Nuclei of a radioactive element `A` are being produced at a constant rate `alpha`. The element has a decay constant `lambda`. At time `t=0`, there are `N_0` nuclei of the element.
(a) Calculate the number `N` of nuclei of `A` at time `t`.
(b) If `alpha=2N_0lambda`, calculate the number of nuclei of A after one half-life of A, and also the limiting value of N as `trarroo`.

Answer

`N=(1)/l(ambda)[alpha-alpha lambda N_0]e^(-lambda t)`, (b) `2N_(0)`

Explanation

(a) At `t=0,N=N_(0)`
Rate of decay = -`lambda N`, and rate of foramtion = `alpha` Thus, accumualtion rate of element is
`(dN)/(dt) =alpha-lambdaN`
or (dN)/(alpha-lambdaN) =dt`
Integrating this expression, we get
`log_e[(alpha-lambdaN)/(alpha-lambda N_(0))] = -lambda t`
or `(alpha-lambdaN)/(alpha-lambda N_0) = -lambda t`
or `alpha-lambdaN=(alpha-lambda N_(0) =e^(-lambda t)`
or `N=(1)/(lambda)[alpha-(alpha-lambda N_(0)) = e^(-lambda t)]`
(b) If `alpha =2 N_(0) lambda`
`N=(1)/(lambda)[2 N_(0) lambda -(2 N_(0) lambda - N_(0) lambda)e^(-lambda t)]`
`=N_(0)(2-e^(-lambda t))`
At the time of half-life, `T=0.693//lambda`. So,
`N=N_(0)[2-e^(0.693)] -(3N_(0))/(2)`
Limiting value of `(as t rarr infty)` is
`N =N_(0)[2 -e^(-infty)]=2N_(0)`