Answer
$^{211}_{82}Pb$.
Work Step by Step
The original activity is $R_0$ and the activity after 4.00 hours is $ (0.01050)R_0$.
Use equation 30-5.
$$ (0.01050)R_0=R_0e^{-\lambda t}$$
$$ ln 0.01050= -\lambda t=-\frac{ln2}{T_{1/2}}t$$
Solve for the half-life.
$$ T_{1/2}= -\frac{ln2}{ ln 0.01050}(4.00h)=0.6085h$$
This is about 36.5 minutes. Appendix B shows us that $^{211}_{82}Pb$ has a half-life of 36.1 minutes, so that is the radioactive isotope of lead in the sample that is being produced. The lead itself is further decaying to Bi-211 and a less-excited version of Pb-211.
