Determine the amount of  ${ }_{84}P{o}^{210}$  necessary to provide a source of $\alpha$ particles of 10 milli-curie strength if mean life of  ${ }_{84}P{o}^{210}$ is $1.72\times {10}^7 sec$

$\begin{array}{rcl}Activity. A& =& \left|\frac{dN}{dt}\right|=\lambda N\\ Given : A & =& 10 milli curie = 10\times {10}^{-3}\times 3.7\times {10}^{10} dps\\ \lambda & =& \frac{1}{T_m}=\frac{1}{1.72\times {10}^7}se{c}^{-1}\\ we know that number of atoms & =& N=\frac{mass in grams xavagadro number}{molar mass}\\ No of atoms & =& N=\frac{A}{\lambda }=10\times {10}^{-3}\times 3.7\times {10}^{10}\times 1.72\times {10}^7 \\ & \Rightarrow & N=6.36\times {10}^{15}\\ & & \\ Mass of all these atoms & =& moles\times molar mass = \frac{N}{N_A}\times M\\ & \Rightarrow & Mass =\frac{210}{6\times {10}^{23}}\times 6.36\times {10}^{15}=2.2\times {10}^{-6}gm\\ & & \end{array}$