756\: \!128\: \!174\: \!2}}$$

or the artifact was made ≈ 3560 years ago.

Alan  Jul 13, 2014
Sort:

#1
+26547
+8

The radioactive decay equation is N = N0e-ln(2)*t/τ, where τ is the half-life, N0 is the number of atoms at time t = 0 and N is the number at time t.

Rewrite this as (N/N0) = e-ln(2)*t/τ, where, here, N/N0 is 0.65  (i.e. 65% expressed as a fraction). So:

0.65 =  e-ln(2)*t/5730.

Take logs of both sides:

ln(0.65) = -ln(2)*t/5730.

Rearrange:

t = -5730*ln(0.65)/ln(2) years

$${\mathtt{t}} = {\mathtt{\,-\,}}{\frac{{\mathtt{5\,730}}{\mathtt{\,\times\,}}{ln}{\left({\mathtt{0.65}}\right)}}{{ln}{\left({\mathtt{2}}\right)}}} \Rightarrow {\mathtt{t}} = {\mathtt{3\,561.128\: \!398\: \!756\: \!128\: \!174\: \!2}}$$

or the artifact was made ≈ 3560 years ago.

Alan  Jul 13, 2014

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