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

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#### Solution

Half life of `""_38^90"Sr", "t"_(1/2)`= 28 years

= 28 × 365 × 24 × 60 × 60

= 8.83 × 10^{8} s

Mass of the isotope, m = 15 mg

90 g of `""_38^90"Sr"` atom contains 6.023 × 10^{23 }(Avogadro’s number) atoms.

Therefore, 15 mg of `""_38^90"Sr"` contains:

`(6.023 xx 10^23 xx 15 xx 10^(-3))/90 "i.e." 1.0038 xx 10^20` number of atoms

Rate of disintegration, `"dN"/"dt" = lambda"N"`

Where,

λ = Decay constant = `0.693/(8.83 xx 10^8) "s"^(-1)`

`= "dN"/"dt" = (0.693 xx 1.0038 xx 10^20)/(8.83 xx 10^8) = 7.878 xx 10^10 "atoms/s"`

Hence, the disintegration rate of 15 mg of the given isotope is 7.878 × 10^{10} atoms/s.

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