Question

The ratio of carbon-14 to carbon-12 in a piece of charcoal from a fire pit in an archaeological excavation is found to be 12.5% of that in a sample of modern wood. Approximately how old is the site? (Carbon-14 half-life is 5730 years.) Answer as a whole number with no units.

Answer 1

Answer:

17190 years          

Explanation:

The exponential decay equation is:

$N_{t} = N_{0}e^{-\lambda t}$          

$\frac{N_{t}}{N_{0}} = e^{-\lambda t}$

Where:

N(t) is the quantity at time t

N₀ is the initial amount

λ is the decay constant = ln(2)/t(1/2)

t(1/2) is the half-life    

Since the ratio of carbon-14 to carbon-12 is 12.5%, we have that:

$\frac{N_{t}}{N_{0}} = e^{-\lambda t}$

$\frac{0.125N_{0}}{N_{0}} = e^{-\lambda t}$

$ln(0.125) = -\lambda t$

By solving the above equation for t:

$t = \frac{ln(0.125)}{-\lambda} = \frac{ln(0.125)}{-ln(2)/t_{1/2}} = \frac{5730 y* ln(0.125)}{-ln(2)} = 17190 y$

                                               

Therefore, the site is 17190 years old.  

                                                                       

I hope it helps you!