Modern Physics Ques 126
- For a radioactive material, its activity $A$ and rate of change of its activity $R$ are defined as $A=-\frac{d N}{d t}$ and $R=-\frac{d A}{d t}$, where $N(t)$ is the number of nuclei at time $t$. Two radioactive source $P($ mean life $\tau)$ and $Q$ (mean life $2 \tau)$ have the same activity at $t=0$. Their rate of change of activities at $t=2 \tau$ are $R _P$ and $R _Q$, respectively. If $\frac{R _P}{R _Q}=\frac{n}{e}$, then the value of $n$ is
(2015 Adv.)
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Answer:
Correct Answer: 126.$(2)$
Solution:
Formula:
- Let initial numbers are $N _1$ and $N _2$.
$ \frac{\lambda _1}{\lambda _2}=\frac{\tau _2}{\tau _1}=\frac{2 \tau}{\tau}=2=\frac{T _2}{T _1} $
$(T=$ Half life $)$
$ A=\frac{-d N}{d t}=\lambda N $
Initial activity is same
$\therefore \quad \lambda _1 N _1=\lambda _2 N _2$
Activity at time $t$,
$A =\lambda N=\lambda N _0 e^{-\lambda t} $
$A _1 =\lambda _1 N _1 e^{-\lambda _1 t} $
$\Rightarrow \quad R _1- =\frac{d A _1}{d t}=\lambda _1^{2} N _1 e^{-\lambda _1 t} $
$\text { Similarly, } \quad R _2 =\lambda _2^{2} N _2 e^{-\lambda _2 t}$
After $t=2 \tau$
$ \begin{aligned} & \lambda _1 t=\frac{1}{\tau _1}(t)=\frac{1}{\tau}(2 \tau)=2 \\ & \lambda _2 t=\frac{1}{\tau _2}(t)=1=\frac{1}{2 \tau}(2 \tau)=1 \\ & \frac{R _P}{R _Q}=\frac{\lambda _1^{2} N _1 e^{-\lambda _1 t}}{\lambda _2^{2} N _2 e^{-\lambda _2 t}} \\ & \frac{R _P}{R _Q}=\frac{\lambda _1}{\lambda _2} (\frac{e^{-2}}{e^{-1}})=\frac{2}{e} \end{aligned} $
BETA
