Question

When charcoal is prepared from a living tree, it shows a disintegration rate of 15.3 disintegrations of ¹⁴C per gram per minute. A sample from an ancient piece of charcoal shows ¹⁴C activity to be 12.3 disintegrations per gram per minute. How old is this sample? Half-life of ¹⁴C is 5730 y.

Answer 1

Given:
Initial activity of charcoal, A₀ = 15.3 disintegrations per gram per minute
Half-life of charcoal, $T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$= 5730 years
Final activity of charcoal after a few years, A = 12.3 disintegrations per gram per minute

Disintegration constant, $\lambda =\frac{0.693}{T_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$ = $\frac{0.693}{5370}{\mathrm{y}}^{-1}$

Let the sample take a time of t years for the activity to reach 12.3 disintegrations per gram per minute.

Activity of the sample, $A=A_0{e}^{-\lambda t}$
$$\begin{gathered} A=A_0{e}^{-\frac{0.693}{5730}\times t} \\ \Rightarrow \mathrm{In}\frac{12.3}{15.3}=\frac{-0.693}{5730}t \\ \Rightarrow 0.218253=\frac{0.693}{5730}\times t \\ \Rightarrow t=1804.3\mathrm{years} \end{gathered}$$