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the half life of c14 is 5730 years. Suppose that wood found at an archeological excavation site contains about 35% as much C14 as does living plant material. Determine when the wood was cut

User Xram
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Answer:

The wood was cut approximately 8679 years ago.

Explanation:

At first we assume that examination occured in 2020. The decay of radioactive isotopes are represented by the following ordinary differential equation:


(dm)/(dt) = -(m)/(\tau) (Eq. 1)

Where:


(dm)/(dt) - First derivative of mass in time, measured in miligrams per year.


\tau - Time constant, measured in years.


m - Mass of the radioactive isotope, measured in miligrams.

Now we obtain the solution of this differential equation:


\int {(dm)/(m) } = -(1)/(\tau)\int dt


\ln m = -(1)/(\tau) + C


m(t) = m_(o)\cdot e^{-(t)/(\tau) } (Eq. 2)

Where:


m_(o) - Initial mass of isotope, measured in miligrams.


t - Time, measured in years.

And time is cleared within the equation:


t = -\tau \cdot \ln \left[(m(t))/(m_(o)) \right]

Then, time constant can be found as a function of half-life:


\tau = (t_(1/2))/(\ln 2) (Eq. 3)

If we know that
t_(1/2) = 5730\,yr and
(m(t))/(m_(o)) = 0.35, then:


\tau = (5730\,yr)/(\ln 2)


\tau \approx 8266.643\,yr


t = -(8266.643\,yr)\cdot \ln 0.35


t \approx 8678.505\,yr

The wood was cut approximately 8679 years ago.

User Mangokitty
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