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The half life of ""(38) Sr^(90) isotope ...

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) )`

Text Solution

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Given : T = 28 yrs
w.k.t : ` 1 = lambda = (0.693)/(T)`
i.e., ` lambda = (0.693)/(28) = 0.024 75 yr^(-1)`
but activity of the sample ` A = (dN)/(dt) = lambda N `
` 90g " of " sr^(50) " has 6.023 xx 10^(23) ` atoms
Hence ` 15 xx 10^(-3) g " of " sr ^(50) ` hr .N. atoms .
` therefore N = (15xx 10^(-3) xx 6.023 xx 10^(23))/(90) `
i.e., ` N = 1.00 xx 10^(20) `
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