Explanation

(c) Decay rate $R = \lambda N$
$R = R_{0} = \ \lambda N_{0}$
$\lambda = (\ln{2)}/T_{1/2} = (\ln{2)}/(78\ h)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 8.89\ \times \ 10^{- 3}\ h^{- 1}\ then\ N_{0}\ = \ M/m$
Now $m = (67u)(1.661\ \times \ 10^{- 24}\ g/u)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ 1.113\ \times \ 10^{- 22}\ g$
and $\ \ \ \ \ \ \ \ N_{0}\ = (3.4\ g)/(1.113\ \times \ 10^{- 22}\ g)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3.05\ \times \ 10^{22}$
$Thus,\ \ \ \ {\ \ R}_{0} = \ (8.89\ \times 10^{- 3}\ h^{- 1})(3.05\ \times \ 10^{22})$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.71\ \times \ 10^{20}\ h^{- 1} = \ 7.53\ \times \ 10^{16}s^{- 1}$