A radioactive isotope is being produced at a constant rate `dN//dt=R` in an experiment. The isotope has a half-life `t_(1//2)`.Show that after a time `t gtgt t_(1//2)`,the number of active nuclei will become constant. Find the value of this constant.

Answer

A, B, C

Explanation

Given Half life period =` t_(1/2)`
Rate of Radio active decay
` = (dN)/(dt) = R `
` rArr R = (dN)/(dt) `
After time` t gtgt t _(1/2) ` the number of active
nuclei will became constant
` i.e. ((dN)/(dt)) _(present) = R = ((DN)/(Dt)) _(decay)`
` :. R = ((DN)/(Dt)) _(decay)`
rArr ` R = lambda N`
(Where` lambda = ` Radioactive decay constant ,
` N = ` constant number ]
` rArr R ' = (0.693)/(t_(1/2) ) (N) `
`rArr Rt_(1/2) = 0.693 N`
` rArr N = (Rt _(1/2))/(0.693). `