Here rate of disintegration requiredR = 8.0 mCi = 8.0 × $10^{-3}$ × 3.7 × $10^{10}\,\text{diss}^{-1}$ = 29.6 × $10^{7}\,\text{diss}^{-1}$ Half life $T_{1/2}$ = 5.3 years = 5.3 × 3.16 × $10^{7}$ s But R = λN = $\frac{0.693}{T_{1/2}}$ . N No. of atoms for given rate required, N = $\frac{RT_{1/2}}{0.693}$ = = 7.15 × $10^{16}$ atoms As 1 mole i.e., 60 g of cobalt contains 6.023 × $10^{23}$ atoms, so, the mass of cobalt required for given rate of disintegration = $\frac{60\times7.15\times10^{16}}{6.023\times10^{23}}$ = 7.123 × $10^{-6}$ g