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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.

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#### Solution

The strength of the radioactive source is given as:

`"dN"/"dt"` = 8.0 mCi

= 8 × 10^{−3} × 3.7 × 10^{10}

= 29.6 × 10^{−7} decay/s

N = Required number of atoms

Half-life of `""_27^60"Co"`, `"T"_(1/2)` = 5.3 years

= 5.3 × 365 × 24 × 60 × 60

= 1.67 × 10^{8} s

For decay constant λ, we have the rate of decay as:

`"dN"/"dt" = lambda"N"`

Where, `lambda = 0.693/"T"_(1/2) = 0.693/(1.67 xx 10^8) "s"^(-1)`

∴ N = `1/lambda ("dN")/"dt"`

= `(29.6 xx 10^7)/(0.639/(1.67 xx 10^8))`

= 7.133 × 10^{16} atom

For `""_27^60"Co"`

Mass of 6.023 × 10^{23 }(Avogadro’s number) atoms = 60 g

∴ Mass of 7.133 × 10^{16} atom

= `(60 xx 7.133 xx 10^16)/(6.023 xx 10^23)`

= 7.106 × 10^{−6} g

Hence, the amount of For `""_27^60"Co"` necessary for the purpose is 7.106 × 10^{−6} g.

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