The half-life of Au 198 is 3 days. If atomic weight of Au 198 is...

$A = \lambdaN$

$\lambda = \dfrac{\ln 2}{{t}_{1/2}} = \dfrac{\ln 2}{3 \times 24 \times 60 \times 60}\sec^{- 1} = 2 . 67 \times {10}^{- 6}\sec^{- 1}$

$N =$ Number of atoms in $2mgAu$

$= \dfrac{2 \times {10}^{- 3}}{198} \times 6 \times {10}^{23} = 6 . 06 \times {10}^{15}$

$A = \lambdaN = 1 . 618 \times {10}^{13} = 16 . 18 \times {10}^{12}dps$