Obtain the amount of `""_(27)^(60)Co` necessary to provide a radioactive source of 8.0 mCi strength. The half-life of `""_(27)^(60)Co` is 5.3 years.

Explanation

Activity, `A = 8 mCi = 8 xx 10^(-3) Ci = 8 xx 10^(-3) xx 3.7 xx 10^(10) = 2.96 xx 10^(8) s^(-1)`
`T_(1//2) = 5.3` years
`= 5.3 xx 365 xx 24 xx 60 xx 60 = 1.67 xx 10^8 s`
`lambda = (0.693)/(T_(1//2)) = (0.693)/(1.67 xx 10^8) = 4.15 xx 10^(-9) s^(-1)`
`A = lambda N`
`N = A/(lambda) = (2.96 xx 10^8)/(4.15 xx 10^(-9)) = 7.13 xx 10^(16)`
Mass of 60 g of Co atoms = `(60 xx 7.13 xx 10^(16))/(6.02 xx 10^(23)) = 7.1 xx 10^(-6) g`.