Question

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 1

Given: Rate of formation of  X = α

          At time t=0, number of nuclei of X are zero.

To Find:

(a) The maximum number of nuclei of X.

(b) Net rate of formation of X at time t

Solution:

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

Rate of formation of  X=α

Rate of disintegration = λ N

Number of nuclei of X will increase until both the rates will become equal. Therefore,

                       $\alpha = \lambda N_(max)$

                      $N_(max)= \frac{\alpha}{\lambda}$

(b) Net rate of formation of X at time t is

                    $\frac{dN}{dt} = \alpha - \lambda N$

                    $\frac{dN}{\alpha - \lambda N } = dt$

Integrating with proper limits, we have

                           $\int\limits^N_0 {\frac{dN}{\alpha - \lambda N }} \, dt = \int\limits^t_0 {t} \, dt$

or                      $N= \frac{\alpha }{\lambda} (1- e^{{-\lambda t}} )$

This expression shows that number of nuclei of X are increasing exponentially from $0$ to $\frac{\alpha }{\lambda}$.