Question

"Carbon 14 (C-14), a radioactive isotope of carbon, has a half-life of 5730 ± 40 years. Measuring the amount of this isotope left in the remains of animals and plants is how anthropologists determine the age of samples. Example: The skeletal remains of the so-called Pittsburgh Man, unearthed in Pennsylvania, had lost 82% of the C-14 they originally contained. Determine the approximate age of the bones assuming a half-life of 5730."

Answer 1

Answer:

The age of the bones is approximately 14172 years.

Explanation:

The age of the bones can be determinated using the following decay equation:

$N_{(t)} = N_{0}e^{-\lambda t}$   (1)

Where:

N(t): is the quantity of C-14 at time t

No: is the initial quantity of C-14  

λ: is the decay rate      

t: is the time

First, we need to find λ:

$\lambda = \frac{ln(2)}{t_{1/2}}$

Where:

t(1/2): is the half-life of C-14 = 5730 y

$\lambda = \frac{ln(2)}{5730 y} = 1.21 \cdot 10^{-04} y^{-1}$

Now, we can calculate the age of the bones by solving equation (1) for t:

$t = \frac{-ln(\frac{N_{(t)}}{N_{0}})}{\lambda}$

We know that the bones have lost 82% of the C-14 they originally contained, so:

$N_{t} = (1 - 0.82)N_{0} = 0.18N_{0}$

$t = \frac{-ln(0.18)}{1.21 \cdot 10^{-04} y^{-1}}$

$t = 14172 y$

Therefore, the age of the bones is approximately 14172 years.

I hope it helps you!