Given :
The half life of Carbon-14 , [tex]t_{0.5}=5730[/tex] .
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 , [tex]\dfrac{C}{2408}[/tex] .
Rate constant is :
[tex]k=\dfrac{0.693}{t_{0.5}}\\\\k=\dfrac{0.693}{5730}\\\\k=1.2\times 10^{-4}\ years^{-1}[/tex]
By first order equation :
[tex]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[/tex]
Therefore , the rock is 64887.9 years old .
Hence , this is the required solution .
