Chapter 32 - Nuclear Physics and Nuclear Radiation - Problems and Conceptual Exercises - Page 1155: 77

$1.7\ Ci$

We know that $\lambda =\frac{\ln 2}{T_{\frac{1}{2}}}=\frac{\ln 2}{4.33\times 10^{-4}y^{-1}}$ $N=\frac{M}{m_{th}}$ $N=(\frac{0.0017Kg}{226.025406u})(\frac{1u}{1.66\times 10^{-24}Kg})$ $N=4.5\times 10^{24}$ Now $R=\lambda N$ We plug in the known values to obtain: $R=4.33\times 10^{-4}y^{-1}(4.5\times 10^{21})(\frac{1\ yr}{3.16\times 10^7s})$ $R=6.2\times 10^{10}Bq(\frac{1\ Ci}{3.7\times 10^7Bq})=1.7\ Ci$