An unstable element is produced in nuclear reactor at a constant rate R. If its half-life β^--decay is T_1/2, how much time, in terms of T_1/2, is required to produce 50% of the equilibrium quantity?

We have, 

Rate of increase of element = $\frac{\mathrm{number} \mathrm{of} \mathrm{nuclei} \mathrm{by} \mathrm{reactor}}{1 \mathrm{second}}-\frac{\mathrm{number} \mathrm{of} \mathrm{nuclei} \mathrm{decaying}}{1 \mathrm{second}}$

That is, 
                 $\frac{\mathrm{dN}}{\mathrm{dt}} = \mathrm{R} - \mathrm{\lambda N} \mathrm{or} \frac{\mathrm{dN}}{\mathrm{dt}}+\mathrm{\lambda N} = \mathrm{R}$ 
                      
The solution to this is the sum of the homogeneous solution, 
              ${\mathrm{N}}_{\mathrm{h}} = {\mathrm{ce}}^{-\mathrm{\lambda t}},$ where c is a constant, and

a particular solution, ${\mathrm{N}}_{\mathrm{l}} = \frac{\mathrm{R}}{\mathrm{\lambda }}.$ 

Therefore, the required solution is, 

                       $\mathrm{N} = {\mathrm{N}}_{\mathrm{h}}+{\mathrm{N}}_{\mathrm{p}} = {\mathrm{ce}}^{-\mathrm{\lambda t}}+\frac{\mathrm{R}}{\mathrm{\lambda }}$ 

The constant c is obtained from the requirement that the initial number of nuclei be zero, 
                           $\mathrm{N}\left(0\right) = 0 = \mathrm{c}+\frac{\mathrm{R}}{\mathrm{\lambda }}$
$\Rightarrow \mathrm{c} = -\frac{\mathrm{R}}{\mathrm{\lambda }}$
so that,                $\mathrm{N} = \frac{\mathrm{R}}{\mathrm{\lambda }}(1-{\mathrm{e}}^{-\mathrm{\lambda t}})$ 

The equilibrium value is $(\mathrm{t}\to \infty ) = \mathrm{R}/\mathrm{\lambda }.$ 

Setting N equal to 1/2 of this value gives, 

                  $\frac{1}{2}\left(\frac{\mathrm{R}}{\mathrm{\lambda }}\right) = \frac{\mathrm{R}}{\mathrm{\lambda }}(1-{\mathrm{e}}^{-\mathrm{\lambda t}})$ 

                       ${\mathrm{e}}^{-\mathrm{\lambda t}} = \frac{1}{2}$

$\Rightarrow$         $\mathrm{t} = \frac{\ln 2}{\mathrm{\lambda }} = {\mathrm{T}}_{1/2}$ 

The result is independent of R.