Plutonium has atomic mass 210 and a decay constant equal to $5.8 \mathrm{x} {10}^{-8} {\sec }^{-1}.$ The number of $\mathrm{\alpha }$-particles emitted per second by 1 mg is:
(Avogadro’s constant = $6.0 \mathrm{x} {10}^{23}$)

Number of  $\mathrm{\alpha }$-particles per second = activity A 

$(-\mathrm{dN}/\mathrm{dt})=\mathrm{N\lambda }$ is required, where

$$\begin{gathered} \mathrm{N}=\frac{6.0 \mathrm{x} {10}^{23}}{210} \mathrm{x} 1 \mathrm{x} {10}^{-3} \\ \mathrm{And} \mathrm{\lambda }=5.8 \mathrm{x} {10}^{-8} \mathrm{per} \sec \end{gathered}$$

$$\begin{gathered} \mathrm{So}, \mathrm{A}= \mathrm{N\lambda } \\ =\frac{6.0 \mathrm{x} {10}^{23}}{210} \mathrm{x} 1 \mathrm{x} {10}^{-3} \mathrm{x} 5.8 \mathrm{x} {10}^{-8} \\ =1.7 \mathrm{x} {10}^{11}. \end{gathered}$$