A cobalt-60 source for cancer treatment has an activity of 3.7 × 10¹³ Bq, which is 1000 Ci…

Step 1. The decay constant. $T = 5.27 \times 3.156 \times 10^{7}\ \text{s} = 1.663 \times 10^{8}$ s, so $\lambda = \dfrac{0.693}{1.663 \times 10^{8}\ \text{s}} = 4.17 \times 10^{-9}\ \text{s}^{-1}$.

Step 2. The number of atoms. $N = \dfrac{R}{\lambda} = \dfrac{3.7 \times 10^{13}\ \text{s}^{-1}}{4.17 \times 10^{-9}\ \text{s}^{-1}} = 8.88 \times 10^{21}$.

Step 3. The mass. The number of moles is $\dfrac{8.88 \times 10^{21}}{6.022 \times 10^{23}\ \text{mol}^{-1}} = 0.01474$ mol. The mass is $0.01474\ \text{mol} \times 60\ \text{g/mol} = 0.885$ g.