Question

An archaeologist analysis of the wood in a prehistoric structure reveals that the ratio of ${}^{14}C$to${}^{12}C$ is only one-eighth in the cells in living plants. What is the age of the wood? Given: Half life of ${}^{14}C$ is 5700 years.

Answer 1

$\underset{–}{Given:}$

Half life of ${}^{14}C$= 5700 year

Therefore, decay constant, $\lambda$= $\frac{0.693}{t_{1/2}}$

= $\frac{0.693}{5700year}$$\Rightarrow \lambda$=1.21$\times$10${}^{-4}$year${}^{-1}$

We know that, at any instant of time `t', the number of nuclei undecayed is N${}_t$ = $N_0{e}^{-\lambda t}$

where N${}_0$$\to$ No. of nuclei at t = 0

And N${}_t$$\to$ No: of nuclei at t = t

Substituting the values in

$\frac{N_t}{N_0}$= $e^{-\lambda t}$, we get,

$\frac{\left(\frac{1}{8}{N}_0\right)}{N_0}$=$e^{-\lambda t}\Rightarrow e^{\lambda t}$ = 8

(or) $\lambda$t = ln(8)

(or) t = $\frac{ln\left(8\right)}{\lambda }$ = $\frac{2.079}{1.21\times {10}^{-4}}$ =1.71$\times$10${}^4$ year