Obtain the amount of `.^60Co` necessary to provide a radioactive source of `8.0 Ci` strength. The half-life of `.^60Co` is `5.3` years?

Explanation

The strength of the radioactive source is given as:
`(dN)/(dt)=8.0mCi`
`8xx10^(-3)xx3.7xx10^(10)`
`=29.6xx10^(7)decay//s`
Where,
N = Required number of atoms
Half-life of `_(27)^(60)Co,T_(1/2)=5.3" "years`
`=5.3xx365xx24xx60xx60`
`=1.67xx10^(8)s`
For decay constant `lambda`, we have the rate of decay as:
`(dN)/(dh)=lambdaN`
Where,`lambda=0.693/T_(1/2)=(0.693)/(1.67xx10)s^(-1)`
`therefore N=1/lambda (dn)/(dt)`
`=(29.6xx10^(7))/((0.693)/(1.67xx10^(8)))=7.133xx10^(16) " atoms"`
`"For " _(27)Co^(60):`
`"For " _(27)Co^(60):`
Mass of `6.023xx10^(23)`
(Avogadro’s number) atoms `= 60 g`
`therefore "Mass of "7.133xx10^(16) " atoms "=(60xx7.133xx10^(16))/(6.023xx10^(23))=7.106xx10^(-6)g`
Hence, the amount of `_(27)Co^(60)` necessary for the purpose is `7.106xx10−6 g`.