How old is a skeleton that has lost 27% of its carbon-14? Use the exponential decay model for carbon-14, A=A0e^-0.000121t Round the answer to the nearest whole number

Respuesta :

Answer:

2,601 years

Step-by-step explanation:

Given the decay model for carbon-14

[tex]A=A_0e^{-0.000121t}[/tex]

We want to determine the age of a skeleton that has lost 27% of its C-14.

Initial Value of C-14=100%=1

Present Amount, A(t)=100%-27%=73%=0.73

Substituting these into the model, we have:

[tex]0.73=1e^{-0.000121t}\\$Take the natural logarithm of both sides\\ln(0.73)=-0.000121t\\Divide both sides by -0.000121\\t=ln (0.73) \div (-0.000121)\\t=2600.9 \approx 2601$ years (to the nearest whole number)[/tex]

The skeleton is approximately 2,601 years old.