Question:
The half-life of ${ }^{198} \mathrm{Au}$ is 3 days. If atomic weight of ${ }^{198} \mathrm{Au}$ is $198 \mathrm{~g} / \mathrm{mol}$ then the activity of $2 \mathrm{mg}$ of ${ }^{198} \mathrm{Au}$ is [in disintegration/second] :
Correct Option: , 4
Solution:
$\mathrm{A}=\lambda \mathrm{N}$
$\lambda=\frac{\ln 2}{t_{1 / 2}}=\frac{\ln 2}{3 \times 24 \times 60 \times 60} \mathrm{sec}^{-1}=2.67 \times 10^{-6} \mathrm{sec}^{-1}$
$\mathrm{N}=$ Number of atoms in $2 \mathrm{mg} \mathrm{Au}$
$=\frac{2 \times 10^{-3}}{198} \times 6 \times 10^{23}=6.06 \times 10^{15}$
$A=\lambda N=1.618 \times 10^{13}=16.18 \times 10^{12} \mathrm{dps}$