Question

An archaeologist analyses the wood in a prehistoric structure and finds that $C^{14}$ (Half life $=5700years$) to $C^{123}$ is only one-fourth of that found in the cells of buried plants. The age of the wood is about:

• A. $5700years$
• B. $2850years$
• C. $11,400years$
• D. $22,800years$

Answer 1

Answer: (C)

$11,400years$

The law of radioactive decay predicts how the number of the not decayed nuclei of a given radioactive substance decreases in the course of time.

Relation is $N=N_0{\left(1/2\right)}^{\frac{t}{\tau }}$

$N$ is undecayed part

$N_0$ is initial part

$\tau$ is half-life and t is time taken.

$N_0=C^{123}$

$N=C^{14}$

$N=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{\tau }}$

$C^{14}=C^{123}{\left(\frac{1}{2}\right)}^{\frac{t}{\tau }}\Rightarrow \frac{C^{14}}{C^{123}}={\left(\frac{1}{2}\right)}^{\frac{t}{\tau }}\Rightarrow \frac{1}{4}={\left(\frac{1}{2}\right)}^{\frac{t}{\tau }}\Rightarrow {\left(\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^{\frac{t}{\tau }}\Rightarrow 2=\frac{t}{\tau }=\frac{t}{5700}\Rightarrow t=5700\times 2=11400$

Age of $C^{14}$ is 11,400yrs