Question

A fossil is found to have a 14 C 14C level of 70.0 70.0 % compared to living organisms. How old is the fossil?

Answer 1

Answer:

The age of the fossil be $2.1987\times 10^{4} years$.

Explanation:

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

where,

$N_o$ = initial mass of isotope  C-14 = x

N = mass of the parent isotope left after the time, (t)  = 70.0% of x=0.07x

$t_{\frac{1}{2}}$ = half life of the isotope  C-14 = 5730 years

$\lambda$ = rate constant

Let the age of the fossil be t.

Now put all the given values in this formula, we get  t :

$0.07x=x\times e^{-(\frac{0.693}{5730 years})\times t}$

$t=2.1987\times 10^{4} years$

The age of the fossil be $2.1987\times 10^{4} years$.