Answer: 54678 years
Step-by-step explanation:
This can be solved by the following equation:
[tex]N_{t}=N_{o}e^{-\lambda t}[/tex] (1)
Where:
[tex]N_{t}=54\%=0.54[/tex] is the quantity of atoms of carbon-14 left after time [tex]t[/tex]
[tex]N_{o}=1[/tex] is the initial quantity of atoms of C-14 in the mammal hide
[tex]\lambda[/tex] is the rate constant for carbon-14 radioactive decay
[tex]t[/tex] is the time elapsed
On the other hand, [tex]\lambda[/tex] has a relation with the half life [tex]h[/tex] of the C-14, which is [tex]5730 years[/tex]:
[tex]\lambda=\frac{ln(2)}{h}=\frac{ln(2)}{5730 years}=1.21(10)^{-4} years^{-1}=0.000121 years^{-1}[/tex] (2)
Substituting (2) in (1):
[tex]0.54=1e^{-(0.000121 years^{-1}) t}[/tex] (3)
Applying natural logarithm on both sides of the equation:
[tex]ln(0.54)=ln(1e^{-(0.000121 years^{-1}) t})[/tex] (4)
[tex]-0.616=-(0.000121 years^{-1}) t[/tex] (5)
Isolating [tex]t[/tex]:
[tex]t=\frac{-0.616}{-0.000121 years^{-1}}[/tex] (6)
[tex]t=54677.68 years \approx 54678 years[/tex] (7) This is the age of the mammal hide
