Question

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

Answer 1

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.