Question

The half life of Carbon-14 is 5730 years. You measure a rock sample and find that it has trace amounts of Carbon-14. 1/2408 remains. How old is the rock?

Answer 1

Given :

The half life of Carbon-14 , $t_{0.5}=5730$ .

Trace amounts of Carbon-14 1/2408 remains.

To Find :

How old is the rock .

Solution :

Let , initial concentration of Carbon-14 is C .

Quantity remains , $\dfrac{C}{2408}$ .

Rate constant is :

$k=\dfrac{0.693}{t_{0.5}}\\\\k=\dfrac{0.693}{5730}\\\\k=1.2\times 10^{-4}\ years^{-1}$

By first order equation :

$kt=-ln(\dfrac{[A_t]}{[A_o]})\\\\t=-\dfrac{ln(\dfrac{[A_t]}{[A_o]})}{k}\\\\t=-\dfrac{ln(\dfrac{[A_o]}{[A_o]\times 2408})}{1.2\times 10^{-4}}\ years\\\\t=-\dfrac{ln(\dfrac{1}{2408})}{1.2\times 10^{-4}}\ years\\\\t=64887.9\ years$

Therefore , the rock is 64887.9 years old .

Hence , this is the required solution .