Question

An archeologist finds a skull in Piltdown Park in England. He discovers that only 3% of the original amount of carbon-14 is still present in the skull. If the decay constant for carbon-14 is 11×10−5 , estimate the age of the skull.

Answer 1

Answer:

$3.19\times10^4\text{ seconds}$ or 8.86 hours

Step-by-step explanation:

The quantity of material left after a time, t, during a radioactive decay is given by

$N = N_0 e^{-\lambda t}$

$N_0$ is the initial quantity, $\lambda$ is the decay constant

$\dfrac{N}{N_0} = e^{-\lambda t}$

$\dfrac{N}{N_0}$ is the fraction left.

Substituting given values from the question,

$0.03 = e^{-\lambda t}$

$\ln 0.03 =-\lambda t$

$t = \dfrac{\ln 0.03}{-\lambda}$

$t = \dfrac{-3.507}{-11\times10^{-5}} = 3.19\times10^4\text{ seconds}$

This is about 8.86 hours