Number of atoms present initially

                 $=\frac{6.023\times {10}^{23}\times 2.2\times {10}^{-3}}{11}$

                 $=1.2\times {10}^{20}\text{atoms}$

No. of atoms present in $5 \mathrm{\mu g}$ of the sample

                 $N=\frac{6.023\times {10}^{23}\times 5\times {10}^{-6}}{11}=2.74\times {10}^{17}\text{atoms}$

Activity of the sample$=\lambda N$

                           $=\frac{0.693}{T_{\frac{1}{2}}}\times N$

                          $=\frac{0.693\times 2.74\times {10}^{17}}{1224}$

                         $=1.55\times {10}^{14}\text{disintegrations/second}$