Nuclei - Result Question 75

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:

  1. (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} $