A certain radionuclide is being manufactured in a cyclotron at a constant rate R. It is also decaying with disintegration constant$\mathit{\lambda }$. Assume that the production process has been going on for a time that is much longer than the half-life of the radionuclide. (a) Show that the numbers of radioactive nuclei present after such time remains constant and is given by$\mathit{N}\mathbf{=}\frac{\mathbf{R}}{\mathbf{\lambda }}$. (b) Now show that this result holds no matter how many radioactive nuclei were present initially. The nuclide is said to be in secular equilibriumwith its source; in this state its decay rate is just equal to its production rate.

The time taken by the production process $t>T_{\frac{1}{2}}\left(\mathrm{halflife}\mathrm{of}\mathrm{the}\mathrm{radionuclide}\right)$,

The radioactive decay is due to the loss of the elementary particles from an unstable nucleus to convert them into a more stable one. The radioactive decay constant or the disintegration constant represents the fraction of radioactive atoms that disintegrates in a unit of time. The production rate of the atoms of a given isotope about the disintegration constant and time will give us the required relation. At times that are long compared to the half-life, the rate of production equals the rate of decay, and N is a constant. The nuclide is in secular equilibrium with its source.

Formula:

The rate of undecayed nuclei for the given time is as follows:

$\frac{dN}{dt}=R-\lambda N$ …… (i)

Here,Nis the number of undecayed nuclei present at time .

R is the rate of production by the cyclotron,

$\lambda$is the disintegration constant.

The second term gives the rate of decay t.

Rearranging equation (i) and integrating it as per the problem, we can get the equation to number of nuclei as follows:

$$\begin{gathered} {\int }_{{\mathrm{N}}_0}^{\mathrm{N}}\frac{dN}{R-\lambda N}={\int }_0^{t}dt \\ -\frac{1}{\lambda }In\frac{R-\lambda N}{R-\lambda N_0}=t \\ N=\frac{R}{\lambda }+\left(N_0-\frac{R}{\lambda }\right)e^{-\lambda t} \end{gathered}$$

After many half-lives, the exponential is small and the second term can be neglected.

Thus, the above equation becomes,$N=\frac{R}{\lambda }$

The result $N=\frac{R}{\lambda }$holds regardless of the initial value $N_0$, because the dependence on $N_0$shows up only in the second term, which is exponentially suppressed at large t.