The half life of `._38Sr^(90)` is 28 years. What is disintegration rate of 15g of this isotope?

Explanation

Half life of `_(38)^(90)Sr,t_(1/2)=28" years "`
`=28xx365xx24xx60xx60`
`=8.83xx10^(8)s`
Mass of the isotope, `m = 15 mg`
90g of `_(38)^(90)Sr` contains:
`(6.023xx10^(23)xx15xx10^(-3))/(90),i.e,1.0038xx10^(20)" number of atoms "`
Rate of disintegration, `(dN)/(dt)=lambdaN`
Where,
` lambda`=decay constant `=(0.693)/(8.83xx10^(8))s^(-1)`
`therefore (dN)/(dt)=(0.693xx1.0038xx10^(20))/(8.83xx10^(8))=7.878xx10^(10)" atoms/s "`
Hence, the disintegration rate of 15 mg of the given isotope is `7.878xx10^(10)" atoms/s "`