Answer
$m = 1.023 \times 10^{-3} g = 1.023 \space mg$
Work Step by Step
The equation of radioactive decay is given by $R = \lambda N$ and $\lambda = \frac{ln2}{T_{1/2}}$
Rearrange the equation to solve for number of nuclei, N,
$N = \frac{R}{\lambda} = \frac{R}{ln2/T_{1/2}} = \frac{R (T_{1/2} )}{ln 2}$
From the question,
$T_{1/2} = 2.70 d \times 86400 s/d = 233280 s $
$ R = 250 Ci \times 3.70\times 10^{10} Bq/Ci = 9.25 \times 10^{12} Bq$
So
$N = \frac{9.25 \times 10^{12} Bq (233280 s )}{0.693}$
$N = 3.11 \times 10^{18} $
The mass of ^{198}Au is
$m_{Au} = (198 u)(1.661 \times 10^{-24} g/u)$
$m_{Au} = 3.29 \times 10^{-22} g$
So the mass required will be the mass of a single gold atom multiplied by number of gold nuclei
$m = Nm_{Au}$
$m = (3.11 \times 10^{18})(3.29 \times 10^{-22} g)$
$m = 1.023 \times 10^{-3} g = 1.023 \space mg$
