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.

Respuesta :

Answer:

17190 years          

Explanation:

The exponential decay equation is:

[tex] N_{t} = N_{0}e^{-\lambda t} [/tex]          

[tex] \frac{N_{t}}{N_{0}} = e^{-\lambda t} [/tex]

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:

[tex] \frac{N_{t}}{N_{0}} = e^{-\lambda t} [/tex]

[tex] \frac{0.125N_{0}}{N_{0}} = e^{-\lambda t} [/tex]

[tex] ln(0.125) = -\lambda t [/tex]

By solving the above equation for t:

[tex] 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 [/tex]

                                               

Therefore, the site is 17190 years old.  

                                                                       

I hope it helps you!

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