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. `10^11` Bq
B. `10^10` Bq
C. `10^7` Bq
D. `10^9` Bq

Answer

b

Explanation

Here, `T_(1//2)`=28 years=`28xx3.154xx10^7` s
As number of atoms in 90 g of `._38^90Sr=6.023xx10^23`
`therefore ` Number of atoms in 15 mg of `._38^90Sr =(6.023xx10^23)/90xx15/1000`
i.e., `N=1.0038xx10^20`
Rate of disintegration, `=0.693/T N=(0.693xx1.0038xx10^20)/(28xx3.15xx10^7)`
`=7.877xx10^10 "Bq" approx 10^10` Bq