What should be the 14 C 6 content (in percent of 0) of a fossilized tree that is claimed to be 3000 years?
Answers
Answer:
the current mass fraction of [tex]^(14)
Step-by-step explanation:
The current mass fraction of ^{14}_{6} C614C should be approximately 68.8 percent.
Explanation:
^{14}_{6} C614C 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}dtdm=−τm
Where:
mm - Current mass of the isotope, measured in grams.
\tauτ - Time constant, measured in years.
The solution of this equation is of the form:
m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }m(t)=mo⋅e−τt
Where:
tt - Time, measured in years.
m_{o}mo - Initial mass of the isotope, measured in grams.
The time constant can be found as a function of halflife (t_{1/2}t1/2 ):
\tau = \frac{t_{1/2}}{\ln 2}τ=ln2t1/2
If t_{1/2} = 5568\,yrst1/2=5568yrs and t = 3000\,yrst=3000yrs , the mass fraction of ^{14}_{6} C614C is:
\tau = \frac{5568\,yrs}{\ln 2}τ=ln25568yrs
\tau \approx 8032.926\,yrsτ≈8032.926yrs
\frac{m(3000\,yrs)}{m_{o}} = e^{-\frac{3000\,yrs}{8032.926\,yrs} }mom(3000yrs)=e−8032.926yrs3000yrs
\frac{m(3000\,yrs)}{m_{o}} \approx 0.688mom(3000yrs)≈0.688
The current mass fraction of ^{14}_{6} C614C should be approximately 68.8 percent.
Here is your answer please mark me as brainliest