Question

A radioactive isotope is being produced at a constant rate dN/dt = R in an experiment. The isotope has a half-life t_1/2. Show that after a time t >> t_1/2 the number of active nuclei will become constant. Find the value of this constant.

Answer 1

Given:
Half life period of isotope = t_1/2

Disintegration constant, $\lambda =\frac{0.693}{t_{{\displaystyle 1/2}}}$

Rate of Radio active decay $\left(R\right)$ is given by,
$R=\frac{d\mathrm{N}}{dt}$
We are to show that after time t >> t_1/2 the number of active nuclei is constant.
$$\begin{gathered} {\left(\frac{dN}{dt}\right)}_{Present}=R={\left(\frac{dN}{dt}\right)}_{decay} \\ \therefore R={\left(\frac{dN}{dt}\right)}_{decay} \end{gathered}$$
Rate of radioactive decay, $R=\lambda N$
Here, λ = Radioactive decay constant
N = Constant number
$$\begin{gathered} R=\frac{0.693}{t_{1/2}}\times N \\ \Rightarrow R{t}_{1/2}=0.693N \\ \Rightarrow N=\frac{R{t}_{1/2}}{0.693} \end{gathered}$$
This value of N should be constant.