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Obtain the amount of " "(27)^(60)Co nece...

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.

Text Solution

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Here activity R= 8.0 mCi =`8.0 xx 10^(-3) xx 3.7 xx 10^(10)` Bq=`2.96 xx 10^(8)`Bq and disintegration constant
`lambda=0.6931/T_(1/2)=0.6931/(5.3xx3.154xx10^(7))s^(-1)`
From the relation R=`lambda`N, we find `N=R/lambda = (2.96xx10^(8) xx5.3xx3.154 xx 10^(7))/0.6931 =7.14 xx 10^(16)` atoms
`therefore` Amount of `" "_(27)^(60)Co approx =(Nxx60)/(6.023xx10^(23))g =(60xx7.14xx10^(16))/(6.023xx10^(23))g = 7.126 xx 10^(-6)`g
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