Question

What should the ^14_6C content (in percent of y0) of a fossilized tree that is claimed to be 3000 years old?

Answer 1

Answer:

The current mass fraction of $^{14}_{6} C$ should be approximately 68.8 percent.

Explanation:

$^{14}_{6} C$ is a radioactive isotope with a halflife of 5568 years. The decay of any radioisotope is modelled after the following ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$

Where:

$m$ - Current mass of the isotope, measured in grams.

$\tau$ - Time constant, measured in years.

The solution of this equation is of the form:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where:

$t$ - Time, measured in years.

$m_{o}$ - Initial mass of the isotope, measured in grams.

The time constant can be found as a function of halflife ($t_{1/2}$):

$\tau = \frac{t_{1/2}}{\ln 2}$

If $t_{1/2} = 5568\,yrs$ and $t = 3000\,yrs$, the mass fraction of $^{14}_{6} C$ is:

$\tau = \frac{5568\,yrs}{\ln 2}$

$\tau \approx 8032.926\,yrs$

$\frac{m(3000\,yrs)}{m_{o}} = e^{-\frac{3000\,yrs}{8032.926\,yrs} }$

$\frac{m(3000\,yrs)}{m_{o}} \approx 0.688$

The current mass fraction of $^{14}_{6} C$ should be approximately 68.8 percent.