Answer
a) $6.42\times10^{-5}\,s^{-1}$
b) $3.86\times10^{19}\,decays/s$
Work Step by Step
a) Half-life $t_{1/2}=3.00\,h=3.00\,h\times\frac{3600\,s}{1\,h}=10800\,s$
Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{10800\,s}=6.42\times10^{-5}\,s^{-1}$
b) $k=6.416667\times10^{-5}\,s^{-1}$
$\text{Number of particles N}=\text{Number of moles}\times N_{A}$
$=1.000\times6.022\times10^{23}=6.022\times10^{23}$
Decay rate= $kN=6.416667\times10^{-5}\,s^{-1}\times6.022\times10^{23}$
$=3.86\times10^{19}\,decays/s$
