Answer:
The wood was cut approximately 8679 years ago.
Step-by-step explanation:
At first we assume that examination occured in 2020. The decay of radioactive isotopes are represented by the following ordinary differential equation:
[tex]\frac{dm}{dt} = -\frac{m}{\tau}[/tex] (Eq. 1)
Where:
[tex]\frac{dm}{dt}[/tex] - First derivative of mass in time, measured in miligrams per year.
[tex]\tau[/tex] - Time constant, measured in years.
[tex]m[/tex] - Mass of the radioactive isotope, measured in miligrams.
Now we obtain the solution of this differential equation:
[tex]\int {\frac{dm}{m} } = -\frac{1}{\tau}\int dt[/tex]
[tex]\ln m = -\frac{1}{\tau} + C[/tex]
[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex] (Eq. 2)
Where:
[tex]m_{o}[/tex] - Initial mass of isotope, measured in miligrams.
[tex]t[/tex] - Time, measured in years.
And time is cleared within the equation:
[tex]t = -\tau \cdot \ln \left[\frac{m(t)}{m_{o}} \right][/tex]
Then, time constant can be found as a function of half-life:
[tex]\tau = \frac{t_{1/2}}{\ln 2}[/tex] (Eq. 3)
If we know that [tex]t_{1/2} = 5730\,yr[/tex] and [tex]\frac{m(t)}{m_{o}} = 0.35[/tex], then:
[tex]\tau = \frac{5730\,yr}{\ln 2}[/tex]
[tex]\tau \approx 8266.643\,yr[/tex]
[tex]t = -(8266.643\,yr)\cdot \ln 0.35[/tex]
[tex]t \approx 8678.505\,yr[/tex]
The wood was cut approximately 8679 years ago.
