# Carbon-14 dating assumes that the carbon dioxide on Earth today has the same radioactive content as it did centuries ago. If this is true, the amount of absorbed by a tree that grew several...

Carbon-14 dating assumes that the carbon dioxide on Earth today has the same radioactive content as it did centuries ago. If this is true, the amount of absorbed by a tree that grew several centuries ago should be the same as the amount of absorbed by a tree growing today. A piece of ancient charcoal contains only 15% as much of the radioactive carbon as a piece of modern charcoal. How long ago was the tree burned to make the ancient charcoal? (The half-life of Carbon-14 is 5715 years.)

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It is impossible to predict when a particular atom will decay. However, it is equally likely to decay at any instant in time. Therefore, given a sample of a particular radioisotope, the number of decay events `−dN` expected to occur in a small interval of time `dt` is proportional to the number of atoms present `N,` i.e.

`-(dN)/(dt)propto N`

For different atoms different decay constants apply.

`-(dN)/(dt)=\lambda N`

The above differential equation is easily solved by separation of variables.

`N=N_0e^(-lambda t)`

where `N_0` is the number of undecayed atoms at time `t=0.`

We can now calculate decay constant `lambda` for carbon-14 using the given half-life.

`N_0/2=N_0e^(-lambda 5715)`

`e^(-5715lambda)=1/2`

`-5715lambda=ln(1/2)`

`lambda=-(ln(1/2))/5715`

`lambda=1.21 times 10^-4`

Note that the above constant is usually measured in seconds rather than years.

Now we can return to the problem at hand. Since the charcoal contains only 15% (`0.15N_0` ) of the original carbon-14, we have

`0.15N_0=N_0e^(-1.21times10^-4t)`

Now we solve for `t.`

`e^(-1.21times10^-4t=0.15)`

`1.21times10^-4=-ln 0.15`

`t=-(ln0.15)/(1.21times10^-4)`

`t=15678.68`

According to our calculation **the tree was burned approximately 15679 years ago.**