Modern Physics Ques 193
- Two radioactive substances $A$ and $B$ have decay constants $5 \lambda$ and $\lambda$, respectively. At $t=0$, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become $\frac{1}{e}^{2}$ will be
(Main 2019, 10 April II)
(a) $2 / \lambda$
(b) $1 / 2 \lambda$
(c) $1 / 4 \lambda$
(d) $1 / \lambda$
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Answer:
Correct Answer: 193.(b)
Solution:
Formula:
- Number of active nuclei remained after time $t$ in a sample of radioactive substance is given by
$$ N=N _0 e^{-\lambda t} $$
where, $N _0=$ initial number of nuclei at $t=0$ and $\lambda=$ decay constant.
Here, at $t=0$,
Number of nuclei in sample $A$ and $B$ are equal, i.e.
Also,
$$ N _{0 _A}=N _{0 _B}=N _0 $$
So, after time $t$, number of active nuclei of $A$ and $B$ are
$$ N _A=N _0 e^{-5 \lambda t} \text { and } N _B=N _0 e^{-\lambda t} $$
If $\frac{N _A}{N _B}=\frac{1}{e^{2}}$, then
$$ \frac{N _A}{N _B}=\frac{N _0 e^{-5 \lambda t}}{N _0 e^{-\lambda t}}=\frac{1}{e^{2}} \Rightarrow e^{(\lambda-5 \lambda) t}=e^{-2} $$
Comparing the power of $e$ on both sides, we get
$$ \begin{aligned} 4 \lambda t & =2 \\ \Rightarrow \quad t & =\frac{1}{2 \lambda} \end{aligned} $$
BETA
