Question

A radioactive isotope is being produced at a constant rate $A$. The isotope has a half-life $T$. Initially, there are no decayed nuclei. After a time $t(t>>>T)$, the number of decayed nuclei becomes constant. The value of this constant is,

• A. $AT$
• B. $\left(\frac{A}{T}\right)\ln \left(2\right)$
  
• C. $\left(\frac{A}{T\ln \left(2\right)}\right)$
  
• D. $\frac{AT}{\ln \left(2\right)}$
  

Answer 1

The correct option is D $\frac{AT}{\ln \left(2\right)}$


The activity of a radioactive substance is given by,
$A=\lambda N=\lambda N_0{e}^{-\lambda t}$

The number of decayed nuclei remains constant, that means $N\approx N_0$.
This happens when $t$ is very large compared to $T$

i.e. As $t\to \infty ,N\to N_0$

Now, as $t\to \infty ,e^{-\lambda t}\to 1$

$\Rightarrow N=\frac{A}{\lambda }$

$\therefore N=\frac{AT}{\ln 2}(\because T=\frac{\ln 2}{\lambda })$

Hence, $\left(D\right)$ is the correct answer.