Answer
See answers.
Work Step by Step
a. The fraction of atoms that are C-14 is very small. Use the atomic weight of C-12 to find the total number of carbon atoms in the sample.
$$N_{C}=\frac{72g}{12g/mol}(6.02\times10^{23}nuclei/mol)=3.612\times10^{24}$$
Now find the number of C-14 nuclei. See the textbook, section 30-11 for the natural fraction of C-14 in carbon.
$$N_{C-14}=\frac{1.3}{10^{12}}(3.612\times10^{24})=4.6956\times10^{12}$$
Now find the time for just one C-14 nucleus to remain, using equation 30-4. The half-life for carbon-14, 5730 years, is found in Appendix B.
$$N=N_0e^{-(ln2)t/(T_{1/2})}$$
$$t=-\frac{T_{1/2}}{ln2}ln\frac{N}{N_0}$$
$$t=-\frac{5730y}{ln2}ln\frac{1}{4.6956\times10^{12}}=2.4\times10^5 y$$
b. Repeat, for an initial mass of 340 g.
$$N_{C}=\frac{340g}{12g/mol}(6.02\times10^{23}nuclei/mol)=1.7057\times10^{25}$$
Now find the number of C-14 nuclei. See the textbook, section 30-11 for the natural fraction of C-14 in carbon.
$$N_{C-14}=\frac{1.3}{10^{12}}(1.7057\times10^{25})=2.2174\times10^{13}$$
Now find the time for just one C-14 nucleus to remain, using equation 30-4. The half-life for carbon-14, 5730 years, is found in Appendix B.
$$N=N_0e^{-(ln2)t/(T_{1/2})}$$
$$t=-\frac{T_{1/2}}{ln2}ln\frac{N}{N_0}$$
$$t=-\frac{5730y}{ln2}ln\frac{1}{2.2174\times10^{13}}=2.5\times10^5 y$$
We see that for times on the order of $10^5$ years, the amount of sample barely changes the determination of the age. Another way to say this is that the activity is very low. C-14 dating is not too effective for ages on the order of $10^5$ years.
