The half life of `._38^90Sr` is 28 years. The disintegration rate of 15 mg of this isotope is of the order of

Explanation

Given, half-life of `""_(38)Sr^(90), t_(1//2)=28` years
`=28 xx 365 xx 24 xx 3600 s`
Decay constant, `lambda = 0.693/t_(1//2)`
Since 90 g of Sr contains `6.023 xx 10^(23)` atoms
`therefore 15 xx 10^(-3) g` of Sr contains
`N = (6.023 xx 10^(23))/90 xx 15 x 10^(-3)` atoms
i.e. `N =(6.023 xx 15 xx 10^(20))/90`
Rate of disintegration,
`(dN)/(dt)=lambdaN`,
`=(0.693 xx 6.023 xx 15 xx 10^(20))/(90 xx 28 xx 365 xx 86,400)` Bq
`=7.877 xx 10^(10)` Bq
`=(7.877)/3.7 xx 10^(10) Ci = 2.13` Ci