Answer
a) $^{218}_{84}Po$
b) $\alpha$ decay with a half life of 3.1min and a bit of $\beta$ decay
c) Chemically reactive
d) Initially, $8\times10^6 Bq$. After one month, $3.2×10^4Bq$
Work Step by Step
a) $^{222}_{86}Rn\rightarrow^{218}_{84}Po+^{4}_{2}He$
b) As shown on Fig 30-11, it is $\alpha$ decay with a half life of 3.1min and a bit of $\beta$ decay
c) Because it is in the same group as Oxygen, it will be chemically reactive
d) $\frac{\Delta N}{\Delta t}=\lambda N$
$N=1.4\times10^{-9}g\times\frac{1 atom}{222.02u}\times\frac{1 u}{1.6605\times10^{-24}g}=3.7975\times10^{12}$atoms
$\frac{\Delta N}{\Delta t}=(\frac{0.693}{3.8235d\times\frac{24h}{d}\times\frac{3600s}{h}})(3.7975\times10^{-12}atoms)=8\times10^6Bq$
$\frac{\Delta N}{\Delta t}=(\frac{\Delta N}{\Delta t})_o(\frac{1}{2})^{\frac{t}{T_{0.5}}}=3.2\times10^4Bq$