Modern Physics Ques 138
- Nuclei of a radioactive element $A$ are being produced at a constant rate $\alpha$. The element has a decay constant $\lambda$. At time $t=0$, there are $N _0$ nuclei of the element.
$(1998,8$ M)
(a) Calculate the number $N$ of nuclei of $A$ at time $t$.
(b) If $\alpha=2 N _0 \lambda$, calculate the number of nuclei of $A$ after one half-life of $A$ and also the limiting value of $N$ as $t \rightarrow \infty$.
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Answer:
Correct Answer: 138.(a) $\frac{1}{\lambda}\left[\alpha-\left(\alpha-\lambda N _0\right) e^{-\lambda t}\right]$
(b) (i) $\frac{3}{2} N _0$ (ii) $2 N _0$
Solution:
Formula:
- (a) Let at time $t$, number of radioactive nuclei are $N$. Net rate of formation of nuclei of $A$
$ \begin{aligned} \frac{d N}{d t} & =\alpha-\lambda N \\ \text { or } \quad \frac{d N}{\alpha-\lambda N} & =d t \\ \text { or } \quad \int _{N _0}^{N} \frac{d N}{\alpha-\lambda N} & =\int _0^{t} d t \end{aligned} $
Solving this equation, we get
$ N=\frac{1}{\lambda}\left[\alpha-\left(\alpha-\lambda N _0\right) e^{-\lambda t}\right] $ $\quad$ …….(i)
(b) (i) Substituting $\alpha=2 \lambda N _0$ and $t=t _{1 / 2}=\frac{\ln (2)}{\lambda}$ in Eq. (i)
we get, $N=\frac{3}{2} N _0$
(ii) Substituting $\alpha=2 \lambda N _0$ and $t \rightarrow \infty$ in Eq. (i), we get
$ N=\frac{\alpha}{\lambda}=2 N _0 \quad \text { or } \quad N=2 N _0 $
BETA
