Modern Physics 3 Question 39
38. 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: 38. (a) $\frac{1}{\lambda}\left[\alpha-\left(\alpha-\lambda N _0\right) e^{-\lambda t}\right]$ (b) (i) $\frac{3}{2} N _0$ (ii) $2 N _0$ 39. (a) $14.43 s$ (b) $40 s 40.5 .95 L$
Solution:
- (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
$$ N=\frac{1}{\lambda}\left[\alpha-\left(\alpha-\lambda N _0\right) e^{-\lambda t}\right] $$
(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
$$ N=\frac{\alpha}{\lambda}=2 N _0 \text { or } \quad N=2 N _0 $$
