The half-life of `._(38)^(90)Sr` is 28 years. What is the disintegration rate of 15 mg of this isotope ?

Explanation

Here T = 28 years `= 28xx3.154xx10^(7)s`
As number of atoms in 90 g of `._(38)Sr^(90)`
`= 6.023xx10^(23)`
`therefore` Number of atoms in 15 mg of `._(38)Sr^(90)`
`= (6.023xx10^(23))/(90)xx(15)/(1000)`
i.e., `N=1.0038xx10^(20)`
Rate of disintegration `(dN)/(dt)=lambda N`
`= (0.693)/(T)N`
`= (0.693xx1.0038xx10^(20))/(28xx3.154xx10^(7))`
`= 7.87xx10^(10)Bq`