Modern Physics Ques 204
- Two radioactive materials $A$ and $B$ have decay constants $10 \lambda$ and $\lambda$, respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $A$ to that of $B$ will be $1 / e$ after a time
(a) $\frac{1}{11 \lambda}$
(b) $\frac{11}{10 \lambda}$
(c) $\frac{1}{9 \lambda}$
(d) $\frac{1}{10 \lambda}$
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Answer:
Correct Answer: 204.(c)
Solution:
Formula:
- Given, $\lambda _A=10 \lambda$ and $\lambda _B=\lambda$
Number of nuclei (at any instant) present in material is
$$ N=N _0 e^{-\lambda t} $$
So, for materials $A$ and $B$, we can write
$$ \frac{N _A}{N _B}=\frac{e^{-\lambda _A t}}{e^{-\lambda _B t}}=e^{-\left(\lambda _A-\lambda _B\right) t} $$
Given,
$$ \frac{N _A}{N _B}=\frac{1}{e} $$
Equating Eqs. (i) and (ii), we get
$$ \begin{aligned} \frac{1}{e} & =e^{-\left(\lambda _A-\lambda _B\right) t} \\ \Rightarrow \quad e^{-1} & =e^{-\left(\lambda _A-\lambda _B\right) t} \end{aligned} $$
Comparing the power of ’ $e$ ’ on both sides, we get
$$ \begin{array}{ll} \text { or } & \left(\lambda _A-\lambda _B\right) t=1 \\ \Rightarrow & t=\frac{1}{\lambda _A-\lambda _B} \end{array} $$
By putting values of $\lambda _A$ and $\lambda _B$ in the above equation, we get
$$ t=\frac{1}{10 \lambda-\lambda} \Rightarrow t=\frac{1}{9 \lambda} $$
BETA
