A radio nuclide X is produced at constant rate `alpha`. At time `t=0`, number of nuclei of X are zero. Find
(a) the maximum number of nuclei of X.
the number of nuclei at time t.
Decay constant of X is `lambda`.

Answer

A, B, D

Explanation

(a) Let N be the number of nuclei of X at time t.

Rate of formation of `X=alpha`
Rate of disintegration `=lambdaN`
Number of nuclei of X will increase until both the rates will become equal. Therefore,
`alpha = lambdaN_(max)`
:. `N_(max)=alpha/lambda` (b) Net rate of formation of X at time t is

`(dN)/(dt) = alpha - lambdaN`
:. `(dN)/(alpha-lambdaN)=dt`
Integrating with proper limits, we have
`int_0^N(dN)/(alpha-lambdaN)=int_0^tdt`
or `N=alpha/lambda(1-e^(-lambdat))`
This expression shows that number of nuclei of X are increasing exponentially from 0 to `alpha/lambda`.