Answer:
[tex]\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659[/tex] and [tex]\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661[/tex]
Step-by-step explanation:
The equation of the isotope decay is:
[tex]\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }[/tex]
14-Carbon has a half-life of 5568 years, the time constant of the isotope is:
[tex]\tau = \frac{5568\,years}{\ln 2}[/tex]
[tex]\tau \approx 8032.926\,years[/tex]
The decay time is:
[tex]t = 1315\,years + 2007\,years \pm 13\,years[/tex] (There is no a year 0 in chronology).
[tex]t = 3335 \pm 13\,years[/tex]
Lastly, the relative amount is estimated by direct substitution:
[tex]\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }[/tex]
[tex]\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }[/tex]
[tex]\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659[/tex]
[tex]\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }[/tex]
[tex]\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661[/tex]
