Answer
0.2 decays/s.
Work Step by Step
The fraction of atoms that are C-14 is very small. Use the atomic weight of C-12 to find the original total number of carbon atoms in the sample.
$$N_{C}=\frac{1000g}{12g/mol}(6.02\times10^{23}nuclei/mol)=5.017\times10^{25}$$
Now find the number of C-14 nuclei at that time, 60000 years ago. See the textbook, section 30-11.
$$N_{C-14}=\frac{1.3}{10^{12}}(5.017\times10^{25})=6.522\times10^{13}$$
Now find the number of C-14 nuclei now, using equation 30-4. The half-life for carbon-14, 5730 years, is found in Appendix B
$$N=N_0e^{-(ln2)t/(T_{1/2})}$$
$$N=6.522\times10^{13}e^{-(ln2)(60000y)/(5730y)}= 4.594\times10^{10}$$
Calculate the current activity using equation 30–3b.
$$\frac{\Delta N}{\Delta t}=\lambda N=\frac{ln 2}{T_{1/2}}N$$
$$|\frac{dN}{dt}|=\frac{ln 2}{(5730y)(3.156\times10^7s/y)}( 4.594\times10^{10}\;nuclei)$$
$$=0.176 counts/s\approx 0.2 counts/s$$
For samples that are about 60000 years old, the number of counts per second is only about 0.2, which limits the technique.
