Question

A fossil was analyzed and determined to have a carbon-14 level that is 80 % that of living organisms. the half-life of c-14 is 5730 years. how old is the fossil? express your answer with the appropriate units.

Answer 1

Answer:

The fossil is 1860 years old.

Step-by-step explanation:

The equation for the amount of fossil has the following format:

$Q(t) = Q(0)e^{rt}$

In which Q(t) is the amount after t years, Q(0) is the initial amount and r is the rate of change.

Half-life of c-14 is 5730 years.

This means that $Q(5730) = 0.5Q(0)$

So

$Q(t) = Q(0)e^{rt}$

$0.5Q(0) = Q(0)e^{5730r}$

$e^{5730r} = 0.5$

$\ln{e^{5730r}} = \ln{0.5}$

$5730r = \ln{0.5}$

$r = \frac{\ln{0.5}}{5730}$

$r = -0.00012$

So

$Q(t) = Q(0)e^{-0.00012t}$

How old is the fossil?

This is t for which

$Q(t) = 0.8Q(0)$

So

$Q(t) = Q(0)e^{-0.00012t}$

$0.8Q(0) = Q(0)e^{-0.00012t}$

$e^{-0.00012t} = 0.8$

$\ln{e^{-0.00012t}} = \ln{0.8}$

$-0.00012t = \ln{0.8}$

$t = -\frac{\ln{0.8}}{-0.00012}$

$t = 1860$

The fossil is 1860 years old.