A radioactive nuclide is produced at the constant rate of n per second (say, by bombarding a target with neutrons). The expected number `N` of nuclei in existence `t` s after the number is `N_(0)` is given by

Options

A. `N=N_(0)e^(-lambda t)`
B. `N=N_(0)e^(-lambda t)`
C. `N=n/lambda+(N_0+n/lambda)e^(-lambda t)`
D. `N=n/lambda+(N_0+n/lambda)e^(-lambda t)`

Answer

c

Explanation

`(dN)/(dt) =n -lambda N`
because the population `N` is simultaneously incereasing at rate n and decreasing due to decay at rate `lambda N`.
` underset(N_(0))overset(N) (int)(dN)/(n-lambdaN)=underset(0) overset(t)(int)dt`
`(1)/(lambda)In ((n-lambdaN_(0))/(n-lambdaN))=t`
`N=(n)/(lambda)+(N_(0)-(n)/(lambda))e^(-lambdat)`.