Answer
a) $3.60\times10^{12}Bq$
b) $3.58\times10^{12}Bq$
c) $1.33\times10^{9}Bq$
Work Step by Step
$\lambda=\frac{\ln(2)}{T_{0.5}}$. The half life of $^{131}_{53}I$ is 8.0252 days, so the decay constant is: $\lambda=9.997\times10^{-7}s^{-1}$
The number of nuclei is: $N_0=782\times10^{-6}g\times\frac{1 mol}{130.906g}\times6.02\times10^{23}\frac{atoms}{mol}$
$=3.5962\times10^{18}$ nuclei
a) $\frac{\Delta N}{\Delta t}=\lambda N_0l^{-\lambda t}=(9.997\times10^{-7}s^{-1})(3.5962\times10^{18} nuclei)e^{0}=3.60\times10^{12}Bq$
b) $\frac{\Delta N}{\Delta t}=\lambda N_0l^{-\lambda t}=(9.997\times10^{-7}s^{-1})(3.5962\times10^{18} nuclei)e^{1.50h\times\frac{3600s}{1h}}=3.58\times10^{12}Bq$
c) $\frac{\Delta N}{\Delta t}=\lambda N_0l^{-\lambda t}=(9.997\times10^{-7}s^{-1})(3.5962\times10^{18} nuclei)e^{3.0m\times30\frac{d}{m}\frac{24h}{d}\frac{3600s}{h}}=1.33\times10^{9}Bq$
