Answer
a) $4.2\times10^{16}$ nuclei
b) $1.3\times10^{15}$ nuclei
c) $2.8\times10^{13}Bq$
d) $t=26$min
Work Step by Step
$\lambda=\frac{\ln(2)}{T_{0.5}}=0.0225s^{-1}$
a) $8.7\times10^{-6}g\times\frac{1 mol}{124g}\times6.02\times10^{23} \frac{atoms}{mol}=4.2\times10^{16}$ nuclei
b) $N=N_0e^{-\lambda t}=(4.2237\times10^{16})e^{-0.0225s^{-1}\times\frac{60s}{min}}$
$=1.3\times10^{15}$ nuclei
c) $\frac{\Delta N}{\Delta t}=\lambda N=(0.0225s^{-1})(1.262\times10^{15})=2.8\times10^{13}Bq$
d) $\frac{\Delta N}{\Delta t}=(\frac{\Delta N}{\Delta t})_oe^{-\lambda t}$
$-\lambda t=\ln(\frac{\frac{\Delta N}{\Delta t}}{\lambda N_o})$
$t=26$min
