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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

Verified by Experts

`R= lambda N`
`N=1.004 xx 10^(20) `
`lambda = (0.693) /(T)`
`lambda = 7.848 xx 10^(-10) s^(-1)`
`R=7.879 xx 10^(10) Bq`
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The half life of a radioactive sample ""_(38) Sr ^(30) is 28 years . Calculate the rate of disintegration of 15 mg of this isotope . Given Avogadro number = 6.023 x 10^(23) .

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Knowledge Check

  • The half life period of a particular isotope is 10 years. Its decay constant is:

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