Answer
18.4 Bq
Work Step by Step
Number of moles n=$\frac{6.00\times10^{-1}\,g}{40\,g/mol}=0.0150\,mol$
Number of particles=$n\times N_{A}$
$=0.0150\times6.022\times10^{23}=9.033\times10^{21}$
Among these, only 0.0117% is radioactive.
Therefore,
$N=9.033\times10^{21}\times0.000117$
$=1.05686\times10^{18}$
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{1.26\times10^{9}\times365\times24\times3600\,s}$
$=1.744\times10^{-17}\,s^{-1}$
Radioactivity=$kN=1.744\times10^{-17}\,s^{-1}\times1.05686\times10^{18}$
$=18.4\,decays/s=18.4\,Bq$
