The half life of `""_(38) Sr^(90)` isotope is 28 years. What is the rate of disintegration of 15 mg of this isotope? (Given Avogadro No `=6.023 xx 10^(23) )`

Explanation

Given : T=28 yrs
w.k.t : `I=lambda = 0.693/T`
i.e., `lambda=0.693/28=0.02475 yr^(-1)`
but activity of the sample
`A=(dN)/(dt)=lambdaN`
90 g of `Sr^90` has `6.023xx10^(23)` atoms
Hence , `15xx10^(-3)g` of `Sr^90` hr .N. atoms.
`therefore N=(15xx10^(-3) xx6.023xx10^23)/90`
i.e., `N=1.00xx10^20`
Hence , `A=(dN)/(dt)=IN` may be obtained
i.e. , `A=0.02475xx1xx10^20`
`=2.475xx18` dis. `yr^(-1)`
but 1 yr = 365 x 86400 =`3.15xx10^7` s
Hence, `A=(2.475xx10^18)/(3.15xx10^7)`
`=7.86xx10^10` dis `s^(-1)`
but 1 curie = `3.7xx10^10` dis `s^(-1)`
So `A=(7.86xx10^10)/(3.7xx10^10)`=2.12 Ci
Activity of the sample = 2.12 Cu.