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प्रश्न
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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उत्तर
The strength of the radioactive source is given as:
`"dN"/"dt"` = 8.0 mCi
= 8 × 10−3 × 3.7 × 1010
= 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 × 108 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 × 1016 atom
For `""_27^60"Co"`
Mass of 6.023 × 1023 (Avogadro’s number) atoms = 60 g
∴ Mass of 7.133 × 1016 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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