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Obtain the amount of .^60Co necessary to...

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?

Text Solution

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Here, mass of `._27Co^(60)=?`
Strength of source, `(dN)/(dt) =8.0mci =8.0xx3.7xx10^(7)"disint".//sec`
Half life, `T=5.3years =5.3xx365xx24xx60xx60sec=1.67xx10^(8)sec`
`lambda=0.693/T=0.693/(1.67xx10^8)=4.14xx10^(-9)s^-1`.
As `(dN)/(dt)=lambdaN :. N=(dN//dt)/lambda=(8xx3.7xx10^7)/(4.14xx10^(-9))=7.15xx10^(16)`
By defination of Avogadro's number, Mass of `6.023xx10^(23)` atom of `._27C0^(60)=60g`
Mass of `7.15xx10^(16)` atoms of `._27C0^(60)=(60xx7.15xx10^(16))/(6.023xx10^(23))=7.12xx10^(-6)g`
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