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

Options

A. `7.88xx10^(10)Bq`
B. `8.88xx10^(10)Bq`
C. `9.88xx10^(10)Bq`
D. `6.88xx10^(10)Bq`

Answer

A

Explanation

Here, `T_(1//2)="28 years"=28xx3.154xx10^(7)s`
Number of atoms in `"90 g of "_(38)Sr^(90)=6.023xx10^(23)`
`therefore" Number of atoms in 15 mg of "_(38)Sr^(90)`
`N=(6.023xx106(23))/(90)xx(15)/(1000), or N=1.0038xx10^(20)`
`"Rate of disintegration, "|(dN)/(dt)|=lambdaN`
`=(0.693)/(T_(1//2))N=(0.693xx1.0038xx10^(20))/(28xx3.154xx10^(7))=7.88xx10^(10)Bq`