Answer:
The correct option is: 13305 years
Step-by-step explanation:
Formula for radioactive decay is: [tex]N(t)= N_{0}(\frac{1}{2})^\frac{t}{t_{1/2}}[/tex] , where [tex]N_{0}=[/tex] Initial amount, [tex]N(t)=[/tex] Final amount after [tex]t[/tex] years and [tex]t_{1/2}=[/tex] Half-life in years.
Given that, the sample from a fossil has 2 grams of carbon 14 and it had 10 grams of carbon 14 at the moment of its demise.
That means, [tex]N_{0}= 10 grams, N(t)= 2 grams[/tex]
The half-life of carbon 14 is 5730 years. So, [tex]t_{1/2}= 5730[/tex] years.
Now according to the above formula, we will get......
[tex]2= 10(\frac{1}{2})^\frac{t}{5730} \\ \\ \frac{2}{10}=(\frac{1}{2})^\frac{t}{5730} \\ \\ \frac{1}{5}=(\frac{1}{2})^\frac{t}{5730}[/tex]
Taking logarithm on both sides......
[tex]log(\frac{1}{5})=log(\frac{1}{2})^\frac{t}{5730}\\ \\ log(\frac{1}{5})=\frac{t}{5730}log(\frac{1}{2})\\ \\ \frac{t}{5730}=\frac{log(\frac{1}{5})}{log(\frac{1}{2})}\\ \\ t=5730*\frac{log(\frac{1}{5})}{log(\frac{1}{2})}= 13304.647... \approx 13305[/tex]
Thus, the age of the fossil is 13305 years.
