How many `beta`-particles are emitted during one hour by `1.0 mu g` of `Na^(24)` radionuclide whose half-life is `15` hours? [Take `e^((-0.693//15))=0.955=0.9555`, and avagadro number `=6xx10^(23)`]

Answer

`(6xx10^(23)xx10^(-6))/(24)[1-e^(-0.693//15)]=1.128xx10^(15)`

Explanation

No. of particle in time `t` is
`N=N_(0)(1-e^(-lambdat))=N_(0)(1-e-t(ln2)/(T_(1//2)))`
`N_(0) =` No. of nuclei in `1mug` of `Na^(24)`
`N_(0)=(6xx10^(23)xx10^(-6))/(24)`
`:. N=(6xx10^(23)xx10^(-6))/(24)xx(1-e^(-(0.693)/(15)xx1))=1.128xx10^(15)`