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

In order to find the number of half-lives that have occurred, we can make use of the given information about the charcoal containing less than \(1/1000\) the normal amount of \(^{14}\mathrm{C}\). We're also given the value of \(2^{10}=1024\), which we can use as an approximation. Let's consider that \(n\) is the number of half-lives elapsed, and we know that at each half-life, the amount of \(^{14}\mathrm{C}\) reduces by a factor of 2. Therefore, after \(n\) half-lives, the remaining amount of \(^{14}\mathrm{C}\) will be: \[ \frac{1}{2^n} \] Since the amount of \(^{14}\mathrm{C}\) is less than \(1/1000\), we can estimate that: \[ \frac{1}{2^n} < \frac{1}{1000} \] Let's use the provided approximation \(2^{10} = 1024\): \[ \frac{1}{2^n} < \frac{1}{2^{10}} \] Which implies that: \[ n > 10 \] This means that more than 10 half-lives have elapsed.

Now that we know there have been more than 10 half-lives, we can multiply this number by the half-life of \(^{14}\mathrm{C}\) (which is 5730 years) to find the minimum age of the charcoal. In this case: Minimum Age = Number of half-lives \(\times\) Half-life of \(^{14}\mathrm{C}\) \[ \text{Minimum Age} > 10 \times 5730 \] Multiplying 10 by 5730 gives us: Minimum Age \(> 57300\) years So, the estimated minimum age of the charcoal is more than 57,300 years.