Question

An archeologist analyses the wood in a prehistoric structure and finds that ${\text{C}}^{14}$ $(\text{Half life}=5700\text{years})$ to ${\text{C}}^{12}$ is only one-fourth of that in the cells of living plants. The age of the wood is about

• A. $5700\text{years}$
• B. $11,400\text{years}$
• C. $2850\text{years}$
• D. $22,800\text{years}$

Answer 1

The correct option is B $11,400\text{years}$

${}^{14}\text{C}$ is a highly unstable carbon isotope.

The plants keep absorbing ${}^{14}\text{C}$ from the atmosphere in the form of carbon dioxide and decays into ${}^{12}\text{C}$.

The absorption of ${}^{14}\text{C}$ stops when the plant dies and the disintegration rate keeps decreasing.

The ratio of ${}^{14}\text{C}$ to ${}^{12}\text{C}$ in a sample describes the disintegration rate.

The activity of the wood at the moment it died will be the same as that of the living plant,

If $A_0$ is the activity of the living plant, and $A$ is the activity of the wood now,

Given that, $\frac{A}{A_0}=\frac{1}{4}$

Also, we know that,

$\frac{A}{A_0}=\frac{1}{2^n}$

$\Rightarrow \frac{1}{4}=\frac{1}{2^n}$

where, $n=\frac{t}{t_{1/2}}$

Here, $n=2$

$\Rightarrow t=2\times 5700=11400\text{years}$

Hence, $\left(C\right)$ is the correct answer.