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

Here, mass of `._(27)Co^(60) = ?`
Strength of source, `(dN)/(dt) = 8.0` mci
`= 8.0xx3.7xx10^(7)` disint/sec
Half life, T = 5.3 years
`= 5.3xx365xx24xx60xx60` sec
`= 1.67xx10^(8)` sec
`lambda = (0.693)/(T)=(0.693)/(1.67xx10^(8))=4.14xx10^(-9)s^(-1)`
As `(dN)/(dt)=2N`
`therefore N = (dN//dt)/(lambda)=(8xx3.7xx10^(7))/(4.14xx10^(-9))`
`= 7.15xx10^(16)`
By definition of Avogadro's number, Mass of `6.023xx10^(23)` atoms of `._(27)Co^(60)`
`= (60xx7.15xx10^(16))/(6.023xx10^(23))=7.12xx10^(-6)g`