Calculation of true thickness of the bed

A calcareous fossil is found to have ¼ of its original 14c activity( Half life of 14c is 5730 years), its age is therefore

A) 5730 years

B) 1432.5 years

C) 11460 years

D) 22920 years

Solution:

Given that N = No/4

Half-life ($t_{\frac{1}{2}}$) = 5730 years

Age of the fossil (t) =?

Formula

$\frac{N}{No}= \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Substitute the given values in the above equation

$\frac{N}{No}= \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

$\frac{No}{4No}= \frac{1}{2}^{\frac{t}{5730}}$

$\frac{1}{4} = \frac{1}{2}^{\frac{t}{5730}}$

$\log(\frac{1}{4}) = \log(\frac{1}{2})^{\frac{t}{5730}}$

$\frac{t}{5730}=\frac{\log(\frac{1}{4})}{\log(\frac{1}{2})}$

$\frac{t}{5730}=2$

$t= 5730*2$

$t = 11460 years$