Question

A radioactive isotope is being produced at a constant rate of $dN/dt=R$ in an experiment. The isotope has a half-life of $t_{1/2}$. After a time $t>>t_{1/2}$, the number of active nuclei will become constant. The value of this constant is :

• A. $R$
• B. $1$
• C. $R/\lambda$
• D. $\lambda /R$

Answer 1

The correct option is C $R/\lambda$

Given :

Half life period = $t_{1/2}$

Rate of radioactive decay = $\frac{dN}{dt}=R$

$\Rightarrow R=\frac{dN}{dt}$

After time t > > $t_{1/2}$, the number of active nuclei will become constant..

$\Rightarrow (\frac{dN}{dt}{)}_{present}=R=(\frac{dN}{dt}{)}_{decay}$

$\therefore R=(\frac{dN}{dt}{)}_{deacy}$

$\Rightarrow$ R = $\lambda N$

(Where $\lambda =$ Radioactive decay constant , $N=$ constant number )

R = $\frac{0.693}{t_{1/2}}$(N)

$\Rightarrow R{t}_{1/2}=0.693N$

$\Rightarrow N=\frac{R{t}_{1/2}}{0.693}$

As we know ;

$\lambda =\frac{0.693}{t_{1/2}}$

$\therefore$ N = $\frac{R}{\lambda }$

Therefore, the correct answer is option (C).