Answer
See answers.
Work Step by Step
Sr is a Group II element in the Periodic Table, just below calcium (Ca). Ingesting Sr is probably a bad idea because if the body treats it like Ca, the Sr might end up in one’s bones.
Use equation 30–4.
$$N=N_0e^{-\lambda t}$$
Now apply equation 30-6, and the given information that the final number of Sr atoms is only 0.01 times the number at the start.
$$N=N_0e^{-(ln2)t/(T_{1/2})}=0.01N_0$$
$$0.01=e^{-(ln2)t/(T_{1/2})}$$
Solve for t.
$$t=-\frac{ T_{1/2}ln(0.01)}{ln 2}=-\frac{(29y)ln(0.01)}{ln2}=1.9\times10^2y $$
Here is the decay reaction. If they have too many neutrons, assume Sr and its daughter decay by emitting beta particles.
$$^{90}_{38}Sr \rightarrow\;^{90}_{39}Y +\;^{0}_{-1}\beta +\overline{\nu}$$
$$^{90}_{39}Y \rightarrow\;^{90}_{40}Zr +\;^{0}_{-1}\beta +\overline{\nu}$$
