Question

When charcoal is prepared from a living tree, it shows a disintegration rate of $15.3$ disintegrations of ${}^{14}\text{C}$ per gram per minute. A sample from an ancient piece of charcoal shows ${}^{14}\text{C}$ activity to be $12.3$ disintegrations per gram per minute. How old is this sample? (Half - life of ${}^{14}\text{C}$ is $5730\text{years and}\ln 0.8039=-0.2182)$

• A. $1804\text{years}$
• B. $2309\text{years}$
• C. $546\text{years}$
• D. $1345\text{years}$

Answer 1

The correct option is A $1804\text{years}$

The disintegration per unit mass per unit time remains the same for a given sample.

For a freshly prepared charcoal, if the rate of disintegration decreases from $15.3$ to $12.3$ in time $t$, then the time $t$ will be the age of this ancient piece.

$A_0=15.3{\text{min}}^{-1};A=12.3{\text{min}}^{-1};T_{1/2}=5730\text{year}$

$\Rightarrow \lambda =\frac{0.6931}{T_{1/2}}=\frac{0.6931}{5730}{\text{year}}^{-1}$

We know that, by the law of radioactivity,

$A=A_0{e}^{-\lambda t}$

$\Rightarrow \frac{12.3}{15.3}=e^{-\lambda t}$

$\Rightarrow \ln 0.8039=-\lambda t$

$\Rightarrow -0.2182=-\frac{0.6931}{5730}\times t$

$\Rightarrow t=1804.3\text{years}$.

Hence, option $\left(A\right)$ is correct.