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The Half-life of 9038sr3890sr is 28 Years. What is the Disintegration Rate of 15 Mg of this Isotope? - Physics

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 `""_38^90Sr`, `t_(1/2)`= 28 years

= 28 × 365 × 24 × 60 × 60

= 8.83 × 108 s

Mass of the isotope, m = 15 mg

90 g of `""_38^90Sr` atom contains 6.023 × 1023 (Avogadro’s number) atoms.

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

`(6.023 xx 10^23 xx 15 xx 10^(-3))/90 ie 1.0038 xx 10^20` number of atoms

Rate of disintegration, `"dN"/dt = lambdaN`

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 × 1010 atoms/s.

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APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 13 Nuclei
Q 10 | Page 463
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