Nuclei Ques 76
76. Two radioactive substances $A$ and $B$ have decay constants $5 \lambda$ and $\lambda$ respectively. At $t=0$ they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be $(1 / e)^{2}$ after a time interval
[2007]
(a) $4 \lambda$
(b) $2 \lambda$
(c) $1 / 2 \lambda$
(d) $1 / 4 \lambda$
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Answer:
Correct Answer: 76.(c)
Solution:
- (c) $\lambda_A=5 \lambda$ and $\lambda_B=\lambda$
At $t=0,(N_0) _{A}=(N_0) _{B}$
Given, $\frac{N_A}{N_B}=(\frac{1}{e})^{2}$
According to radioactive decay,
$\frac{N}{N_0}=e^{-\lambda t}$ $\therefore \frac{N_A}{(N_0) _{A}}=e^{-\lambda A^{t}}$
$\frac{N_B}{(N_0) _{B}}=e^{-\lambda_B t}$
From (1) and (2),
$\frac{N_A}{N_B}=e^{-(5 \lambda-\lambda) t}$
$ \begin{aligned} & \Rightarrow(\frac{1}{e})^{2}=e^{-4 \lambda t}=(\frac{1}{e})^{4 \lambda t} \\ & \Rightarrow 4 \lambda t=2 \quad \therefore t=\frac{1}{2 \lambda} . \end{aligned} $
BETA
