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.
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