Modern Physics 5 Question 2

2. 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: 2. (b)

Solution:

  1. 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

$$ \begin{aligned} 4 \lambda t & =2 \\ \Rightarrow \quad t & =\frac{1}{2 \lambda} \end{aligned} $$



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