Assume a constant \(1^{14} \mathrm{C} /^{12} \mathrm{C}\) ratio of 13.6 counts per minute per gram of living matter. A sample of a petrified tree was found to give 1.2 counts per minute per gram. How old is the tree? (For $^{14} \mathrm{C}, t_{1 / 2}=5730$ years.)

To find the decay constant, we will use the half-life formula: \[t_{1/2} = \frac{0.693}{\lambda}\] where \(t_{1/2}\) is the half-life and \(\lambda\) is the decay constant. We are given the half-life value for \(^{14}C\), which is 5730 years. We can find the decay constant as follows: \[\lambda = \frac{0.693}{5730}\] Now let's compute the decay constant: \[\lambda \approx 1.21 \times 10^{-4}\, \mathrm{year}^{-1}\]

In the problem, we are given that the \(^{14}C/^{12}C\) ratio in living matter is 13.6 counts per minute per gram, and that the ratio in the tree sample is 1.2 counts per minute per gram. We need to determine the ratio of the sample to that of living matter: \[\frac{\mathrm{ratio_{sample}}}{\mathrm{ratio_{living}}} = \frac{1.2}{13.6}\] Now let's calculate the ratio: \[\frac{\mathrm{ratio_{sample}}}{\mathrm{ratio_{living}}} \approx 0.0882\]

Now we will use the decay formula: \[N_t = N_0 \times e^{-\lambda t}\] where \(N_t\) is the number of radioactive isotopes at time \(t\), \(N_0\) is the initial number of radioactive isotopes, \(e\) is the base of the natural logarithm, \(\lambda\) is the decay constant, and \(t\) is the time elapsed (in years) since the tree died. We found the decay constant \(\lambda\) in step 1 and the ratio of \(^{14}C\) in the sample to that in living matter in step 2. We can use this information to rewrite the decay formula in terms of the ratios: \[\frac{\mathrm{ratio_{sample}}}{\mathrm{ratio_{living}}} = e^{-\lambda t}\] Rearrange the formula to find \(t\): \[t = - \frac{\ln (\frac{\mathrm{ratio_{sample}}}{\mathrm{ratio_{living}}})}{\lambda}\] Now let's substitute the values we found in steps 1 and 2 to get the age of the tree: \[t = -\frac{\ln(0.0882)}{1.21 \times 10^{-4}\, \mathrm{year}^{-1}}\] Finally, calculate the age of the tree: \[t \approx 18,350\, \mathrm{years}\] So, the petrified tree is approximately 18,350 years old.