Q-44PE - College Physics (Urone)

Question

An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than \(1/1000\) the normal amount of \({}^{{\rm{14}}}{\rm{C}}\). Estimate the minimum age of the charcoal, noting that \({2^{10}} = 1024\).

Step-by-Step Solution

Answer

The charcoal's minimum age is \(5.71 \times {10^4}\,{\rm{y}}\).

Define Charcoal

Charcoal is a carbon-based porous substance that is black in colour. It's a low-density material. The mechanical strength of charcoal is poor.

Evaluating the minimum age

The equation for radioactive decay is,

\(N = {N_0}{e^{ - \lambda t}}\)

Where \({\rm{\lambda }}\) is determined by,

\(\lambda  = \frac{{In(2)}}{{{t_{1/2}}}}\)

Where \({{\rm{t}}_{{\rm{1/2}}}}\) is the half-life, \({t_{1/2}} = 5730\,{\rm{y}}\) is the half-life for\({}^{{\rm{14}}}{\rm{C}}\) . We may conclude the following from equation,

\(\begin{array}{c}\frac{N}{{{N_0}}} = {e^{ - \lambda t}}\\ - \lambda t = In\left( {\frac{N}{{{N_0}}}} \right)\\t = \frac{{In(N/{N_0})}}{{ - \lambda }}\\ = \frac{{In(N/{N_0})}}{{ - In(2)}}\left( {{t_{1/2}}} \right)\\ = \frac{{In(1/1000)}}{{ - In(2)}}(5730\,{\rm{y}})\\ = 5.71 \times {10^4}\,{\rm{y}}\end{array}\)

Therefore, the charcoal's minimum age is\(5.71 \times {10^4}\,{\rm{y}}\).