Answer
(a) $1.316\times 10^3decay/s$
(b) by a factor of 2.
Work Step by Step
(a) We can find the required activity as follows:
Number of atoms of thorium $N=\frac{mass\space of \space thorium}{mass\space of\space thorium\space atom}$
$N=\frac{\frac{0.325g}{atomic\space mass}}{Avogadro\space no}$
$N=\frac{\frac{0.325g}{232.04g}}{6.022\times 10^{23}/mol}$
$N=8.43\times 10^{20}atoms$
Now the activity is given as
$R=(\frac{\ln 2}{t_{\frac{1}{2}}})N$
$R=(\frac{\ln 2}{1.405\times 10^{10}y})(8.43\times 10^{20}atoms)(\frac{1y}{3.16\times 10^7s})$
$R=1.316\times 10^3decay/s$
(b) Since the activity is inversely proportional to the half life, hence if the half life of thorium had been doubled then the activity would be decreased by a factor of 2.
